· 3 DIFFERENTIATION RULES 3.1 Derivatives ofPolynomials and Exponential Functions 1. (a) 0is the...

104
3 DIFFERENTIATION RULES 3.1 Derivatives of Polynomials and Exponential Functions 1. (a) is the number such that lim 0 1 =1. (b) 27 1 0001 09928 00001 09932 0001 09937 00001 09933 28 1 0001 10291 00001 10296 0001 10301 00001 10297 From the tables (to two decimal places), lim 0 27 1 =099 and lim 0 28 1 =103. Since 099 1 103, 27 28. 2. (a) The function value at =0 is 1 and the slope at =0 is 1. (b) ()= is an exponential function and ()= is a power function. ( )= and ( )= 1 . (c) ()= grows more rapidly than ()= when is large. 3. () = 1865 is a constant function, so its derivative is 0, that is, 0 ()=0. 4. ()= 30 is a constant function, so its derivative is 0, that is, 0 ()=0. 5. ()=2 2 3 0 ()=0 2 3 = 2 3 6. ()= 3 4 8 0 ()= 3 4 (8 7 )=6 7 7. ()= 3 4 +6 0 ()=3 2 4(1) + 0 = 3 2 4 8. ()= 1 2 6 3 4 + 0 ()= 1 2 (6 5 ) 3(4 3 )+1=3 5 12 3 +1 9. ()= 1 4 ( 4 + 8) 0 ()= 1 4 ( 4 + 8) 0 = 1 4 (4 41 + 0) = 3 10. ()=( 2)(2 + 3) = 2 2 6 0 () = 2(2) 1 0=4 1 11. ()= 12 5 = 12 5 0 ()= 12(5 6 ) = 60 6 or 60 6 12. ()= 6 0 ()= (6 7 )= 6 7 13. ()=2 34 0 () = 2( 3 4 74 )= 3 2 74 14. ()= 4 4 = 14 4 0 ()= 1 4 34 4( )= 1 4 34 4 15. =3 + 4 3 =3 +4 13 0 = 3( ) + 4( 1 3 ) 43 =3 4 3 43 167 NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.

Transcript of  · 3 DIFFERENTIATION RULES 3.1 Derivatives ofPolynomials and Exponential Functions 1. (a) 0is the...

Page 1:  · 3 DIFFERENTIATION RULES 3.1 Derivatives ofPolynomials and Exponential Functions 1. (a) 0is the number such that lim 0 0 1 =1. (b) 2˘7 1 0˘001 0˘9928 0˘0001 0˘9932 0˘001

3 DIFFERENTIATION RULES3.1 Derivatives of Polynomials and Exponential Functions

1. (a) 0 is the number such that lim��0

0� � 1�

= 1.

(b)

�2�7� � 1

�0�001 0�9928

�0�0001 0�9932

0�001 0�9937

0�0001 0�9933

�2�8� � 1

�0�001 1�0291

�0�0001 1�0296

0�001 1�0301

0�0001 1�0297

From the tables (to two decimal places),

lim��0

2�7� � 1�

= 0�99 and lim��0

2�8� � 1�

= 1�03.

Since 0�99 � 1 � 1�03, 2�7 � 0 � 2�8.

2. (a) The function value at � = 0 is 1 and the slope at � = 0 is 1.

(b) �(�) = 0� is an exponential function and �(�) = �� is a power function. �

��(0�) = 0� and �

��(��) = 0���1.

(c) �(�) = 0� grows more rapidly than �(�) = �� when � is large.

3. �(�) = 186�5 is a constant function, so its derivative is 0, that is, � 0(�) = 0.

4. �(�) =�30 is a constant function, so its derivative is 0, that is, � 0(�) = 0.

5. �(�) = 2� 23 � � � 0(�) = 0� 2

3 = �23

6. ! (�) = 34�8 � ! 0(�) = 3

4(8�7) = 6�7

7. �(�) = �3 � 4�+ 6 � � 0(�) = 3�2 � 4(1) + 0 = 3�2 � 48. �(�) = 1

2�6 � 3�4 + � � � 0(�) = 1

2(6�5)� 3(4�3) + 1 = 3�5 � 12�3 + 1

9. �(�) = 14 (�

4 + 8) � � 0(�) = 14 (�

4 + 8)0 = 14 (4�

4�1 + 0) = �3

10. �(�) = (�� 2)(2�+ 3) = 2�2 � �� 6 � �0(�) = 2(2�)� 1� 0 = 4�� 1

11. �(�) = �12�5= �12��5 � �0(�) = �12(�5��6) = 60��6 or 60��6

12. �() = +�6 � �0() = +(�6�7) = �6+�7

13. �(�) = 2��3�4 � �0(�) = 2(�34 ��7�4) = � 3

2 ��7�4

14. �(�) = 4��� 40� = �1�4 � 40� � �0(�) = 1

4��3�4 � 4(0�) = 1

4��3�4 � 40�

15. = 30� + 43��= 30� + 4��1�3 � 0 = 3(0�) + 4(� 1

3)��4�3 = 30� � 4

3��4�3

167

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168 ¤ CHAPTER 3 DIFFERENTIATION RULES

16. =�� (�� 1) = �3�2 � �1�2 � 0 = 3

2�1�2 � 1

2��1�2 = 1

2��1�2(3�� 1) [factor out 1

2��1�2]

or 0 =3�� 12��

.

17. ! (�) =�12��5=�12

�5�5 = 1

32�5 � ! 0(�) = 1

32(5�4) = 5

32�4

18. �(�) = �2 � 3�+ 1�2

= 1� 3

�+1

�2= 1� 3��1 + ��2 �

� 0(�) = 0� 3(�1)��2 + (�2)��3 = 3��2 � 2��3 or 3

�2� 2

�3or 3�� 2

�3

19. = �2 + 4�+ 3��

= �3�2 + 4�1�2 + 3��1�2 �

0 = 32�1�2 + 4

�12

���1�2 + 3

�� 12

���3�2 = 3

2

��+

2��� 3

2���

�note that �3�2 = �2�2 · �1�2 = �

���

The last expression can be written as 3�2

2���+

4�

2���� 3

2���=3�2 + 4�� 32���

.

20. �(�) =�2�+

�3� =

�2�+

�3�� � �0(�) =

�2 (1) +

�3�12�

�1�2�=�2 +

�3

2��

21. = 4�2 � 0 = 0 since 4�2 is a constant.

22. = �0� +

)+

+

)2= �0� + )�1 + +)�2 � 0 = �0� � )�2 � 2+)�3 = �0� �

)2� 2+

)3

23. � = 5��+ 4

��5 = �1�5 + 4�5�2 � �0 = 1

5��4�5 + 4

�52�3�2

�= 1

5��4�5 + 10�3�2 or 1�

�5

5��4�+ 10

��3

24. ) =��

�+13��

�2=��

��2+ 2

�� · 1

3��+

�13��

�2= �+ 2�1�2�1�3 + 1��2�3 = �+ 2�1�6 + ��2�3 �

)0 = 1+ 2�16��5�6

�� 2

3��5�3 = 1 + 1

3��5�6 � 2

3��5�3 or 1 +

1

36��5� 2

33��5

25. 8 = �

10+�0� = ��10 +�0� � 80 = �10��11 +�0� = �10�

11+�0�

26. = 0�+1 + 1 = 0�01 + 1 = 0 · 0� + 1 � 0 = 0 · 0� = 0�+1

27. = 4�� = �1�4 � 0 = 1

4��3�4 =

1

44��3

. At (1� 1), 0 = 14 and an equation of the tangent line is

� 1 = 14(�� 1) or = 1

4�+ 3

4.

28. = �4 + 2�2 � � � 0 = 4�3 + 4�� 1. At (1� 2), 0 = 7 and an equation of the tangent line is

� 2 = 7(�� 1) or = 7�� 5.

29. = �4 + 20� � 0 = 4�3 + 20�. At (0� 2), 0 = 2 and an equation of the tangent line is � 2 = 2(�� 0)or = 2�+ 2. The slope of the normal line is � 1

2 (the negative reciprocal of 2) and an equation of the normal line is

� 2 = � 12(�� 0) or = � 1

2�+ 2.

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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 169

30. = (1 + 2�)2 = 1 + 4�+ 4�2 � 0 = 4 + 8�. At (1� 9), 0 = 12 and an equation of the tangent line is

� 9 = 12(�� 1) or = 12�� 3. The slope of the normal line is � 112

(the negative reciprocal of 12) and an equation of the

normal line is � 9 = � 112 (�� 1) or = � 1

12�+10912 .

31. = 3�2 � �3 � 0 = 6�� 3�2.

At (1� 2), 0 = 6� 3 = 3, so an equation of the tangent line is

� 2 = 3(�� 1) or = 3�� 1.

32. = ���� � 0 = 1� 1

2��1�2 = 1� 1

2��

.

At (1� 0), 0 = 12 , so an equation of the tangent line is

� 0 = 12(�� 1) or = 1

2�� 1

2.

33. �(�) = 0� � 5� � � 0(�) = 0� � 5.

Notice that � 0(�) = 0 when � has a horizontal tangent, � 0 is positive

when � is increasing, and � 0 is negative when � is decreasing.

34. �(�) = 3�5 � 20�3 + 50� � � 0(�) = 15�4 � 60�2 + 50.

Notice that � 0(�) = 0 when � has a horizontal tangent and that � 0 is

an even function while � is an odd function.

35. �(�) = 3�15 � 5�3 + 3 � � 0(�) = 45�14 � 15�2.

Notice that � 0(�) = 0 when � has a horizontal tangent, � 0 is positive

when � is increasing, and � 0 is negative when � is decreasing.

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170 ¤ CHAPTER 3 DIFFERENTIATION RULES

36. �(�) = �+ 1�� = �+ ��1 � � 0(�) = 1� ��2 = 1� 1��2.

Notice that � 0(�) = 0 when � has a horizontal tangent, � 0 is positive

when � is increasing, and � 0 is negative when � is decreasing.

37. To graphically estimate the value of � 0(1) for �(�) = 3�2 � �3, we’ll graph � in the viewing rectangle [1� 0�1� 1 + 0�1] by

[�(0�9)� �(1�1)], as shown in the �gure. [When assigning values to the window variables, it is convenient to use Y1(0�9) for

Ymin and Y1(1�1) for Ymax.] If we have suf�ciently zoomed in on the graph of � , we should obtain a graph that looks like a

diagonal line; if not, graph again with 1� 0�01 and 1 + 0�01, etc.

Estimated value:

� 0(1) 2�299� 1�7011�1� 0�9 =

0�589

0�2= 2�99.

Exact value: �(�) = 3�2 � �3 � � 0(�) = 6�� 3�2,

so � 0(1) = 6� 3 = 3.

38. See the previous exercise. Since � is a decreasing function, assign Y1(3�9) to Ymax and Y1(4�1) to Ymin.

Estimated value: � 0(4) 0�49386� 0�506374�1� 3�9 =

�0�012510�2

= �0�06255.

Exact value: �(�) = ��1�2 � � 0(�) = � 12��3�2, so � 0(4) = � 1

2(4�3�2) = � 1

2

�18

�= � 1

16= �0�0625.

39. (a) (b) From the graph in part (a), it appears that � 0 is zero at �1 �1�25, �2 0�5,

and �3 3. The slopes are negative (so � 0 is negative) on (��� �1) and

(�2� �3). The slopes are positive (so � 0 is positive) on (�1� �2) and (�3��).

(c) �(�) = �4 � 3�3 � 6�2 + 7�+ 30 �

� 0(�) = 4�3 � 9�2 � 12�+ 7

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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 171

40. (a) (b) From the graph in part (a), it appears that � 0 is zero at �1 0�2 and �2 2�8.

The slopes are positive (so � 0 is positive) on (��� �1) and (�2��). The slopes

are negative (so � 0 is negative) on (�1� �2).

(c) �(�) = 0� � 3�2 � �0(�) = 0� � 6�

41. �(�) = 10�10 + 5�5 � � � � 0(�) = 100�9 + 25�4 � 1 � � 00(�) = 900�8 + 100�3

42. #(�) =�� +

3�� � # 0(�) = 1

2��1�2 + 1

3��2�3 � # 00(�) = � 1

4��3�2 � 2

9��5�3

43. �(�) = 2�� 5�3�4 � � 0(�) = 2� 154��1�4 � � 00(�) = 15

16��5�4

Note that � 0 is negative when � is decreasing and positive when � is

increasing. � 00 is always positive since � 0 is always increasing.

44. �(�) = 0� � �3 � � 0(�) = 0� � 3�2 � � 00(�) = 0� � 6�

Note that � 0(�) = 0 when � has a horizontal tangent and that � 00(�) = 0

when � 0 has a horizontal tangent.

45. (a) � = �3 � 3� � )(�) = �0 (�) = 3�2 � 3 � �(�) = )0(�) = 6�

(b) �(2) = 6(2) = 12 m�s2

(c) )(�) = 3�2 � 3 = 0 when �2 = 1, that is, � = 1 and �(1) = 6 m�s2.

46. (a) � = �4 � 2�3 + �2 � � �)(�) = �0(�) = 4�3 � 6�2 + 2�� 1 �

�(�) = )0(�) = 12�2 � 12�+ 2

(c)

(b) �(1) = 12(1)2 � 12(1) + 2 = 2m� s2

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172 ¤ CHAPTER 3 DIFFERENTIATION RULES

47. �(�) = 5�� 0� � � 0(�) = 5� 0�. � 0(�) 0 � 5� 0� 0 � 0� � 5 � � � ln 5 1�61.

� is increasing when � 0 is positive; that is, on (��� ln 5).

48. �(�) = �3 � 4�2 + 5� � � 0(�) = 3�2 � 8�+ 5 � � 00(�) = 6�� 8.

� 00(�) 0 � 6�� 8 0 � � 43

. � is concave upward when � 00(�) 0; that is, on�43���

.

49. The curve = 2�3 + 3�2 � 12�+ 1 has a horizontal tangent when 0 = 6�2 + 6�� 12 = 0 � 6(�2 + �� 2) = 0 �

6(�+ 2)(�� 1) = 0 � � = �2 or � = 1. The points on the curve are (�2� 21) and (1��6).

50. �(�) = �3 + 3�2 + �+ 3 has a horizontal tangent when � 0(�) = 3�2 + 6�+ 1 = 0 �

� =�6±�36� 12

6= �1± 1

3

�6.

51. = 6�3 + 5�� 3 � � = 0 = 18�2 + 5, but �2 � 0 for all �, so � � 5 for all �.

52. = ��� = �3�2 � 0 = 3

2�1�2. The slope of the line = 1 + 3� is 3, so the slope of any line parallel to it is also 3.

Thus, 0 = 3 � 32�1�2 = 3 �

�� = 2 � � = 4, which is the �-coordinate of the point on the curve at which the

slope is 3. The -coordinate is = 4�4 = 8, so an equation of the tangent line is � 8 = 3(�� 4) or = 3�� 4.

53. The slope of the line 12�� = 1 (or = 12�� 1) is 12, so the slope of both lines tangent to the curve is 12.

= 1+ �3 � 0 = 3�2. Thus, 3�2 = 12 � �2 = 4 � � = ±2, which are the �-coordinates at which the tangent

lines have slope 12. The points on the curve are (2� 9) and (�2��7), so the tangent line equations are � 9 = 12(�� 2)

or = 12�� 15 and + 7 = 12(�+ 2) or = 12�+ 17.

54. The slope of = 1 + 20� � 3� is given by � = 0 = 20� � 3.

The slope of 3�� = 5 � = 3�� 5 is 3.

� = 3 � 20� � 3 = 3 � 0� = 3 � � = ln 3.

This occurs at the point (ln 3� 7� 3 ln 3) (1�1� 3�7).

55. The slope of = �2 � 5�+ 4 is given by � = 0 = 2�� 5. The slope of �� 3 = 5 � = 13�� 5

3is 1

3,

so the desired normal line must have slope 13

, and hence, the tangent line to the parabola must have slope�3. This occurs if

2�� 5 = �3 � 2� = 2 � � = 1. When � = 1, = 12 � 5(1) + 4 = 0, and an equation of the normal line is

� 0 = 13(�� 1) or = 1

3�� 1

3.

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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 173

56. = �(�) = �� �2 � � 0(�) = 1� 2�.

So � 0(1) = �1, and the slope of the normal line is the negative reciprocal of that

of the tangent line, that is,�1�(�1) = 1. So the equation of the normal line at

(1� 0) is � 0 = 1(�� 1) � = �� 1. Substituting this into the equation of

the parabola, we obtain �� 1 = �� �2 � � = ±1. The solution � = �1 is

the one we require. Substituting � = �1 into the equation of the parabola to �nd

the -coordinate, we have = �2. So the point of intersection is (�1��2), as shown in the sketch.

57. Let��� �2

�be a point on the parabola at which the tangent line passes

through the point (0��4). The tangent line has slope 2� and equation

� (�4) = 2�(�� 0) � = 2��� 4. Since��� �2

�also lies on the

line, �2 = 2�(�)� 4, or �2 = 4. So � = ±2 and the points are (2� 4)

and (�2� 4).

58. (a) If = �2 + �, then 0 = 2�+ 1. If the point at which a tangent meets the parabola is��� �2 + �

�, then the slope of the

tangent is 2�+ 1. But since it passes through (2��3), the slope must also be ���

=�2 + �+ 3

�� 2 .

Therefore, 2�+ 1 = �2 + �+ 3

�� 2 . Solving this equation for � we get �2 + �+ 3 = 2�2 � 3�� 2 �

�2 � 4�� 5 = (�� 5)(�+ 1) = 0 � � = 5 or �1. If � = �1, the point is (�1� 0) and the slope is �1, so the

equation is � 0 = (�1)(�+ 1) or = ��� 1. If � = 5, the point is (5� 30) and the slope is 11, so the equation is

� 30 = 11(�� 5) or = 11�� 25.

(b) As in part (a), but using the point (2� 7), we get the equation

2�+ 1 =�2 + �� 7�� 2 � 2�2 � 3�� 2 = �2 + �� 7 � �2 � 4�+ 5 = 0.

The last equation has no real solution (discriminant = �16 � 0), so there is no line

through the point (2� 7) that is tangent to the parabola. The diagram shows that the

point (2� 7) is “inside” the parabola, but tangent lines to the parabola do not pass

through points inside the parabola.

59. � 0(�) = lim��0

�(�+ �)� �(�)

�= lim

��0

1

�+ �� 1

��

= lim��0

�� (�+ �)

��(�+ �)= lim

��0

����(�+ �)

= lim��0

�1�(�+ �)

= � 1

�2

60. (a) �(�) = � � � 0(�) = (��1 � � 00(�) = ( ((� 1)��2 � · · · �

� ()(�) = (((� 1)((� 2) · · · 2 · 1�� = (!

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174 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b) �(�) = ��1 � � 0(�) = (�1)��2 � � 00(�) = (�1)(�2)��3 � · · · �

� ()(�) = (�1)(�2)(�3) · · · (�()��(+1) = (�1)(!��(+1) or (�1) (!

�+1

61. Let � (�) = ��2 + �+ +. Then � 0(�) = 2��+ and � 00(�) = 2�. � 00(2) = 2 � 2� = 2 � � = 1.

� 0(2) = 3 � 2(1)(2) + = 3 � 4 + = 3 � = �1.

� (2) = 5 � 1(2)2 + (�1)(2) + + = 5 � 2 + + = 5 � + = 3. So � (�) = �2 � �+ 3.

62. = ��2 +��+ � � 0 = 2��+� � 00 = 2�. We substitute these expressions into the equation

00 + 0 � 2 = �2 to get

(2�) + (2��+�)� 2(��2 +��+�) = �2

2�+ 2��+� � 2��2 � 2��� 2� = �2

(�2�)�2 + (2�� 2�)�+ (2�+� � 2�) = (1)�2 + (0)�+ (0)

The coef�cients of �2 on each side must be equal, so �2� = 1 � � = � 12

. Similarly, 2�� 2� = 0 �

� = � = � 12

and 2�+� � 2� = 0 � �1� 12� 2� = 0 � � = � 3

4.

63. (a) At this stage, we would guess that an antiderivative of �2 must have �3 in it. Differentiating �3 gives us 3�2, so we know

that we must divide �3 by 3. That gives us ! (�) = 13�3. Checking, we have ! 0(�) = 1

3(3�2) = �2 = �(�). Because we

can add an arbitrary constant � to ! without changing its derivative, we have an in�nite number of antiderivatives of the

form ! (�) = 13�3 +�.

(b) As in part (a), antiderivatives of �(�) = �3 and �(�) = �4 are ! (�) = 14�

4 + � and ! (�) = 15�

5 +�.

(c) Similarly, an antiderivative for �(�) = � is ! (�) = 1

(+ 1�+1 + �, since then

! 0(�) =1

(+ 1[((+ 1)�] = � = �(�) for ( 6= �1.

64. (a) �(�) =�� = �1�2 � ! (�) = 1

(1�2)+ 1�(1�2)+ 1 + � = 2

3�3�2 + �

(b) �(�) = 0� + 8�3 � ! (�) = 0� + 8 · 13+1

�3+ 1 +� = 0� + 2�4 + �

65. Substituting � = 1 and = 1 into = ��2 + � gives us �+ = 1 (1). The slope of the tangent line = 3�� 2 is 3 and the

slope of the tangent to the parabola at (�� ) is 0 = 2��+ . At � = 1, 0 = 3 � 3 = 2�+ (2). Subtracting (1) from

(2) gives us 2 = � and it follows that = �1. The parabola has equation = 2�2 � �.

66. = �4 + ��3 + �2 + +�+ � � (0) = �. Since the tangent line = 2�+ 1 is equal to 1 at � = 0, we must

have � = 1. 0 = 4�3+3��2+2�+ + � 0(0) = +. Since the slope of the tangent line = 2�+1 at � = 0 is 2, we

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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 175

must have + = 2. Now (1) = 1+ �+ + ++ � = �+ + 4 and the tangent line = 2� 3� at � = 1 has -coordinate�1,

so �+ + 4 = �1 or �+ = �5 (1). Also, 0(1) = 4 + 3�+ 2+ + = 3�+ 2+ 6 and the slope of the tangent line

= 2� 3� at � = 1 is �3, so 3�+ 2+ 6 = �3 or 3�+ 2 = �9 (2). Adding �2 times (1) to (2) gives us � = 1 and

hence, = �6. The curve has equation = �4 + �3 � 6�2 + 2�+ 1.

67. = �(�) = ��3 + �2 + +�+ � � � 0(�) = 3��2 + 2�+ +. The point (�2� 6) is on � , so �(�2) = 6 �

�8�+ 4� 2++ � = 6 (1). The point (2� 0) is on � , so �(2) = 0 � 8�+ 4+ 2++ � = 0 (2). Since there are

horizontal tangents at (�2� 6) and (2� 0), � 0(±2) = 0. � 0(�2) = 0 � 12�� 4+ + = 0 (3) and � 0(2) = 0 �12�+ 4+ + = 0 (4). Subtracting equation (3) from (4) gives 8 = 0 � = 0. Adding (1) and (2) gives 8+ 2� = 6,

so � = 3 since = 0. From (3) we have + = �12�, so (2) becomes 8�+ 4(0) + 2(�12�) + 3 = 0 � 3 = 16� �

� = 316

. Now + = �12� = �12� 316

�= � 9

4and the desired cubic function is = 3

16�3 � 9

4�+ 3.

68. The slope of the curve = +�� is 0 = +

2��

and the slope of the tangent line = 32�+ 6 is 3

2. These must be equal at the

point of tangency��� +

���

, so +

2��=3

2� + = 3

��. The -coordinates must be equal at � = �, so

+�� = 3

2�+ 6 �

�3����

� = 32�+ 6 � 3� = 3

2�+ 6 � 3

2� = 6 � � = 4. Since + = 3

��, we have

+ = 3�4 = 6.

69. = �(�) = ��2 � � 0(�) = 2��. So the slope of the tangent to the parabola at � = 2 is � = 2�(2) = 4�. The slope

of the given line, 2�+ = � = �2�+ , is seen to be �2, so we must have 4� = �2 � � = � 12

. So when

� = 2, the point in question has -coordinate � 12· 22 = �2. Now we simply require that the given line, whose equation is

2�+ = , pass through the point (2��2): 2(2) + (�2) = � = 2. So we must have � = � 12

and = 2.

70. (a) � = + � =+

�. Let � =

���+

�. The slope of the tangent line at � = � is 0(�) = � +

�2. Its equation is

� +

�= � +

�2(�� �) or = � +

�2�+

2+

�, so its -intercept is 2+

�. Setting = 0 gives � = 2�, so the �-intercept is 2�.

The midpoint of the line segment joining�0�2+

�and (2�� 0) is

���+

�= � .

(b) We know the �- and -intercepts of the tangent line from part (a), so the area of the triangle bounded by the axes and the

tangent is 12(base)(height) = 1

2� = 1

2(2�)(2+��) = 2+, a constant.

71. Solution 1: Let �(�) = �1000. Then, by the de�nition of a derivative, � 0(1) = lim��1

�(�)� �(1)

�� 1 = lim��1

�1000 � 1�� 1 .

But this is just the limit we want to �nd, and we know (from the Power Rule) that � 0(�) = 1000�999, so

� 0(1) = 1000(1)999 = 1000. So lim��1

�1000 � 1�� 1 = 1000.

[continued]

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176 ¤ CHAPTER 3 DIFFERENTIATION RULES

Solution 2: Note that (�1000 � 1) = (�� 1)(�999 + �998 + �997 + · · ·+ �2 + �+ 1). So

lim��1

�1000 � 1�� 1 = lim

��1

(�� 1)(�999 + �998 + �997 + · · ·+ �2 + �+ 1)

�� 1 = lim��1

(�999 + �998 + �997 + · · ·+ �2 + �+ 1)

= 1 + 1 + 1 + · · ·+ 1 + 1 + 1� �� � = 1000, as above.

1000 ones

72. In order for the two tangents to intersect on the -axis, the points of tangency must be at

equal distances from the -axis, since the parabola = �2 is symmetric about the -axis.

Say the points of tangency are��� �2

�and

���� �2�, for some � 0. Then since the

derivative of = �2 is ���� = 2�, the left-hand tangent has slope �2� and equation

� �2 = �2�(�+ �), or = �2��� �2, and similarly the right-hand tangent line has

equation � �2 = 2�(�� �), or = 2��� �2. So the two lines intersect at�0���2�. Now if the lines are perpendicular,

then the product of their slopes is�1, so (�2�)(2�) = �1 � �2 = 14

� � = 12

. So the lines intersect at�0�� 1

4

�.

73. = �2 � 0 = 2�, so the slope of a tangent line at the point (�� �2) is 0 = 2� and the slope of a normal line is �1�(2�),

for � 6= 0. The slope of the normal line through the points (�� �2) and (0� +) is �2 � +

�� 0 , so �2 � +

�= � 1

2��

�2 � + = � 12 � �2 = +� 1

2 . The last equation has two solutions if + 12 , one solution if + = 1

2 , and no solution if

+ � 12 . Since the -axis is normal to = �2 regardless of the value of + (this is the case for � = 0), we have three normal lines

if + 12

and one normal line if + � 12

.

74. From the sketch, it appears that there may be a line that is tangent to both

curves. The slope of the line through the points � (�� �2) and

/(� 2 � 2+ 2) is 2 � 2+ 2� �2

� �. The slope of the tangent line at �

is 2� [0 = 2�] and at / is 2� 2 [0 = 2�� 2]. All three slopes are

equal, so 2� = 2� 2 � � = � 1.

Also, 2� 2 = 2 � 2+ 2� �2

� �� 2� 2 = 2 � 2+ 2� (� 1)2

� (� 1) � 2� 2 = 2 � 2+ 2� 2 +2� 1 �

2 = 3 � = 32

and � = 32� 1 = 1

2. Thus, an equation of the tangent line at � is � �

12

�2= 2

�12

���� 1

2

�or

= �� 14

.

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APPLIED PROJECT BUILDING A BETTER ROLLER COASTER ¤ 177

APPLIED PROJECT Building a Better Roller Coaster

1. (a) �(�) = ��2 + �+ + � � 0(�) = 2��+ .

The origin is at � : �(0) = 0 � + = 0

The slope of the ascent is 0�8: � 0(0) = 0�8 � = 0�8

The slope of the drop is �1�6: � 0(100) = �1�6 � 200�+ = �1�6

(b) = 0�8, so 200�+ = �1�6 � 200�+ 0�8 = �1�6 � 200� = �2�4 � � = � 2�4200

= �0�012.

Thus, �(�) = �0�012�2 + 0�8�.

(c) Since $1 passes through the origin with slope 0�8, it has equation = 0�8�.

The horizontal distance between � and / is 100, so the -coordinate at / is

�(100) = �0�012(100)2 + 0�8(100) = �40. Since $2 passes through the

point (100��40) and has slope�1�6, it has equation + 40 = �1�6(�� 100)or = �1�6�+ 120.

(d) The difference in elevation between � (0� 0) and /(100��40) is 0� (�40) = 40 feet.

2. (a)Interval Function First Derivative Second Derivative

(��� 0) $1(�) = 0�8� $01(�) = 0�8 $001 (�) = 0

[0� 10) �(�) = ��3 + 9�2 +��+ ( �0(�) = 3��2 + 29�+� �00(�) = 6��+ 29

[10� 90] �(�) = ��2 + �+ + �0(�) = 2��+ �00(�) = 2�

(90� 100] �(�) = ��3 + ��2 + ��+ � �0(�) = 3��2 + 2��+ � �00(�) = 6��+ 2�

(100��) $2(�) = �1�6�+ 120 $02(�) = �1�6 $002 (�) = 0

There are 4 values of � (0, 10, 90, and 100) for which we must make sure the function values are equal, the �rst derivativevalues are equal, and the second derivative values are equal. The third column in the following table contains the value ofeach side of the condition — these are found after solving the system in part (b).

At � = Condition Value Resulting Equation

0 �(0) = $1(0) 0 ( = 0

�0(0) = $01(0)45

� = 0�8

�00(0) = $001 (0) 0 29 = 0

10 �(10) = �(10) 689

1000� + 1009 + 10�+ ( = 100�+ 10+ +

�0(10) = �0(10) 23

300� + 209+� = 20�+

�00(10) = �00(10) � 275

60� + 29 = 2�

90 �(90) = �(90) � 2209

729,000�+ 8100� + 90� + � = 8100�+ 90+ +

�0(90) = �0(90) � 2215

24,300�+ 180� + � = 180�+

�00(90) = �00(90) � 275

540�+ 2� = 2�

100 �(100) = $2(100) �40 1,000,000�+ 10,000� + 100� + � = �40�0(100) = $02(100) � 8

530,000�+ 200� + � = �1�6

�00(100) = $002 (100) 0 600�+ 2� = 0

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178 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b) We can arrange our work in a 12× 12 matrix as follows.

� + � 9 � ( � � � � constant

0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0�8

0 0 0 0 2 0 0 0 0 0 0 0

�100 �10 �1 1000 100 10 1 0 0 0 0 0

�20 �1 0 300 20 1 0 0 0 0 0 0

�2 0 0 60 2 0 0 0 0 0 0 0

�8100 �90 �1 0 0 0 0 729,000 8100 90 1 0

�180 �1 0 0 0 0 0 24,300 180 1 0 0

�2 0 0 0 0 0 0 540 2 0 0 0

0 0 0 0 0 0 0 1,000,000 10,000 100 1 �400 0 0 0 0 0 0 30,000 200 1 0 �1�60 0 0 0 0 0 0 600 2 0 0 0

Solving the system gives us the formulas for �, �, and �.

� = �0�013 = � 175

= 0�93 = 1415

+ = �0�4 = � 49

)��*��+ �(�) = � 1

75�2 + 14

15�� 4

9

� = �0�0004 = � 12250

9 = 0

� = 0�8 = 45

( = 0

)����*����+�(�) = � 1

2250�3 + 4

5�

� = 0�0004 = 12250

� = �0�13 = � 215

� = 11�73 = 17615

� = �324�4 = � 29209

)����*����+�(�) = 1

2250�3 � 2

15�2 + 176

15�� 2920

9

(c) Graph of $1, �, �, �, and $2: The graph of the �ve functions as a piecewise-de�ned function:

This is the piecewise-de�ned function assignment on a TI-83

Plus calculator, where Y2 = $1, Y6 = �, Y5 = �, Y7 = �,

and Y3 = $2.

A comparison of the graphs in part 1(c) and part 2(c):

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 179

3.2 The Product and Quotient Rules

1. Product Rule: �(�) = (1 + 2�2)(�� �2) �

� 0(�) = (1 + 2�2)(1� 2�) + (�� �2)(4�) = 1� 2�+ 2�2 � 4�3 + 4�2 � 4�3 = 1� 2�+ 6�2 � 8�3.

Multiplying �rst: �(�) = (1 + 2�2)(�� �2) = �� �2 + 2�3 � 2�4 � � 0(�) = 1� 2�+ 6�2 � 8�3 (equivalent).

2. Quotient Rule: ! (�) = �4 � 5�3 +���2

=�4 � 5�3 + �1�2

�2�

! 0(�) =�2(4�3 � 15�2 + 1

2��1�2)� (�4 � 5�3 + �1�2)(2�)

(�2)2=4�5 � 15�4 + 1

2�3�2 � 2�5 + 10�4 � 2�3�2

�4

=2�5 � 5�4 � 3

2�3�2

�4= 2�� 5� 3

2��5�2

Simplifying �rst: ! (�) = �4 � 5�3 + ��

�2= �2 � 5�+ ��3�2 � ! 0(�) = 2�� 5� 3

2��5�2 (equivalent).

For this problem, simplifying �rst seems to be the better method.

3. By the Product Rule, �(�) = (�3 + 2�)0� �

� 0(�) = (�3 + 2�)(0�)0 + 0�(�3 + 2�)0 = (�3 + 2�)0� + 0�(3�2 + 2)

= 0�[(�3 + 2�) + (3�2 + 2)] = 0�(�3 + 3�2 + 2�+ 2)

4. By the Product Rule, �(�) =�� 0� = �1�20� � �0(�) = �1�2(0�) + 0�

�12�

�1�2�= 1

2��1�20�(2�+ 1).

5. By the Quotient Rule, = 0�

�2� 0 =

�2�

��(0�)� 0�

��(�2)

(�2)2=

�2 (0�)� 0�(2�)

�4=

�0�(�� 2)�4

=0�(�� 2)

�3.

6. By the Quotient Rule, = 0�

1 + �� 0 =

(1 + �) 0� � 0�(1)

(1 + �)2=

0� + �0� � 0�

(�+ 1)2=

�0�

(�+ 1)2.

The notationsPR� and

QR� indicate the use of the Product and Quotient Rules, respectively.

7. �(�) = 3�� 12�+ 1

QR� �0(�) =(2�+ 1)(3)� (3�� 1)(2)

(2�+ 1)2=6�+ 3� 6�+ 2(2�+ 1)2

=5

(2�+ 1)2

8. �(�) = 2�

4 + �2QR� � 0(�) =

(4 + �2)(2)� (2�)(2�)(4 + �2)2

=8 + 2�2 � 4�2(4 + �2)2

=8� 2�2(4 + �2)2

9. ! () =�1

2� 3

4

�( + 53) = (�2 � 3�4)( + 53) PR�

! 0() = (�2 � 3�4)(1 + 152) + ( + 53)(�2�3 + 12�5)

= (�2 + 15� 3�4 � 45�2) + (�2�2 + 12�4 � 10 + 60�2)

= 5 + 14�2 + 9�4 or 5 + 14�2 + 9�4

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180 ¤ CHAPTER 3 DIFFERENTIATION RULES

10. .(�) =��+ 0�

��3��� � PR�

.0(�) = (�+ 0�)�� 12��1�2

�+�3��� �(1 + 0�)

=�� 12�1�2 � 1

2��1�20�

�+�3 + 30� ������ 0�� = 3+ 30� � 3

2

����� 0� � 0��

�2���

11. = �3

1� �2QR� 0 =

(1� �2) (3�2)� �3(�2�)(1� �2)2

=�2(3� 3�2 + 2�2)

(1� �2)2=

�2(3� �2)

(1� �2)2

12. = �+ 1

�3 + �� 2QR�

0 =(�3 + �� 2)(1)� (�+ 1)(3�2 + 1)

(�3 + �� 2)2 =�3 + �� 2� 3�3 � 3�2 � �� 1

(�3 + �� 2)2 =�2�3 � 3�2 � 3(�3 + �� 2)2

or � 2�3 + 3�2 + 3

(�� 1)2(�2 + �+ 2)2

13. = �2 + 2

�4 � 3�2 + 1QR�

0 =(�4 � 3�2 + 1)(2�)� (�2 + 2)(4�3 � 6�)

(�4 � 3�2 + 1)2 =2�[(�4 � 3�2 + 1)� (�2 + 2)(2�2 � 3)]

(�4 � 3�2 + 1)2

=2�(�4 � 3�2 + 1� 2�4 � 4�2 + 3�2 + 6)

(�4 � 3�2 + 1)2 =2�(��4 � 4�2 + 7)(�4 � 3�2 + 1)2

14. = �

(�� 1)2 =�

�2 � 2�+ 1QR�

0 =(�2 � 2�+ 1)(1)� �(2�� 2)

[(�� 1)2]2 =(�� 1)2 � 2�(�� 1)

(�� 1)4 =(�� 1)[(�� 1)� 2�]

(�� 1)4 =��� 1(�� 1)3

15. = (�2 � 2�)0� PR� 0 = (�2 � 2�)(0�) + 0�(2� � 2) = 0�(�2 � 2� + 2� � 2) = 0�(�2 � 2)

16. = 1

�+ �0�QR� 0 =

(�+ �0�)(0)� (1)(1 + �0�)

(�+ �0�)2= � 1 + �0�

(�+ �0�)2

17. = )3 � 2)�))

= )2 � 2�) = )2 � 2)1�2 � 0 = 2) � 2� 12

�)�1�2 = 2) � )�1�2.

We can change the form of the answer as follows: 2) � )�1�2 = 2) � 1�)=2)�) � 1�)

=2)3�2 � 1�

)

18. 8 = *3�2(*+ +0�) = *5�2 + +*3�20� � 80 = 52*3�2 + +

�*3�2 · 0� + 0� · 3

2*1�2

�= 5

2*3�2 + 1

2+*1�20�(2*+3)

19. �(�) = 2�

2 +��

QR� � 0(�) =(2 + �1�2)(2)� 2�

�12 ��1�2

�(2 +

�� )2

=4 + 2�1�2 � �1�2

(2 +�� )2

=4 + �1�2

(2 +�� )2

or 4 +��

(2 +�� )2

20. �(�) = �����1�3

=�

�1�3� �1�2

�1�3= �2�3 � �1�6 � �0(�) = 2

3 ��1�3 � 1

6 ��5�6

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 181

21. �(�) = �

� +�0�QR� � 0(�) =

(� +�0�) · 0��(�0�)

(� +�0�)2= � ��0�

(� +�0�)2

22. �(�) = 1� �0�

�+ 0�QR� � 0(�) =

(�+ 0�)(��0�)0 � (1� �0�)(1 + 0�)

(�+ 0�)2

PR� � 0(�) =(�+ 0�)[�(�0� + 0� · 1)]� (1 + 0� � �0� � �02�)

(�+ 0�)2

=��20� � �0� � �02� � 02� � 1� 0� + �0� + �02�

(�+ 0�)2=��20� � 02� � 0� � 1

(�+ 0�)2

23. �(�) = �

�+ +��� � 0(�) =

(�+ +��)(1)� �(1� +��2)��+

+

�2 =�+ +��� �+ +���

�2 + +

�2 =2+��

(�2 + +)2

�2

· �2

�2=

2+�

(�2 + +)2

24. �(�) = ��+

+�+ �� � 0(�) =

(+�+ �)(�)� (��+ )(+)

(+�+ �)2=

�+�+ ��� �+�� +

(+�+ �)2=

��� +

(+�+ �)2

25. �(�) = �40� � � 0(�) = �40� + 0� · 4�3 = ��4 + 4�3

�0�

or �30�(�+ 4)

� 00(�) = (�4 + 4�3)0� + 0�(4�3 + 12�2) = (�4 + 4�3 + 4�3 + 12�2)0�

= (�4 + 8�3 + 12�2)0�or �20�(�+ 2)(�+ 6)

26. �(�) = �5�20� � � 0(�) = �5�20� + 0� · 5

2�3�2 =

��5�2 + 5

2�3�2

�0�

�or 1

2�3�20�(2�+ 5)

��

� 00(�) =��5�2 + 5

2�3�2

�0� + 0�

�52�3�2 + 15

4�1�2

�=��5�2 + 5�3�2 + 15

4�1�2

�0�

�or 1

4�1�20�(4�2 + 20�+ 15)

27. �(�) = �2

1 + 2�� � 0(�) =

(1 + 2�)(2�)� �2(2)

(1 + 2�)2=2�+ 4�2 � 2�2(1 + 2�)2

=2�2 + 2�

(1 + 2�)2�

� 00(�) =(1 + 2�)2(4�+ 2)� (2�2 + 2�)(1 + 4�+ 4�2)0

[(1 + 2�)2]2=2(1 + 2�)2(2�+ 1)� 2�(�+ 1)(4 + 8�)

(1 + 2�)4

=2(1 + 2�)[(1 + 2�)2 � 4�(�+ 1)]

(1 + 2�)4=2(1 + 4�+ 4�2 � 4�2 � 4�)

(1 + 2�)3=

2

(1 + 2�)3

28. �(�) = �

�2 � 1 � � 0(�) =(�2 � 1)(1)� �(2�)

(�2 � 1)2 =�2 � 1� 2�2(�2 � 1)2 =

��2 � 1(�2 � 1)2 �

� 00(�) =(�2 � 1)2(�2�)� (��2 � 1)(�4 � 2�2 + 1)0

[(�2 � 1)2]2 =(�2 � 1)2(�2�) + (�2 + 1)(4�3 � 4�)

(�2 � 1)4

=(�2 � 1)2(�2�) + (�2 + 1)(4�)(�2 � 1)

(�2 � 1)4 =(�2 � 1)[(�2 � 1)(�2�) + (�2 + 1)(4�)]

(�2 � 1)4

=�2�3 + 2�+ 4�3 + 4�

(�2 � 1)3 =2�3 + 6�

(�2 � 1)3

29. = 2�

�+ 1� 0 =

(�+ 1)(2)� (2�)(1)(�+ 1)2

=2

(�+ 1)2.

At (1� 1), 0 = 12

, and an equation of the tangent line is � 1 = 12(�� 1), or = 1

2�+ 1

2.

30. = 0�

�� 0 =

� · 0� � 0� · 1�2

=0�(�� 1)

�2.

At (1� 0), 0 = 0, and an equation of the tangent line is � 0 = 0(�� 1), or = 0.

NOT FOR SALE SECTION 3 2EC THE PRODUCT

INSTRUCTOR USE ONLY AtAt (1(1� 0� 0)),, 00 = 0= 0, and an equation of the tangent line isand an equation of the tangent line is �� 00 = 0((��� 1)1), oror == 00..

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182 ¤ CHAPTER 3 DIFFERENTIATION RULES

31. = 2�0� � 0 = 2(� · 0� + 0� · 1) = 20�(�+ 1).At (0� 0), 0 = 200(0 + 1) = 2 · 1 · 1 = 2, and an equation of the tangent line is � 0 = 2(�� 0), or = 2�. The slope of

the normal line is� 12

, so an equation of the normal line is � 0 = � 12(�� 0), or = � 1

2�.

32. =��

�+ 1� 0 =

(�+ 1)

�1

2��

���� (1)

(�+ 1)2=(�+ 1)� (2�)2�� (�+ 1)2

=1� �

2�� (�+ 1)2

.

At (4� 0�4), 0 = �3100 = �0�03, and an equation of the tangent line is � 0�4 = �0�03(�� 4), or = �0�03�+ 0�52. The

slope of the normal line is 1003 , so an equation of the normal line is � 0�4 = 100

3 (�� 4) � = 1003 �� 400

3 + 25 �

= 1003�� 1994

15.

33. (a) = �(�) =1

1 + �2�

� 0(�) =(1 + �2)(0)� 1(2�)

(1 + �2)2=

�2�(1 + �2)2

. So the slope of the

tangent line at the point��1� 12� is � 0(�1) = 2

22= 1

2and its

equation is � 12= 1

2(�+ 1) or = 1

2�+ 1.

(b)

34. (a) = �(�) =�

1 + �2�

� 0(�) =(1 + �2)1� �(2�)

(1 + �2)2=

1� �2

(1 + �2)2. So the slope of the

tangent line at the point (3� 0�3) is � 0(3) = �8100

and its equation is

� 0�3 = �0�08(�� 3) or = �0�08�+ 0�54.

(b)

35. (a) �(�) = (�3 � �)0� � � 0(�) = (�3 � �)0� + 0�(3�2 � 1) = 0�(�3 + 3�2 � �� 1)

(b) � 0 = 0 when � has a horizontal tangent line, � 0 is negative when � is

decreasing, and � 0 is positive when � is increasing.

36. (a) �(�) = 0�

2�2 + �+ 1�

� 0(�) =(2�2 + �+ 1)0� � 0�(4�+ 1)

(2�2 + �+ 1)2=

0�(2�2 + �+ 1� 4�� 1)(2�2 + �+ 1)2

=0�(2�2 � 3�)(2�2 + �+ 1)2

(b) � 0 = 0 when � has a horizontal tangent line, � 0 is negative when � is

decreasing, and � 0 is positive when � is increasing.

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 183

37. (a) �(�) = �2 � 1�2 + 1

� 0(�) =(�2 + 1)(2�)� (�2 � 1)(2�)

(�2 + 1)2=(2�)[(�2 + 1)� (�2 � 1)]

(�2 + 1)2=

(2�)(2)

(�2 + 1)2=

4�

(�2 + 1)2�

� 00(�) =(�2 + 1)2(4)� 4�(�4 + 2�2 + 1)0

[(�2 + 1)2]2=4(�2 + 1)2 � 4�(4�3 + 4�)

(�2 + 1)4

=4(�2 + 1)2 � 16�2(�2 + 1)

(�2 + 1)4=4(�2 + 1)[(�2 + 1)� 4�2]

(�2 + 1)4=4(1� 3�2)(�2 + 1)3

(b) � 0 = 0 when � has a horizontal tangent and � 00 = 0 when � 0 has a

horizontal tangent. � 0 is negative when � is decreasing and positive when �

is increasing. � 00 is negative when � 0 is decreasing and positive when � 0 is

increasing. � 00 is negative when � is concave down and positive when � is

concave up.

38. (a) �(�) = (�2 � 1)0� � � 0(�) = (�2 � 1)0� + 0�(2�) = 0�(�2 + 2�� 1) �� 00(�) = 0�(2�+ 2) + (�2 + 2�� 1)0� = 0�(�2 + 4�+ 1)

(b) We can see that our answers are plausible, since � has horizontal tangents

where � 0(�) = 0, and � 0 has horizontal tangents where � 00(�) = 0.

39. �(�) = �2

1 + �� � 0(�) =

(1 + �)(2�)� �2(1)

(1 + �)2=2�+ 2�2 � �2

(1 + �)2=

�2 + 2�

�2 + 2�+ 1�

� 00(�) =(�2 + 2�+ 1)(2�+ 2)� (�2 + 2�)(2�+ 2)

(�2 + 2�+ 1)2=(2�+ 2)(�2 + 2�+ 1� �2 � 2�)

[(�+ 1)2]2

=2(�+ 1)(1)

(�+ 1)4=

2

(�+ 1)3,

so � 00(1) = 2

(1 + 1)3=2

8=1

4.

40. �(�) = �

0�� �0(�) =

0� · 1� � · 0�(0�)2

=0�(1� �)

(0�)2=1� �

0��

�00(�) =0� · (�1)� (1� �)0�

(0�)2=

0�[�1� (1� �)]

(0�)2=

�� 20�

�000(�) =0� · 1� (�� 2)0�

(0�)2=

0�[1� (�� 2)](0�)2

=3� �

0��

�(4)(�) =0� · (�1)� (3� �)0�

(0�)2=

0�[�1� (3� �)]

(0�)2=

�� 40�

.

The pattern suggests that �()(�) = (�� ()(�1)0�

. (We could use mathematical induction to prove this formula.)

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184 ¤ CHAPTER 3 DIFFERENTIATION RULES

41. We are given that �(5) = 1, � 0(5) = 6, �(5) = �3, and �0(5) = 2.

(a) (��)0(5) = �(5)�0(5) + �(5)� 0(5) = (1)(2) + (�3)(6) = 2� 18 = �16

(b)��

�0(5) =

�(5)� 0(5)� �(5)�0(5)[�(5)]2

=(�3)(6)� (1)(2)

(�3)2 = �209

(c)��

�0(5) =

�(5)�0(5)� �(5)� 0(5)[�(5)]2

=(1)(2)� (�3)(6)

(1)2= 20

42. We are given that �(2) = �3, �(2) = 4, � 0(2) = �2, and �0(2) = 7.

(a) �(�) = 5�(�)� 4�(�) � �0(�) = 5� 0(�)� 4�0(�), so

�0(2) = 5� 0(2)� 4�0(2) = 5(�2)� 4(7) = �10� 28 = �38.

(b) �(�) = �(�)�(�) � �0(�) = �(�)�0(�) + �(�)� 0(�), so

�0(2) = �(2)�0(2) + �(2)� 0(2) = (�3)(7) + (4)(�2) = �21� 8 = �29.

(c) �(�) = �(�)

�(�)� �0(�) =

�(�)� 0(�)� �(�)�0(�)[�(�)]2

, so

�0(2) =�(2)� 0(2)� �(2)�0(2)

[�(2)]2=4(�2)� (�3)(7)

42=�8 + 2116

=13

16.

(d) �(�) = �(�)

1 + �(�)� �0(�) =

[1 + �(�)] �0(�)� �(�)� 0(�)[1 + �(�)]2

, so

�0(2) =[1 + �(2)] �0(2)� �(2) � 0(2)

[1 + �(�)]2=[1 + (�3)](7)� 4(�2)

[1 + (�3)]2 =�14 + 8(�2)2 =

�64= �3

2.

43. �(�) = 0��(�) � � 0(�) = 0��0(�) + �(�)0� = 0�[�0(�) + �(�)]. � 0(0) = 00[�0(0) + �(0)] = 1(5 + 2) = 7

44. �

��

�(�)

!=

��0(�)� �(�) · 1�2

� �

��

�(�)

!�=2

=2�0(2)� �(2)

22=2(�3)� (4)

4=�104

= �2�5

45. (a) From the graphs of � and �, we obtain the following values: �(1) = 2 since the point (1� 2) is on the graph of � ;

�(1) = 1 since the point (1� 1) is on the graph of �; � 0(1) = 2 since the slope of the line segment between (0� 0) and

(2� 4) is 4� 02� 0 = 2; �0(1) = �1 since the slope of the line segment between (�2� 4) and (2� 0) is 0� 4

2� (�2) = �1.

Now �(�) = �(�)�(�), so �0(1) = �(1)�0(1) + �(1) � 0(1) = 2 · (�1) + 1 · 2 = 0.

(b) )(�) = �(�)��(�), so )0(5) = �(5)� 0(5)� �(5)�0(5)[�(5)]2

=2�� 1

3

�� 3 · 23

22=� 83

4= �2

3

46. (a) � (�) = ! (�)#(�), so � 0(2) = ! (2)#0(2) +#(2)! 0(2) = 3 · 24+ 2 · 0 = 3

2�

(b) /(�) = ! (�)�#(�), so /0(7) =#(7)! 0(7)� ! (7)#0(7)

[#(7)]2=1 · 1

4� 5 · �� 2

3

�12

=1

4+10

3=43

12

47. (a) = ��(�) � 0 = ��0(�) + �(�) · 1 = ��0(�) + �(�)

(b) = �

�(�)� 0 =

�(�) · 1� ��0(�)[�(�)]2

=�(�)� ��0(�)[�(�)]2

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 185

(c) = �(�)

�� 0 =

��0(�)� �(�) · 1(�)2

=��0(�)� �(�)

�2

48. (a) = �2�(�) � 0 = �2� 0(�) + �(�)(2�)

(b) = �(�)

�2� 0 =

�2� 0(�)� �(�)(2�)

(�2)2=

�� 0(�)� 2�(�)�3

(c) = �2

�(�)� 0 =

�(�)(2�)� �2� 0(�)[�(�)]2

(d) = 1 + ��(�)��

0 =

�� [�� 0(�) + �(�)]� [1 + ��(�)]

1

2��

(�� )

2

=�3�2� 0(�) + �1�2�(�)� 1

2��1�2 � 1

2�1�2�(�)

�· 2�

1�2

2�1�2=

��(�) + 2�2� 0(�)� 12�3�2

49. If � (�) denotes the population at time � and �(�) the average annual income, then � (�) = � (�)�(�) is the total personal

income. The rate at which � (�) is rising is given by � 0(�) = � (�)�0(�) +�(�)� 0(�) �

� 0(1999) = � (1999)�0(1999) +�(1999)� 0(1999) = (961,400)($1400�yr) + ($30,593)(9200�yr)

= $1,345,960,000�yr + $281,455,600�yr = $1,627,415,600�yr

So the total personal income was rising by about $1.627 billion per year in 1999.

The term � (�)�0(�) $1.346 billion represents the portion of the rate of change of total income due to the existing

population’s increasing income. The term �(�)� 0(�) $281 million represents the portion of the rate of change of total

income due to increasing population.

50. (a) �(20) = 10,000 means that when the price of the fabric is $20�yard, 10,000 yards will be sold.

� 0(20) = �350 means that as the price of the fabric increases past $20�yard, the amount of fabric which will be sold is

decreasing at a rate of 350 yards per (dollar per yard).

(b) .(�) = ��(�) � .0(�) = �� 0(�) + �(�) · 1 � .0(20) = 20� 0(20) + �(20) · 1 = 20(�350) + 10,000 = 3000.

This means that as the price of the fabric increases past $20�yard, the total revenue is increasing at $3000�($�yard). Note

that the Product Rule indicates that we will lose $7000�($�yard) due to selling less fabric, but this loss is more than made

up for by the additional revenue due to the increase in price.

51. � is increasing when � 0 is positive. �(�) = �30� � � 0(�) = �30� + 0�(3�2) = �20�(�+ 3). Now �2 � 0 and 0� 0

for all �, so � 0(�) 0 when �+ 3 0 and � 6= 0; that is, when � � (�3� 0) (0��). So � is increasing on (�3��).

52. � is concave downward when � 00 is negative. �(�) = �20� � � 0(�) = �20� + 0�(2�) �� 00(�) = �20� + 0�(2�) + 0�(2) + (2�)0� = 0�(�2 + 2�+ 2 + 2�) = 0�(�2 + 4�+ 2). Note that 0� 0 for all �

and � 00(�) = 0 � � = �2±�2. � 00(�) � 0 when � � ��2��2��2 +�2 �.

SECTION 3 2EC THE PRODUCT

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186 ¤ CHAPTER 3 DIFFERENTIATION RULES

53. If = �(�) =�

�+ 1, then � 0(�) = (�+ 1)(1)� �(1)

(�+ 1)2=

1

(�+ 1)2. When � = �, the equation of the tangent line is

� �

�+ 1=

1

(�+ 1)2(�� �). This line passes through (1� 2) when 2� �

�+ 1=

1

(�+ 1)2(1� �) �

2(�+ 1)2 � �(�+ 1) = 1� � � 2�2 + 4�+ 2� �2 � �� 1 + � = 0 � �2 + 4�+ 1 = 0.

The quadratic formula gives the roots of this equation as � =�4±�

42 � 4(1)(1)2(1)

=�4±�12

2= �2±�3,

so there are two such tangent lines. Since

���2±�3 �= �2±�3

�2±�3 + 1 =�2±�3�1±�3 ·

�1��3�1��3

=2± 2�3��3� 3

1� 3 =�1±�3�2 =

1��32

,

the lines touch the curve at ���2 +�3� 1�

�3

2

� (�0�27��0�37)

and ���2��3� 1+

�3

2

� (�3�73� 1�37).

54. = �� 1�+ 1

� 0 =(�+ 1)(1)� (�� 1)(1)

(�+ 1)2=

2

(�+ 1)2. If the tangent intersects

the curve when � = �, then its slope is 2�(�+ 1)2. But if the tangent is parallel to

�� 2 = 2, that is, = 12�� 1, then its slope is 1

2. Thus, 2

(�+ 1)2=1

2�

(�+ 1)2 = 4 � �+ 1 = ±2 � � = 1 or �3. When � = 1, = 0 and the

equation of the tangent is � 0 = 12(�� 1) or = 1

2�� 1

2. When � = �3, = 2 and

the equation of the tangent is � 2 = 12(�+ 3) or = 1

2�+ 7

2.

55. . = �

�� .0 =

�� 0 � ��0

�2. For �(�) = �� 3�3 + 5�5, � 0(�) = 1� 9�2 + 25�4,

and for �(�) = 1 + 3�3 + 6�6 + 9�9, �0(�) = 9�2 + 36�5 + 81�8.

Thus, .0(0) =�(0)� 0(0)� �(0)�0(0)

[�(0)]2=1 · 1� 0 · 0

12=1

1= 1.

56. / = �

�� /0 =

�� 0 � ��0

�2. For �(�) = 1 + �+ �2 + �0�, � 0(�) = 1 + 2�+ �0� + 0�,

and for �(�) = 1� �+ �2 � �0�, �0(�) = �1 + 2�� �0� � 0�.

Thus, /0(0) =�(0)� 0(0)� �(0)�0(0)

[�(0)]2=1 · 2� 1 · (�2)

12=4

1= 4.

57. (a) (���)0 = [(��)�]0 = (��)0�+ (��)�0 = (� 0� + ��0)�+ (��)�0 = � 0��+ ��0�+ ���0

(b) Putting � = � = � in part (a), we have �

��[�(�)]3 = (���)0 = � 0�� + �� 0� + ��� 0 = 3��� 0 = 3[�(�)]2� 0(�).

(c) �

��(03�) =

��(0�)3 = 3(0�)20� = 302�0� = 303�

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 187

58. (a) We use the Product Rule repeatedly: ! = �� � ! 0 = � 0� + ��0 �! 00 = (� 00� + � 0�0) + (� 0�0 + ��00) = � 00� + 2� 0�0 + ��00.

(b) ! 000 = � 000� + � 00�0 + 2 (� 00�0 + � 0�00) + � 0�00 + ��000 = � 000� + 3� 00�0 + 3� 0�00 + ��000 �! (4) = � (4)� + � 000�0 + 3 (� 000�0 + � 00�00) + 3 (� 00�00 + � 0�000) + � 0�000 + ��(4)

= � (4)� + 4� 000�0 + 6� 00�00 + 4� 0�000 + ��(4)

(c) By analogy with the Binomial Theorem, we make the guess:

! () = � ()� + (� (�1)�0 +

�(

2

�� (�2)�00 + · · ·+

�(

�� (��)�(�) + · · ·+ (� 0�(�1) + ��(),

where

�(

�=

(!

�! ((� �)!=

(((� 1)((� 2) · · · ((� � + 1)

�!.

59. For �(�) = �20�, � 0(�) = �20� + 0�(2�) = 0�(�2 + 2�). Similarly, we have

� 00(�) = 0�(�2 + 4�+ 2)

� 000(�) = 0�(�2 + 6�+ 6)

� (4)(�) = 0�(�2 + 8�+ 12)

� (5)(�) = 0�(�2 + 10�+ 20)

It appears that the coef�cient of � in the quadratic term increases by 2 with each differentiation. The pattern for the

constant terms seems to be 0 = 1 · 0, 2 = 2 · 1, 6 = 3 · 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that

� ()(�) = 0�[�2 + 2(�+ (((� 1)].

Proof: Let & be the statement that � ()(�) = 0�[�2 + 2(�+ (((� 1)].1. &1 is true because � 0(�) = 0�(�2 + 2�).

2. Assume that &� is true; that is, � (�)(�) = 0�[�2 + 2��+ �(� � 1)]. Then

� (�+1)(�) =�

��

�� (�)(�)

�= 0�(2�+ 2�) + [�2 + 2��+ �(� � 1)]0�

= 0�[�2 + (2� + 2)�+ (�2 + �)] = 0�[�2 + 2(� + 1)�+ (� + 1)�]

This shows that &�+1 is true.

3. Therefore, by mathematical induction, & is true for all (; that is, � ()(�) = 0�[�2 + 2(�+ (((� 1)] for everypositive integer (.

60. (a) �

��

�1

�(�)

�=

�(�) · ���(1)� 1 · �

��[�(�)]

[�(�)]2[Quotient Rule] =

�(�) · 0� 1 · �0(�)[�(�)]2

=0� �0(�)[�(�)]2

= � �0(�)[�(�)]2

(b) = 1

�+ �0�� 0 = � 1 + �0�

(�+ �0�)2

(c) �

��(��) =

��

�1

�= � (�

)0

(�)2[by the Reciprocal Rule] = �(�

�1

�2= �(��1�2 = �(���1

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188 ¤ CHAPTER 3 DIFFERENTIATION RULES

3.3 Derivatives of Trigonometric Functions

1. �(�) = 3�2 � 2 cos� � � 0(�) = 6�� 2(� sin�) = 6�+ 2 sin�

2. = 2csc�+ 5cos� � 0 = �2 csc� cot�� 5 sin�

3. �(�) = sin�+ 12cot� � � 0(�) = cos�� 1

2csc2 �

4. �(�) =�� sin� � � 0(�) =

�� cos�+ sin�

�12��1�2

�=�� cos�+

sin�

2��

5. = sec � tan � � 0 = sec � (sec2 �) + tan � (sec � tan �) = sec � (sec2 � + tan2 �). Using the identity

1 + tan2 � = sec2 �, we can write alternative forms of the answer as sec � (1 + 2 tan2 �) or sec � (2 sec2 � � 1).

6. �(�) = 0�(tan � � �) � �0(�) = 0�(sec2 � � 1) + (tan � � �)0� = 0�(sec2 � � 1 + tan � � �)

7. = + cos �+ �2 sin � � 0 = +(� sin �) + �2(cos �) + sin � (2�) = �+ sin �+ �(� cos �+ 2 sin �)

8. �(�) = cot �

0�� � 0(�) =

0�(� csc2 �)� (cot �)0�(0�)2

=0�(� csc2 �� cot �)

(0�)2= �csc

2 �+ cot �

0�

9. = �

2� tan� � 0 =(2� tan�)(1)� �(� sec2 �)

(2� tan�)2 =2� tan�+ � sec2 �

(2� tan�)2

10. = 1 + sin�

�+ cos��

0 =(�+ cos�)(cos�)� (1 + sin�)(1� sin�)

(�+ cos�)2=

� cos�+ cos2 �� (1� sin2 �)(�+ cos�)2

=� cos�+ cos2 �� (cos2 �)

(�+ cos�)2=

� cos�

(�+ cos�)2

11. �(�) = sec �

1 + sec ��

� 0(�) =(1 + sec �)(sec � tan �)� (sec �)(sec � tan �)

(1 + sec �)2=(sec � tan �) [(1 + sec �)� sec �]

(1 + sec �)2=sec � tan �

(1 + sec �)2

12. = 1� sec�tan�

0 =tan� (� sec� tan�)� (1� sec�)(sec2 �)

(tan�)2=sec�

�� tan2 �� sec�+ sec2 ��tan2 �

=sec� (1� sec�)

tan2 �

13. Using Exercise 3.2.57(a), �(�) = �0� csc� �

� 0(�) = (�)00� csc�+ �(0�)0 csc�+ �0�(csc�)0 = 10� csc�+ �0� csc�+ �0�(� cot� csc�)= 0� csc� (1 + �� � cot�)

14. Using Exercise 3.2.57(a), �(�) = �2 sin� tan� �

� 0(�) = (�2)0 sin� tan�+ �2(sin�)0 tan�+ �2 sin� (tan�)0 = 2� sin� tan�+ �2 cos� tan�+ �2 sin� sec2 �

= 2� sin� tan�+ �2 sin�+ �2 sin� sec2 � = � sin� (2 tan�+ �+ � sec2 �).

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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 189

15. �

��(csc�) =

��

�1

sin�

�=(sin�)(0)� 1(cos�)

sin2 �=� cos�sin2 �

= � 1

sin�· cos�sin�

= � csc� cot�

16. �

��(sec�) =

��

�1

cos�

�=(cos�)(0)� 1(� sin�)

cos2 �=sin�

cos2 �=

1

cos�· sin�cos�

= sec� tan�

17. �

��(cot�) =

��

�cos�sin�

�=(sin�)(� sin�)� (cos�)(cos�)

sin2 �= � sin

2 �+ cos2 �

sin2 �= � 1

sin2 �= � csc2 �

18. �(�) = cos� �

� 0(�) = lim��0

�(�+ �)� �(�)

�= lim

��0

cos (�+ �)� cos��

= lim��0

cos� cos�� sin� sin�� cos��

= lim��0

�cos�

cos�� 1�

� sin� sin��

�= cos� lim

��0

cos�� 1�

� sin� lim��0

sin�

= (cos�)(0)� (sin�)(1) = � sin�

19. = sec� � 0 = sec� tan�, so 0(�3) = sec �

3tan �

3= 2

�3. An equation of the tangent line to the curve = sec�

at the point��3� 2�

is � 2 = 2�3 ��� �3

�or = 2

�3�+ 2� 2

3

�3�.

20. = 0� cos� � 0 = 0�(� sin�) + (cos�)0� = 0�(cos�� sin�) � the slope of the tangent line at (0� 1) is

00(cos 0� sin 0) = 1(1� 0) = 1 and an equation is � 1 = 1(�� 0) or = �+ 1.

21. = �+cos� � 0 = 1� sin�. At (0� 1), 0 = 1, and an equation of the tangent line is � 1 = 1(�� 0), or = �+1.

22. = 1

sin�+ cos�� 0 = � cos�� sin�

(sin�+ cos�)2[Reciprocal Rule]. At (0� 1), 0 = � 1� 0

(0 + 1)2= �1, and an equation

of the tangent line is � 1 = �1(�� 0), or = ��+ 1.

23. (a) = 2� sin� � 0 = 2(� cos�+ sin� · 1). At��2� ��,

0 = 2��2 cos

�2 + sin

�2

�= 2(0 + 1) = 2, and an equation of the

tangent line is � � = 2��� �

2

�, or = 2�.

(b)

24. (a) = 3�+ 6cos� � 0 = 3� 6 sin�. At (�3 � � + 3),

0 = 3� 6 sin �3= 3� 6

�32= 3� 3�3, and an equation of the

tangent line is � (� + 3) = �3� 3�3 ���� �

3

�, or

=�3� 3�3 �+ 3 + �

�3.

(b)

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190 ¤ CHAPTER 3 DIFFERENTIATION RULES

25. (a) �(�) = sec�� � � � 0(�) = sec� tan�� 1(b)

Note that � 0 = 0 where � has a minimum. Also note that � 0 is negative

when � is decreasing and � 0 is positive when � is increasing.

26. (a) �(�) = 0� cos� � � 0(�) = 0�(� sin�) + (cos�) 0� = 0�(cos�� sin�) �

� 00(�) = 0�(� sin�� cos�) + (cos�� sin�) 0� = 0�(� sin�� cos�+ cos�� sin�) = �20� sin�(b)

Note that � 0 = 0 where � has a minimum and � 00 = 0 where � 0 has a

minimum. Also note that � 0 is negative when � is decreasing and � 00 is

negative when � 0 is decreasing.

27. "(�) = � sin � � "0(�) = � (cos �) + (sin �) · 1 = � cos � + sin � �"00(�) = � (� sin �) + (cos �) · 1 + cos � = �� sin � + 2cos �

28. �(�) = csc � � � 0(�) = � csc � cot � � � 00(�) = �[csc �(� csc2 �) + cot �(� csc � cot �)] = csc �(csc2 �+ cot2 �),

so � 00(�6) = 2(22 +

�32) = 2(4 + 3) = 14.

29. (a) �(�) = tan�� 1sec�

� 0(�) =sec�(sec2 �)� (tan�� 1)(sec� tan�)

(sec�)2=sec�(sec2 �� tan2 �+ tan�)

sec 2 �=1 + tan�

sec�

(b) �(�) = tan�� 1sec�

=

sin�

cos�� 1

1

cos�

=

sin�� cos�cos�1

cos�

= sin�� cos� � � 0(�) = cos�� (� sin�) = cos�+ sin�

(c) From part (a), � 0(�) = 1 + tan�

sec�=

1

sec�+tan�

sec�= cos�+ sin�, which is the expression for � 0(�) in part (b).

30. (a) �(�) = �(�) sin� � �0(�) = �(�) cos� + sin� · � 0(�), so

�0(�3) = �(�

3) cos �

3+ sin �

3· � 0(�

3) = 4 · 1

2+

�32· (�2) = 2��3

(b) �(�) = cos�

�(�)� �0(�) =

�(�) · (� sin�)� cos� · � 0(�)[�(�)]2

, so

�0(�3) =

�(�3) · (� sin �

3)� cos �

3· � 0(�

3)

���3

�2 =4���32

�� �

12

�(�2)

42=�2�3 + 1

16=1� 2�316

31. �(�) = �+ 2 sin� has a horizontal tangent when � 0(�) = 0 � 1 + 2 cos� = 0 � cos� = � 12 �

� = 2�3+ 2�( or 4�

3+ 2�(, where ( is an integer. Note that 4�

3and 2�

3are ±�

3units from �. This allows us to write the

solutions in the more compact equivalent form (2(+ 1)� ± �3

, ( an integer.

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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 191

32. �(�) = 0� cos� has a horizontal tangent when � 0(�) = 0. � 0(�) = 0�(� sin�) + (cos�)0� = 0�(cos�� sin�).� 0(�) = 0 � cos�� sin� = 0 � cos� = sin� � tan� = 1 � � = �

4+ (�, ( an integer.

33. �(�) = �� 2 sin�, 0 � � � 2�. � 0(�) = 1� 2 cos�. So � 0(�) 0 � 1� 2 cos� 0 � �2 cos� �1 �cos� � 1

2� �

3� � � 5�

3� � is increasing on

��3� 5�3

�.

34. �(�) = 2� � tan�, ��2� � � �

2, is concave downward when � 00(�) � 0.

� 0(�) = 2� sec2 � = 2� (sec�)(sec�) � � 00(�) = �[sec�(sec� tan�) + sec�(sec� tan�)] = �2 sec2 � tan�.

� 00(�) � 0 when tan� 0, that is, for 0 � � � �2

.

35. (a) �(�) = 8 sin � � )(�) = �0(�) = 8 cos � � �(�) = �00(�) = �8 sin �

(b) The mass at time � = 2�3

has position ��2�3

�= 8 sin 2�

3= 8

��32

�= 4

�3, velocity )

�2�3

�= 8cos 2�

3= 8

�� 12

�= �4,

and acceleration ��2�3

�= �8 sin 2�

3= �8

��32

�= �4�3. Since )

�2�3

�� 0, the particle is moving to the left.

36. (a) �(�) = 2 cos �+ 3 sin � � )(�) = �2 sin �+ 3cos � ��(�) = �2 cos �� 3 sin �

(b)

(c) � = 0 � �2 2�55. So the mass passes through the equilibrium

position for the �rst time when � 2�55 s.

(d) ) = 0 � �1 0�98, �(�1) 3�61 cm. So the mass travels

a maximum of about 3�6 cm (upward and downward) from its equilibrium position.

(e) The speed |)| is greatest when � = 0, that is, when � = �2 + (�, ( a positive integer.

37. From the diagram we can see that sin � = ��10 � � = 10 sin �. We want to �nd the rate

of change of � with respect to �, that is, �����. Taking the derivative of � = 10 sin �, we get

����� = 10(cos �). So when � = �3

, ����= 10 cos �

3= 10

�12

�= 5 ft�rad.

38. (a) ! = :%

: sin � + cos �� �!

��=(: sin � + cos �)(0)� :% (: cos � � sin �)

(: sin � + cos �)2=

:% (sin � � : cos �)

(: sin � + cos �)2

(b) �!��

= 0 � :% (sin � � : cos �) = 0 � sin � = : cos � � tan � = : � � = tan�1 :

(c) From the graph of ! = 0�6(50)

0�6 sin � + cos �for 0 � � � 1, we see that

�!

��= 0 � � 0�54. Checking this with part (b) and : = 0�6, we

calculate � = tan�1 0�6 0�54. So the value from the graph is consistent

with the value in part (b).

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192 ¤ CHAPTER 3 DIFFERENTIATION RULES

39. �

��(sin�) = cos� � �2

��2(sin�) = � sin� � �3

��3(sin�) = � cos� � �4

��4(sin�) = sin�.

The derivatives of sin� occur in a cycle of four. Since 99 = 4(24) + 3, we have �99

��99(sin�) =

�3

��3(sin�) = � cos�.

40. Let �(�) = � sin� and �(�) = sin�, so �(�) = ��(�). Then � 0(�) = �(�) + ��0(�),

� 00(�) = �0(�) + �0(�) + ��00(�) = 2�0(�) + ��00(�),

� 000(�) = 2�00(�) + �00(�) + ��000(�) = 3�00(�) + ��000(�)� · · · � � ()(�) = (�(�1)(�) + ��()(�).

Since 34 = 4(8) + 2, we have �(34)(�) = �(2)(�) =�2

��2(sin�) = � sin� and �(35)(�) = � cos�.

Thus, �35

��35(� sin�) = 35�(34)(�) + ��(35)(�) = �35 sin�� � cos�.

41. = � sin�+� cos� � 0 = � cos��� sin� � 00 = �� sin��� cos�. Substituting these

expressions for , 0, and 00 into the given differential equation 00 + 0 � 2 = sin� gives us

(�� sin��� cos�) + (� cos��� sin�)� 2(� sin�+� cos�) = sin� �

�3� sin��� sin�+� cos�� 3� cos� = sin� � (�3���) sin�+ (�� 3�) cos� = 1 sin�, so we must have

�3��� = 1 and �� 3� = 0 (since 0 is the coef�cient of cos� on the right side). Solving for � and �, we add the �rst

equation to three times the second to get � = � 110

and � = � 310

.

42. (a) If � 0, then 5� 0, and hence � [= 5�] 0. lim��0

sin 5�

�= lim

��0

sin �15�

= 5 lim��0

sin �

�= 5(1) = 5.

(b) Let �(�) = sin 5�. Then

��(sin 5�) = � 0(�) = lim

��0

�(�+ �)� �(�)

�= lim

��0

sin [5(�+ �)]� sin 5��

= lim��0

sin(5�+ 5�)� sin 5��

= lim��0

sin 5� cos 5�+ cos 5� sin 5�� sin 5��

= lim��0

�sin 5� · cos 5�� 1

�+ cos 5�

sin 5�

= sin 5� lim��0

cos 5�� 1�

+ cos 5� lim��0

sin 5�

The �rst limit can be shown to equal 0 in a manner similar to part (a) and Equation 3. The second limit is equal to 5,

by part (a). Thus, we have � 0(�) = (sin 5�)(0) + (cos 5�)(5) = 5 cos 5�.

43. lim��0

tan 6�

sin 2�= lim

��0

�sin 6�

�· 1

cos 6�· �

sin 2�

�= lim

��0

6 sin 6�

6�· lim��0

1

cos 6�· lim��0

2�

2 sin 2�

= 6 lim��0

sin 6�

6�· lim��0

1

cos 6�· 12lim��0

2�

sin 2�= 6(1) · 1

1· 12(1) = 3

44. lim��0

sin 3� sin 5�

�2= lim

��0

�3 sin 3�

3�· 5 sin 5�

5�

�= lim

��0

3 sin 3�

3�· lim��0

5 sin 5�

5�

= 3 lim��0

sin 3�

3�· 5 lim

��0

sin 5�

5�= 3(1) · 5(1) = 15

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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 193

45. lim��0

sin �

� + tan �=

lim��0

sin �

lim��0

� + tan �

=1

lim��0

�1 +

sin �

�· 1

cos �

� =1

1 + 1 · 1 =1

2

46. (a) Let � = 1

�. Then as � �, � 0, and lim

���� sin

1

�= lim

��0

1

�sin � = lim

��0

sin �

�= 1.

(b) Since �1 � sin (1��) � 1, we have (as illustrated in the �gure)

�|�| � � sin (1��) � |�|. We know that lim��0

(|�|) = 0 and

lim��0

(�|�|) = 0; so by the Squeeze Theorem, lim��0

� sin (1��) = 0.

(c)

47. (a) �

��tan� =

��

sin�

cos�� sec2 � =

cos� cos�� sin� (� sin�)cos2 �

=cos2 �+ sin2 �

cos2 �. So sec2 � = 1

cos2 �.

(b) �

��sec� =

��

1

cos�� sec� tan� =

(cos�)(0)� 1(� sin�)cos2 �

. So sec� tan� = sin�

cos2 �.

(c) �

��(sin�+ cos�) =

��

1 + cot�

csc��

cos�� sin� = csc� (� csc2 �)� (1 + cot�)(� csc� cot�)csc2 �

=csc� [� csc2 �+ (1 + cot�) cot�]

csc2 �

=� csc2 �+ cot2 �+ cot�

csc�=�1 + cot�csc�

So cos�� sin� = cot�� 1csc�

.

48. We get the following formulas for � and � in terms of �:

sin�

2=

10� � = 10 sin

2and cos

2=

10� � = 10 cos

2

Now �(�) = 12��2 and �(�) = 1

2(2�)� = ��. So

lim��0+

�(�)

�(�)= lim

��0+

12��2

��= 1

2� lim��0+

�= 1

2� lim��0+

10 sin(��2)

10 cos(��2)

= 12� lim

��0+tan(��2) = 0

49. By the de�nition of radian measure, � = ��, where � is the radius of the circle. By drawing the bisector of the angle �, we can

see that sin �2=

��2

�� � = 2� sin

2. So lim

��0+

�= lim

��0+

��

2� sin(��2)= lim

��0+

2 · (��2)2 sin(��2)

= lim��0

��2

sin(��2)= 1.

[This is just the reciprocal of the limit lim��0

sin ��= 1 combined with the fact that as � 0, �

2 0 also�]

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194 ¤ CHAPTER 3 DIFFERENTIATION RULES

50. (a) It appears that �(�) = ��1� cos 2� has a jump discontinuity at � = 0.

(b) Using the identity cos 2� = 1� sin2�, we have ��1� cos 2� =

��1� (1� 2 sin2 �) =

��2 sin2�

=��

2 |sin�| . Thus,

lim��0�

��1� cos 2� = lim

��0���

2 |sin�| =1�2lim

��0��

�(sin�)

= � 1�2lim

��0�1

sin���= � 1�

2· 11= �

�2

2

Evaluating lim��0+

�(�) is similar, but | sin�| = +sin�, so we get 12

�2. These values appear to be reasonable values for

the graph, so they con�rm our answer to part (a).Another method: Multiply numerator and denominator by

�1 + cos 2�.

3.4 The Chain Rule

1. Let � = �(�) = 1 + 4� and = �(�) = 3��. Then �

��=

��

��

��= ( 1

3��2�3)(4) =

4

3 3�(1 + 4�)2

.

2. Let � = �(�) = 2�3 + 5 and = �(�) = �4. Then �

��=

��

��

��= (4�3)(6�2) = 24�2(2�3 + 5)3.

3. Let � = �(�) = �� and = �(�) = tan�. Then �

��=

��

��

��= (sec2 �)(�) = � sec2 ��.

4. Let � = �(�) = cot� and = �(�) = sin�. Then �

��=

��

��

��= (cos�)(� csc2 �) = � cos(cot�) csc2 �.

5. Let � = �(�) =�� and = �(�) = 0�. Then �

��=

��

��

��= (0�)

�12��1�2

�= 0

�� · 1

2��=

0��

2��

.

6. Let � = �(�) = 2� 0� and = �(�) =��. Then �

��=

��

��

��= ( 1

2��1�2)(�0�) = � 0�

2�2� 0�

.

7. ! (�) = (�4 + 3�2 � 2)5 � ! 0(�) = 5(�4 + 3�2 � 2)4 · ���

��4 + 3�2 � 2� = 5(�4 + 3�2 � 2)4(4�3 + 6�)

or 10�(�4 + 3�2 � 2)4(2�2 + 3)

8. ! (�) = (4�� �2)100 � ! 0(�) = 100(4�� �2)99 · �

��(4�� �2) = 100(4�� �2)99(4� 2�)

or 200�99(�� 2)(�� 4)99

9. ! (�) =�1� 2� = (1� 2�)1�2 � ! 0(�) = 1

2(1� 2�)�1�2(�2) = � 1�

1� 2�

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SECTION 3.4 THE CHAIN RULE ¤ 195

10. �(�) = (1 + �4)2�3 � � 0(�) = 23

�1 + �4

��1�3�4�3

�=

8�3

3 3�1 + �4

11. �(8) = 1

82 + 1= (82 + 1)�1 � � 0(8) = �1(82 + 1)�2(28) = � 28

(82 + 1)2

12. �(�) = 3�1 + tan � = (1 + tan �)1�3 � � 0(�) = 1

3(1 + tan �)�2�3 sec2 � =

sec2 �

3 3�(1 + tan �)2

13. = cos(�3 + �3) � 0 = � sin(�3 + �3) · 3�2 [�3 is just a constant] = �3�2 sin(�3 + �3)

14. = �3 + cos3 � � 0 = 3(cos�)2(� sin�) [�3 is just a constant] = �3 sin� cos2 �

15. �(�) = �3 � 3� � �0(�) = 3�2 � 3� ln 3 [by Formula 5]

16. = 3cot((�) � 0 = 3[� csc2((�) · (] = �3( csc2((�)

17. = �0��� � 0 = �0���(��)+ 0��� · 1 = 0���(���+ 1)

or (1� ��)0���

18. = 0�2� cos 4� � 0 = 0�2�(� sin 4� · 4) + cos 4�[0�2�(�2)] = �20�2�(2 sin 4�+ cos 4�)

19. = (2�� 5)4(8�2 � 5)�3 �0 = 4(2�� 5)3(2)(8�2 � 5)�3 + (2�� 5)4(�3)(8�2 � 5)�4(16�)= 8(2�� 5)3(8�2 � 5)�3 � 48�(2�� 5)4(8�2 � 5)�4

[This simpli�es to 8(2�� 5)3(8�2 � 5)�4(�4�2 + 30�� 5).]

20. �(�) = (�4 � 1)3(�3 + 1)4 ��0(�) = (�4 � 1)3 · 4(�3 + 1)3(3�2) + (�3 + 1)4 · 3(�4 � 1)2(4�3)

= 12�2(�4 � 1)2(�3 + 1)3 (�4 � 1) + �(�3 + 1)= 12�2(�4 � 1)2(�3 + 1)3(2�4 + �� 1)

21. = 0� cos � � 0 = 0� cos � · ���(� cos�) = 0� cos � [�(� sin�) + (cos�) · 1] = 0� cos�(cos�� � sin�)

22. Using Formula 5 and the Chain Rule, = 101��2 � 0 = 101��2(ln 10) · ���

�1� �2

�= �2�(ln 10)101��2 .

23. =��2 + 1

�2 � 1�3

0 = 3��2 + 1

�2 � 1�2· ���

��2 + 1

�2 � 1�= 3

��2 + 1

�2 � 1�2· (�

2 � 1)(2�)� (�2 + 1)(2�)(�2 � 1)2

= 3

��2 + 1

�2 � 1�2· 2�[�

2 � 1� (�2 + 1)](�2 � 1)2 = 3

��2 + 1

�2 � 1�2· 2�(�2)(�2 � 1)2 =

�12�(�2 + 1)2(�2 � 1)4

24. #() =�

2

+ 1

�5� # 0() = 5

�2

+ 1

�4· ( + 1)(2)� 2(1)

( + 1)2= 5 · 8

( + 1)4· (2 + 2� )

( + 1)2=59( + 2)

( + 1)6

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196 ¤ CHAPTER 3 DIFFERENTIATION RULES

25. = sec2 � + tan2 � = (sec�)2 + (tan�)2 �0 = 2(sec�)(sec� tan�) + 2(tan�)(sec2 �) = 2 sec2� tan�+ 2 sec2� tan� = 4 sec2� tan�

26. = 0� � 0��

0� + 0���

0 =(0� + 0��)(0� � (�0�))� (0� � 0��)(0� + (�0��))

(0� + 0��)2=

02� + 00 + 00 + 0�2� � (02� � 00 � 00 + 0�2�)(0� + 0��)2

=400

(0� + 0��)2=

4

(0� + 0��)2

27. = ���2 + 1

0 =

��2 + 1 (1)� � · 12 (�2 + 1)�1�2(2�)��

�2 + 1�2 =

��2 + 1� �2�

�2 + 1���2 + 1

�2 =

��2 + 1

��2 + 1� �2�

�2 + 1���2 + 1

�2=

��2 + 1

�� �2���2 + 1

�3 =1

(�2 + 1)3�2or (�2 + 1)�3�2

Another solution: Write as a product and make use of the Product Rule. = �(�2 + 1)�1�2 �0 = � · � 1

2(�2 + 1)�3�2(2�) + (�2 + 1)�1�2 · 1 = (�2 + 1)�3�2[��2 + (�2 + 1)1] = (�2 + 1)�3�2(1) = (�2 + 1)�3�2.

The step that students usually have trouble with is factoring out (�2 + 1)�3�2. But this is no different than factoring out �2

from �2 + �5; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, �32

is

smaller than �12

.

28. = 0� tan�� � 0 = 0� tan

�� · �

��

�� tan

���= 0� tan

���� sec2

�� · 1

2��1�2

�=

� sec2��

2��

0� tan��

29. = sin(tan 2�) � 0 = cos(tan 2�) · ���(tan 2�) = cos(tan 2�) · sec2(2�) · �

��(2�) = 2 cos(tan 2�) sec2(2�)

30. �(�) =�

�2 + 4=

��

�2 + 4

�1�2�

� 0(�) =1

2

��

�2 + 4

��1�2· ���

��

�2 + 4

�=1

2

��2 + 4

�1�2· (�

2 + 4)(1)� �(2�)

(�2 + 4)2

=(�2 + 4)1�2

2� 1�2· �

2 + 4� 2�2(�2 + 4) 2

=4� �2

2�1�2(�2 + 4)3�2

31. Using Formula 5 and the Chain Rule, = 2 sin�� �

0 = 2 sin��(ln 2) · ���(sin��) = 2 sin��(ln 2) · cos�� · � = 2 sin��(� ln 2) cos��

32. = sin(sin(sin�)) � 0 = cos(sin(sin�))�

��(sin(sin�)) = cos(sin(sin�)) cos(sin�) cos�

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SECTION 3.4 THE CHAIN RULE ¤ 197

33. = cot2(sin �) = [cot(sin �)]2 �

0 = 2[cot(sin �)] · ���[cot(sin �)] = 2 cot(sin �) · [� csc2(sin �) · cos �] = �2 cos � cot(sin �) csc2(sin �)

34. =��+

��+

�� � 0 = 1

2

��+

��+

����1�2

1 + 12

��+

����1�2�

1 + 12��1�2

�!

35. = cos�sin(tan��) = cos(sin(tan��))1�2 �

0 = � sin(sin(tan��))1�2 · ���(sin(tan��))1�2 = � sin(sin(tan��))1�2 · 1

2(sin(tan��))�1�2 · �

��(sin(tan��))

=� sin�sin(tan��)2�sin(tan��)

· cos(tan��) · ���tan�� =

� sin�sin(tan��)2�sin(tan��)

· cos(tan��) · sec2(��) · �

=�� cos(tan��) sec2(��) sin�sin(tan��)

2�sin(tan��)

36. = 2 3�2 � 0 = 2 3

�2

(ln 2)�

��

�3�

2�= 2 3

�2

(ln 2)3�2

(ln 3)(2�)

37. = cos(�2) � 0 = � sin(�2) · 2� = �2� sin(�2) �00 = �2� cos(�2) · 2�+ sin(�2) · (�2) = �4�2 cos(�2)� 2 sin(�2)

38. = cos2� = (cos�)2 � 0 = 2cos�(� sin�) = �2 cos� sin� �00 = (�2 cos�) cos�+ sin�(2 sin�) = �2 cos2�+ 2 sin2�Note: Many other forms of the answers exist. For example, 0 = � sin 2� and 00 = �2 cos 2�.

39. = 0�� sin;� � 0 = 0�� · ; cos;�+ sin;� · 10�� = 0��(; cos;�+ 1 sin;�) �00 = 0��(�;2 sin;�+ 1; cos;�) + (; cos;�+ 1 sin;�) · 10��

= 0��(�;2 sin;�+ 1; cos;�+ 1; cos;�+ 12 sin;�) = 0��(12 sin;�� ;2 sin;�+ 21; cos;�)

= 0��(12 � ;2) sin;�+ 21; cos;�

40. = 0�

� � 0 = 0�� · (0�)0 = 0�

� · 0� �

00 = 0�� · (0�)0 + 0� ·

�0�

��0= 0�

� · 0� + 0� · 0�� · 0� = 0�� · 0�(1 + 0�) or 0�

�+�(1 + 0�)

41. = (1 + 2�)10 � 0 = 10(1 + 2�)9 · 2 = 20(1 + 2�)9.

At (0� 1), 0 = 20(1 + 0)9 = 20, and an equation of the tangent line is � 1 = 20(�� 0), or = 20�+ 1.

42. =�1 + �3 = (1 + �3)1�2 � 0 = 1

2 (1 + �3)�1�2 · 3�2 = 3�2

2�1 + �3

. At (2� 3), 0 = 3 · 42�9= 2, and an equation of

the tangent line is � 3 = 2(�� 2), or = 2�� 1.

43. = sin(sin�) � 0 = cos(sin�) · cos�. At (�� 0), 0 = cos(sin�) · cos� = cos(0) · (�1) = 1(�1) = �1, and an

equation of the tangent line is � 0 = �1(�� �), or = ��+ �.

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198 ¤ CHAPTER 3 DIFFERENTIATION RULES

44. = sin�+ sin2 � � 0 = cos�+ 2 sin� cos�.

At (0� 0), 0 = 1, and an equation of the tangent line is � 0 = 1(�� 0), or = �.

45. (a) = 2

1 + 0��� 0 =

(1 + 0��)(0)� 2(�0��)

(1 + 0��)2=

20��

(1 + 0��)2.

At (0� 1), 0 = 200

(1 + 00)2=

2(1)

(1 + 1)2=2

22=1

2. So an equation of the

tangent line is � 1 = 12(�� 0) or = 1

2�+ 1.

(b)

46. (a) For � 0, |�| = �, and = �(�) =��2� �2

� 0(�) =

�2� �2 (1)� �

�12

�(2� �2)�1�2(�2�)��

2� �2�2 · (2� �2)1�2

(2� �2)1�2

=(2� �2) + �2

(2� �2)3�2=

2

(2� �2)3�2

So at (1� 1), the slope of the tangent line is � 0(1) = 2 and its equation

is � 1 = 2(�� 1) or = 2�� 1.

(b)

47. (a) �(�) = ��2� �2 = �(2 � �2)1�2 �

� 0(�) = � · 12(2� �2)�1�2(�2�) + (2� �2)1�2 · 1 = (2� �2)�1�2

��2 + (2� �2)=2� 2�2�2� �2

(b) � 0 = 0 when � has a horizontal tangent line, � 0 is negative when � is

decreasing, and � 0 is positive when � is increasing.

48. (a) From the graph of � , we see that there are 5 horizontal tangents, so there

must be 5 zeros on the graph of � 0. From the symmetry of the graph of � ,

we must have the graph of � 0 as high at � = 0 as it is low at � = �. The

intervals of increase and decrease as well as the signs of � 0 are indicated in

the �gure.

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SECTION 3.4 THE CHAIN RULE ¤ 199

(b) �(�) = sin(�+ sin 2�) �

� 0(�) = cos(�+sin 2�) · ���(�+sin 2�) = cos(�+sin 2�)(1+2 cos 2�)

49. For the tangent line to be horizontal, � 0(�) = 0. �(�) = 2 sin�+ sin2 � � � 0(�) = 2 cos�+ 2 sin� cos� = 0 �2 cos�(1 + sin�) = 0 � cos� = 0 or sin� = �1, so � = �

2+ 2(� or 3�

2+ 2(�, where ( is any integer. Now

���2

�= 3 and �

�3�2

�= �1, so the points on the curve with a horizontal tangent are

��2+ 2(�� 3

�and

�3�2+ 2(���1�,

where ( is any integer.

50. �(�) = sin 2�� 2 sin� � � 0(�) = 2 cos 2�� 2 cos� = 4 cos2 �� 2 cos�� 2, and 4 cos2 �� 2 cos�� 2 = 0 �

(cos�� 1)(4 cos�+ 2) = 0 � cos� = 1 or cos� = � 12

. So � = 2(� or (2(+ 1)� ± �3

, ( any integer.

51. ! (�) = �(�(�)) � ! 0(�) = � 0(�(�)) · �0(�), so ! 0(5) = � 0(�(5)) · �0(5) = � 0(�2) · 6 = 4 · 6 = 24

52. �(�) =�4 + 3�(�) � �0(�) = 1

2 (4 + 3�(�))�1�2 · 3� 0(�), so

�0(1) = 12(4 + 3�(1))�1�2 · 3� 0(1) = 1

2(4 + 3 · 7)�1�2 · 3 · 4 = 6�

25= 6

5

53. (a) �(�) = �(�(�)) � �0(�) = � 0(�(�)) · �0(�), so �0(1) = � 0(�(1)) · �0(1) = � 0(2) · 6 = 5 · 6 = 30.

(b) "(�) = �(�(�)) � "0(�) = �0(�(�)) · � 0(�), so "0(1) = �0(�(1)) · � 0(1) = �0(3) · 4 = 9 · 4 = 36.

54. (a) ! (�) = �(�(�)) � ! 0(�) = � 0(�(�)) · � 0(�), so ! 0(2) = � 0(�(2)) · � 0(2) = � 0(1) · 5 = 4 · 5 = 20.

(b) #(�) = �(�(�)) � #0(�) = �0(�(�)) · �0(�), so #0(3) = �0(�(3)) · �0(3) = �0(2) · 9 = 7 · 9 = 63.

55. (a) �(�) = �(�(�)) � �0(�) = � 0(�(�))�0(�). So �0(1) = � 0(�(1))�0(1) = � 0(3)�0(1). To �nd � 0(3), note that � is

linear from (2� 4) to (6� 3), so its slope is 3� 46� 2 = �

1

4. To �nd �0(1), note that � is linear from (0� 6) to (2� 0), so its slope

is 0� 62� 0 = �3. Thus, � 0(3)�0(1) =

�� 14

�(�3) = 3

4.

(b) )(�) = �(�(�)) � )0(�) = �0(�(�))� 0(�). So )0(1) = �0(�(1))� 0(1) = �0(2)� 0(1), which does not exist since

�0(2) does not exist.

(c) *(�) = �(�(�)) � *0(�) = �0(�(�))�0(�). So *0(1) = �0(�(1))�0(1) = �0(3)�0(1). To �nd �0(3), note that � is

linear from (2� 0) to (5� 2), so its slope is 2� 05� 2 =

2

3. Thus, �0(3)�0(1) =

�23

�(�3) = �2.

56. (a) �(�) = �(�(�)) � �0(�) = � 0(�(�))� 0(�). So �0(2) = � 0(�(2))� 0(2) = � 0(1)� 0(2) (�1)(�1) = 1.

(b) �(�) = �(�2) � �0(�) = � 0(�2) · ���

��2�= � 0(�2)(2�). So �0(2) = � 0(2 2)(2 · 2) = 4� 0(4) 4(2) = 8.

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200 ¤ CHAPTER 3 DIFFERENTIATION RULES

57. �(�) = �(�(�)) � �0(�) = � 0(�(�))�0(�). So �0(0�5) = � 0(�(0�5))�0(0�5) = � 0(0�1)�0(0�5).

We can estimate the derivatives by taking the average of two secant slopes.

For � 0 (0�1): �1 =14�8� 12�60�1� 0 = 22, �2 =

18�4� 14�80�2� 0�1 = 36. So � 0(0�1) �1 +�2

2=22 + 36

2= 29.

For �0(0�5): �1 =0�10� 0�170�5� 0�4 = �0�7, �2 =

0�05� 0�100�6� 0�5 = �0�5. So �0(0�5) �1 +�2

2= �0�6.

Hence, �0(0�5) = � 0(0�1)�0(0�5) (29)(�0�6) = �17�4.

58. �(�) = �(�(�)) � �0(�) = � 0(�(�))� 0(�). So �0(1) = � 0(�(1))� 0(1) = � 0(2)� 0(1).

For � 0(2): �1 =3�1� 2�42�0� 1�5 = 1�4, �2 =

4�4� 3�12�5� 2�0 = 2�6. So � 0(2) �1 +�2

2= 2.

For � 0(1): �1 =2�0� 1�81�0� 0�5 = 0�4, �2 =

2�4� 2�01�5� 1�0 = 0�8. So � 0(1) �1 +�2

2= 0�6.

Hence, �0(1) = � 0(2)� 0(1) (2)(0�6) = 1�2.

59. (a) ! (�) = �(0�) � ! 0(�) = � 0(0�)�

��(0�) = � 0(0�)0�

(b) #(�) = 0 (�) � #0(�) = 0 (�)�

���(�) = 0 (�)� 0(�)

60. (a) ! (�) = �(��) � ! 0(�) = � 0(��)�

��(��) = � 0(��)1���1

(b) #(�) = [�(�)]� � #0(�) = 1 [�(�)]��1 � 0(�)

61. �(�) = �(�(�(�))) � �0(�) = � 0(�(�(�))) · �0(�(�)) · �0(�), so

�0(1) = � 0(�(�(1))) · �0(�(1)) · �0(1) = � 0(�(2)) · �0(2) · 4 = � 0(3) · 5 · 4 = 6 · 5 · 4 = 120

62. �(�) = ��(�2) � � 0(�) = ��0(�2) 2�+ �(�2) · 1 = 2�2�0(�2) + �(�2) �� 00(�) = 2�2�00(�2) 2�+ �0(�2) 4�+ �0(�2) 2� = 4�3�00(�2) + 4��0(�2) + 2��0(�2) = 6��0(�2) + 4�3�00(�2)

63. ! (�) = �(3�(4�(�))) �

! 0(�) = � 0(3�(4�(�))) · ���(3�(4�(�))) = � 0(3�(4�(�))) · 3� 0(4�(�)) · �

��(4�(�))

= � 0(3�(4�(�))) · 3� 0(4�(�)) · 4� 0(�), so

! 0(0) = � 0(3�(4�(0))) · 3� 0(4�(0)) · 4� 0(0) = � 0(3�(4 · 0)) · 3� 0(4 · 0) · 4 · 2 = � 0(3 · 0) · 3 · 2 · 4 · 2 = 2 · 3 · 2 · 4 · 2 = 96.

64. ! (�) = �(��(��(�))) �

! 0(�) = � 0(��(��(�))) · ���(��(��(�))) = � 0(��(��(�))) ·

� · � 0(��(�)) · �

��(��(�)) + �(��(�)) · 1

!= � 0(��(��(�))) · [�� 0(��(�)) · (�� 0(�) + �(�) · 1) + �(��(�))] , so

! 0(1) = � 0(�(�(1))) · [� 0(�(1)) · (� 0(1) + �(1)) + �(�(1))] = � 0(�(2)) · [� 0(2) · (4 + 2) + �(2)]

= � 0(3) · [5 · 6 + 3] = 6 · 33 = 198.

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SECTION 3.4 THE CHAIN RULE ¤ 201

65. = 02�(� cos 3�+� sin 3�) �

0 = 02�(�3� sin 3�+ 3� cos 3�) + (� cos 3�+� sin 3�) · 202�

= 02�(�3� sin 3�+ 3� cos 3�+ 2� cos 3�+ 2� sin 3�)= 02�[(2�+ 3�) cos 3�+ (2� � 3�) sin 3�] �

00 = 02�[�3(2�+ 3�) sin 3�+ 3(2� � 3�) cos 3�] + [(2�+ 3�) cos 3�+ (2� � 3�) sin 3�] · 202�

= 02�{[�3(2�+ 3�) + 2(2� � 3�)] sin 3�+ [3(2� � 3�) + 2(2�+ 3�)] cos 3�}= 02�[(�12�� 5�) sin 3�+ (�5�+ 12�) cos 3�]

Substitute the expressions for , 0, and 00 in 00 � 40 + 13 to get

00 � 40 + 13 = 02�[(�12�� 5�) sin 3�+ (�5�+ 12�) cos 3�]� 402�[(2�+ 3�) cos 3�+ (2� � 3�) sin 3�] + 1302�(� cos 3�+� sin 3�)

= 02�[(�12�� 5� � 8� + 12�+ 13�) sin 3�+ (�5�+ 12� � 8�� 12� + 13�) cos 3�]= 02�[(0) sin 3�+ (0) cos 3�] = 0

Thus, the function satis�es the differential equation 00 � 40 + 13 = 0.

66. = 0�� � 0 = �0�� � 00 = �20��. Substituting , 0, and 00 into 00 � 40 + = 0 gives us

�20�� � 4�0�� + 0�� = 0 � 0��(�2 � 4� + 1) = 0. Since 0�� 6= 0, we must have

�2 � 4� + 1 = 0 � � =4±�16� 4

2= 2±�3.

67. The use of ,, ,2, � � �, , is just a derivative notation (see text page 150). In general, ,�(2�) = 2� 0(2�),

,2�(2�) = 4� 00(2�), � � �, ,�(2�) = 2� ()(2�). Since �(�) = cos� and 50 = 4(12) + 2, we have

� (50)(�) = � (2)(�) = � cos�, so ,50 cos 2� = �250 cos 2�.

68. �(�) = �0��, � 0(�) = 0�� � �0�� = (1� �)0��, � 00(�) = �0�� + (1� �)(�0��) = (�� 2)0��. Similarly,

� 000(�) = (3� �)0��, � (4)(�) = (�� 4)0��, � � �, � (1000)(�) = (�� 1000)0��.

69. �(�) = 10 + 14 sin(10��) � the velocity after � seconds is )(�) = �0(�) = 1

4 cos(10��)(10�) =5�2 cos(10��) cm�s.

70. (a) � = � cos(<�+ =) � velocity = �0 = �<� sin(<�+ =).

(b) If � 6= 0 and < 6= 0, then �0 = 0 � sin(<�+ =) = 0 � <�+ = = (� � � =(� � =

<, ( an integer.

71. (a) �(�) = 4�0 + 0�35 sin 2��5�4

� ��

��=

�0�35 cos

2��

5�4

��2�

5�4

�=0�7�

5�4cos

2��

5�4=7�

54cos

2��

5�4

(b) At � = 1, ����

=7�

54cos

2�

5�4 0�16.

72. $(�) = 12 + 2�8 sin�2�365(�� 80)� � $0(�) = 2�8 cos

�2�365(�� 80)�� 2�

365

�.

On March 21, � = 80, and $0(80) 0�0482 hours per day. On May 21, � = 141, and $0(141) 0�02398, which is

approximately one-half of $0(80).

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202 ¤ CHAPTER 3 DIFFERENTIATION RULES

73. �(�) = 20�15� sin 2�� �

)(�) = �0(�) = 2[0�15�(cos 2��)(2�) + (sin 2��)0�15�(�1�5)] = 20�15�(2� cos 2��� 1�5 sin 2��)

74. (a) lim���

�(�) = lim���

1

1 + �0���=

1

1 + � · 0 = 1, since � 0 � ��� �� � 0��� 0.

(b) �(�) = (1 + �0���)�1 � ��

��= �(1 + �0���)�2(���0���) =

��0���

(1 + �0���)2

(c)From the graph of �(�) = (1 + 100�05�)�1, it seems that �(�) = 0�8

(indicating that 80% of the population has heard the rumor) when

� 7�4 hours.

75. By the Chain Rule, �(�) = �)

��=

�)

��

��

��=

�)

��)(�) = )(�)

�)

��. The derivative �)��� is the rate of change of the velocity

with respect to time (in other words, the acceleration) whereas the derivative �)��� is the rate of change of the velocity with

respect to the displacement.

76. (a) The derivative � ��� represents the rate of change of the volume with respect to the radius and the derivative � ���

represents the rate of change of the volume with respect to time.

(b) Since =4

3��3, �

��=

��

��

��= 4��2

��

��.

77. (a) Using a calculator or CAS, we obtain the model / = �� with � 100�0124369 and 0�000045145933.

(b) Use /0(�) = �� ln (from Formula 5) to get /0(0�04) �670�63 :A. The result of Example 2 in Section 2.1 was

�670 :A.

78. (a) � = �� with � = 4�502714× 10�20 and = 1�029953851,

where � is measured in thousands of people. The �t appears to be very good.

(b) For 1800: �1 =5308� 39291800� 1790 = 137�9, �2 =

7240� 53081810� 1800 = 193�2.

So � 0(1800) (�1 +�2)�2 = 165�55 thousand people�year.

For 1850: �1 =23,192� 17,0631850� 1840 = 612�9, �2 =

31,443� 23,1921860� 1850 = 825�1.

So � 0(1850) (�1 +�2)�2 = 719 thousand people�year.

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SECTION 3.4 THE CHAIN RULE ¤ 203

(c) Using � 0(�) = �� ln (from Formula 5) with the values of � and from part (a), we get � 0(1800) 156�85 and

� 0(1850) 686�07. These estimates are somewhat less than the ones in part (b).

(d) � (1870) 41,946�56. The difference of 3�4 million people is most likely due to the Civil War (1861–1865).

79. � = �4 + 1, = �3 + �; � = �1� �

��= 3�2 + 1, ��

��= 4�3, and �

��=

����

�����=3�2 + 1

4�3. When � = �1,

(�� ) = (2��2) and ���� = 4�4 = �1, so an equation of the tangent to the curve at the point corresponding to � = �1

is � (�2) = (�1)(�� 2), or = ��.

80. � = cos � + sin 2�, = sin � + cos 2�; � = 0. �

��=

����

�����=

cos � � 2 sin 2�� sin � + 2cos 2� . When � = 0, (�� ) = (1� 1) and

���� = 12

, so an equation of the tangent to the curve is � 1 = 12(�� 1), or = 1

2�+ 1

2.

81. � = 0��, = �� ln �2; � = 1. �

��= 1� 2�

�2= 1� 2

�, ����=

0��

2��, and �

��=

����

�����=

1� 2��0����2��� · 2�2�=2�� 4��0��

.

When � = 1, (�� ) = (0� 1) and �

��= �2

0, so an equation of the tangent line is � 1 = �2

0(�� 0), or = �2

0�+ 3.

82. � = 2�3 + 3�2 � 12�, = 2�3 + 3�2 + 1.���= 6�2 + 6� = 6�(�+ 1), so �

��= 0 � � = 0 or �1 �

(�� ) = (0� 1) or (13� 2).����= 6�2 +6�� 12 = 6(�+2)(�� 1), so ��

��= 0 � � = �2 or 1 � (�� ) = (20��3) or

(�7� 6). The curve has horizontal tangents at (0� 1) and (13� 2), and vertical tangents at (20��3) and (�7� 6).

83. � = 10� �2, = �3 � 12�.���= 3�2 � 12 = 3(�+ 2)(�� 2), so �

��= 0 � � = ±2 �

(�� ) = (6��16).����= �2�, so ��

��= 0 � � = 0 � (�� ) = (10� 0). The curve has horizontal tangents at (6�±16)

and a vertical tangent at (10� 0).

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204 ¤ CHAPTER 3 DIFFERENTIATION RULES

84. � = sin �, = sin(�+ sin �); (0� 0).

��=

����

�����=cos(�+ sin �)(1 + cos �)

cos �= cos(�+ sin �)

1 + cos �

cos �= cos(�+ sin �)

�1

cos �+ 1

�= (sec �+ 1) cos(�+ sin �).

Now � = sin � is 0 when � = 0 and � = �, so there are two tangents at the point

(0� 0) since both � = 0 and � = � correspond to the origin. The tangent

corresponding to � = 0 has slope (sec 0 + 1) cos (0 + sin 0) = 2 cos 0 = 2,

and its equation is = 2�. The tangent corresponding to � = � has slope

(sec� + 1) cos (� + sin�) = 0, so it is the �-axis; that is, = 0.

85. (a) � = �2, = �3 � 3� � �

��=

����

�����=3�2 � 32�

. At the point (3� 0), � = 3 � �2 = 3 � � = ±�3 �

��=3�±�3 �2 � 32�±�3 � =

6

2�±�3 � = ± 3�

3= ±�3. When � =

�3, an equation of the tangent line is

� 0 = �3 (�� 3) or =�3�� 3�3. When � = ��3, an equation of the tangent line is � 0 = ��3 (�� 3)

or = ��3�+ 3�3.

(b) Horizontal tangent: ���� = 0 � 3�2 � 3 = 0 �3(�2 � 1) = 0 � �2 = 1 � � = ±1. � = 1 corresponds to

the point (�� ) =��2� �3 � 3�� = �

12� 13 � 3 · 1� = (1��2)and � = �1 to (1� 2).

Vertical tangent: ���� is unde�ned � 2� = 0 � � = 0.

The value � = 0 corresponds to the origin; that is, (0� 0).

(c)

86. (a) � = �(� � sin �), = � (1� cos �) � �

��=

����

�����=

� (sin �)

� (1� cos �) =sin �

1� cos � . When � = �3

,

��=

�32

1� 12

=�3, (�� ) =

����3�

�32

�� 12��

, and the tangent is � 12� =

�3��� �

��3�

�32

��.

(b) Horizontal tangent: ���� = 0 � sin � = 0 (and cos � 6= 1) �� = (2(+ 1)�. The corresponding points are ((2(+ 1)��� 2�).

Vertical tangent: ���� is unde�ned � 1� cos � = 0 �cos � = 1 � � = 2(�. The corresponding points are (2(��� 0).

(c)

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SECTION 3.4 THE CHAIN RULE ¤ 205

87. (a) Derive gives �0(�) = 45(�� 2)8(2�+ 1)10

without simplifying. With either Maple or Mathematica, we �rst get

�0(�) = 9(�� 2)8(2�+ 1)9

� 18 (�� 2)9

(2�+ 1)10, and the simpli�cation command results in the expression given by Derive.

(b) Derive gives 0 = 2(�3 � �+ 1)3(2�+ 1)4(17�3 + 6�2 � 9�+ 3) without simplifying. With either Maple or

Mathematica, we �rst get 0 = 10(2�+ 1)4(�3 � �+ 1)4 + 4(2�+ 1)5(�3 � �+ 1)3(3�2 � 1). If we use

Mathematica’s Factor or Simplify, or Maple’s factor, we get the above expression, but Maple’s simplify gives

the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.

88. (a) �(�) =��4 � �+ 1

�4 + �+ 1

�1�2. Derive gives � 0(�) =

(3�4 � 1)��4 � �+ 1

�4 + �+ 1(�4 + �+ 1)(�4 � �+ 1)

whereas either Maple or Mathematica

give � 0 (�) = 3�4 � 1��4 � �+ 1

�4 + �+ 1(�4 + �+ 1)2

after simpli�cation.

(b) � 0(�) = 0 � 3�4 � 1 = 0 � � = ± 4

�13 ±0�7598.

(c) Yes. � 0(�) = 0 where � has horizontal tangents. � 0 has two maxima andone minimum where � has in�ection points.

89. (a) �

��(sin � cos(�) = ( sin�1 � cos� cos(�+ sin � (�( sin(�) [Product Rule]

= ( sin�1 � (cos(� cos�� sin(� sin�) [factor out ( sin�1 �]

= ( sin�1 � cos((�+ �) [Addition Formula for cosine]

= ( sin�1 � cos[((+ 1)�] [factor out �]

(b) �

��(cos � cos(�) = ( cos�1 � (� sin�) cos(�+ cos � (�( sin(�) [Product Rule]

= �( cos�1 � (cos(� sin�+ sin(� cos�) [factor out �( cos�1 �]

= �( cos�1 � sin((�+ �) [Addition Formula for sine]

= �( cos�1 � sin[((+ 1)�] [factor out �]

90. � = 3�2 + 1, = 2�3 + 1, ����= 6�, �

��= 6�2, so �

��=6�2

6�= � [even where � = 0].

So at the point corresponding to parameter value �, an equation of the tangent line is � (2�3 + 1) = �[�� (3�2 + 1)].

If this line is to pass through (4� 3), we must have 3� (2�3 + 1) = �[4� (3�2 + 1)] � 2�3 � 2 = 3�3 � 3� �

�3 � 3�+ 2 = 0 � (�� 1)2(�+ 2) = 0 � � = 1 or �2. Hence, the desired equations are � 3 = �� 4, or

= �� 1, tangent to the curve at (4� 3), and � (�15) = �2(�� 13), or = �2�+ 11, tangent to the curve at (13��15).

91. Since �� =�

�180

�� rad, we have �

��(sin ��) =

��

�sin �

180��= �

180 cos�180� =

�180 cos �

�.

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206 ¤ CHAPTER 3 DIFFERENTIATION RULES

92. (a) �(�) = |�| = ��2 = (�2)1�2 � � 0(�) = 12(�2)�1�2(2�) = ��

��2 = �� |�| for � 6= 0.

� is not differentiable at � = 0.

(b) �(�) = |sin�| =�sin2 � �

� 0(�) = 12(sin2 �)�1�22 sin� cos� =

sin�

|sin�| cos�

=

�cos� if sin� 0

� cos� if sin� � 0

� is not differentiable when � = (�, ( an integer.

(c) �(�) = sin |�| = sin��2 �

�0(�) = cos |�| · �|�| =�

|�| cos� =�cos� if � 0

� cos� if � � 0

� is not differentiable at 0.

93. �2

��2=

��

��

��

�[Leibniz notation for the second derivative]

=�

��

��

��

��

��

�[Chain Rule]

=�

��· ���

���

��

�+��

��· ���

��

��

�[Product Rule]

=�

��· �

2�

��2+��

��· ���

��

��

�· ����

[���� is a function of �]

=�

��

�2�

��2+�2

��2

���

��

�2

Or: Using function notation for = �(�) and � = �(�), we have = �(�(�)), so

0 = � 0(�(�)) · �0(�) [by the Chain Rule] �

(0)0 = [� 0(�(�)) · �0(�)]0 = � 0(�(�)) · �00(�) + �0(�) · � 00(�(�)) · �0(�) = � 0(�(�)) · �00(�) + � 00(�(�)) · [�0(�)]2.

94. = 43��3 � �

��= 4��2

��

��. But �

��is proportional to the surface area, so �

��= � · 4��2 for some constant �.

Therefore, 4��2 ����= � · 4��2 � ��

��= � = constant. An antiderivative of � with respect to � is ��, so � = ��+ �.

When � = 0, the radius � must equal the original radius �0, so � = �0, and � = ��+ �0. To �nd � we use the fact that

when � = 3, � = 3� + �0 and = 12 0 � 4

3�(3� + �0)3 = 1

2 · 43��30 � (3� + �0)3 = 1

2�30 �

3� + �0 =13�2�0 � � = 1

3�0

�13�2� 1

�. Since � = ��+ �0, � = 1

3�0

�13�2� 1

��+ �0. When the snowball

has melted completely we have � = 0 � 13�0

�13�2� 1

��+ �0 = 0 which gives � = 3 3

�2

3�2� 1 . Hence, it takes

3 3�2

3�2� 1 � 3 =

33�2� 1 11 h 33 min longer.

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LABORATORY PROJECT BEZIER CURVES ¤ 207

LABORATORY PROJECT Bézier Curves

1. The parametric equations for a cubic Bézier curve are

� = �0(1� �)3 + 3�1�(1� �)2 + 3�2�2(1� �) + �3�

3

= 0(1� �)3 + 31�(1� �)2 + 32�2(1� �) + 3�

3

where 0 � � � 1. We are given the points �0(�0� 0) = (4� 1), �1(�1� 1) = (28� 48), �2(�2� 2) = (50� 42), and

�3(�3� 3) = (40� 5). The curve is then given by

�(�) = 4(1� �)3 + 3 · 28�(1� �)2 + 3 · 50�2(1� �) + 40�3

(�) = 1(1� �)3 + 3 · 48�(1� �)2 + 3 · 42�2(1� �) + 5�3

where 0 � � � 1. The line segments are of the form � = �0 + (�1 � �0)�,

= 0 + (1 � 0)�:

�0�1 � = 4 + 24�, = 1 + 47�

�1�2 � = 28 + 22�, = 48� 6��2�3 � = 50� 10�, = 42� 37�

2. It suf�ces to show that the slope of the tangent at �0 is the same as that of line segment �0�1, namely 1 � 0�1 � �0

.

We calculate the slope of the tangent to the Bézier curve:

����

�����=�30(1� �)2 + 31

�2�(1� �) + (1� �)2+ 32

��2 + (2�)(1� �)+ 33�

2

�3�20(1� �) + 3�1[�2�(1� �) + (1� �)2] + 3�2[��2 + (2�)(1� �)] + 3�3�2

At point �0, � = 0, so the slope of the tangent is �30 + 31�3�0 + 3�1 =1 � 0�1 � �0

. So the tangent to the curve at �0 passes

through �1. Similarly, the slope of the tangent at point �3 [where � = 1] is �32 + 33�3�2 + 3�3 =3 � 2�3 � �2

, which is also the slope

of line �2�3.

3. It seems that if �1 were to the right of �2, a loop would appear.

We try setting �1 = (110� 30), and the resulting curve does indeed have a loop.

4. Based on the behavior of the Bézier curve in Problems 1–3, we suspect that the

four control points should be in an exaggerated C shape. We try �0(10� 12),

�1(4� 15), �2(4� 5), and �3(10� 8), and these produce a decent C. If you are using

a CAS, it may be necessary to instruct it to make the �- and -scales the same so as

not to distort the �gure (this is called a “constrained projection” in Maple.)

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208 ¤ CHAPTER 3 DIFFERENTIATION RULES

5. We use the same �0 and �1 as in Problem 4, and use part of our C as the top of

an S. To prevent the center line from slanting up too much, we move �2 up to

(4� 6) and �3 down and to the left, to (8� 7). In order to have a smooth joint

between the top and bottom halves of the S (and a symmetric S), we determine

points �4, �5, and �6 by rotating points �2, �1, and �0 about the center of the

letter (point �3). The points are therefore �4(12� 8), �5(12��1), and �6(6� 2).

APPLIED PROJECT Where Should a Pilot Start Descent?

1. Condition (i) will hold if and only if all of the following four conditions hold:

(1) � (0) = 0

(;) � 0(0) = 0 (for a smooth landing)

(>) � 0(7) = 0 (since the plane is cruising horizontally when it begins its descent)

(=) � (7) = �.

First of all, condition 1 implies that � (0) = � = 0, so � (�) = ��3 + �2 + +� � � 0(�) = 3��2 + 2�+ +. But

� 0(0) = + = 0 by condition ;. So � 0(7) = 3�72 + 27 = 7 (3�7+ 2). Now by condition >, 3�7+ 2 = 0 � � = �237

.

Therefore, � (�) = �237�3 + �2. Setting � (7) = � for condition =, we get � (7) = �2

3773 + 72 = � �

�2372 + 72 = � � 1

372 = � � =

3�

72� � = �2�

73. So = � (�) = �2�

73�3 +

3�

72�2.

2. By condition (ii), ����= �) for all �, so � (�) = 7� )�. Condition (iii) states that

�����2��2

���� � �. By the Chain Rule,

we have �

��=

��

��

��= �2�

73�3�2

� ����+3�

72(2�)

��

��=6��2)

73� 6��)

72(for � � 7) �

�2

��2=6�)

73(2�)

��

��� 6�)

72��

��= �12�)

2

73�+

6�)2

72. In particular, when � = 0, � = 7 and so

�2

��2

�����=0

= �12�)2

737+

6�)2

72= �6�)

2

72. Thus,

�����2��2

�����=0

=6�)2

72� �. (This condition also follows from taking � = 0.)

3. We substitute � = 860 mi�h2, � = 35,000 ft× 1 mi5280 ft

, and ) = 300 mi�h into the result of part (b):

6�35,000 · 1

5280

�(300)2

72� 860 � 7 � 300

�6 · 35,0005280 · 860 64�5 miles.

4. Substituting the values of � and 7 in Problem 3 into

� (�) = �2�73�3 +

3�

72�2 gives us � (�) = ��3 + �2,

where � �4�937× 10�5 and 4�78× 10�3.

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 209

3.5 Implicit Differentiation

1. (a) �

��(� + 2�+ 3�2) =

��(4) � (� · 0 + · 1) + 2 + 6� = 0 � �0 = � � 2� 6� �

0 =� � 2� 6�

�or 0 = �6� + 2

�.

(b) � + 2�+ 3�2 = 4 � � = 4� 2�� 3�2 � =4� 2�� 3�2

�=4

�� 2� 3�, so 0 = � 4

�2� 3.

(c) From part (a), 0 = � � 2� 6��

=�(4��� 2� 3�)� 2� 6�

�=�4��� 3�

�= � 4

�2� 3.

2. (a) �

��

�cos�+

��=

��(5) � � sin�+ 1

2�1�2 · 0 = 0 � 1

2�· 0 = sin� � 0 = 2

� sin�

(b) cos�+� = 5 �

� = 5� cos� � = (5� cos�)2, so 0 = 2(5� cos�)0(sin�) = 2 sin�(5� cos�).

(c) From part (a), 0 = 2� sin� = 2

�(5� cos�)2 = 2(5� cos�) sin� [since 5� cos� 0].

3. �

��

��3 + 3

�=

��(1) � 3�2 + 32 · 0 = 0 � 32 0 = �3�2 � 0 = ��

2

2

4. �

��

�2��+

��=

��(3) � 2 · 1

2��1�2 +

1

2�1�2 · 0 = 0 � 1�

�+

0

2�= 0 �

0

2�= � 1�

�� 0 = �2

���

5. �

��(�2 + � � 2) =

��(4) � 2�+ � · 0 + · 1� 2 0 = 0 �

�0 � 2 0 = �2�� � (�� 2) 0 = �2�� � 0 =�2��

�� 2 =2�+

2 � �

6. �

��(2�3 + �2 � �3) =

��(2) � 6�2 + �2 · 0 + · 2�� (� · 320 + 3 · 1) = 0 �

�2 0 � 3�2 0 = �6�2 � 2� + 3 � (�2 � 3�2) 0 = �6�2 � 2� + 3 � 0 =�6�2 � 2� + 3

�2 � 3�2

7. �

��

�4(�+ )

=

��

2(3�� )

� �4(1 + 0) + (�+ ) · 4�3 = 2(3� 0) + (3�� ) · 2 0 �

�4 + �4 0 + 4�4 + 4�3 = 32 � 2 0 + 6� 0 � 22 0 � �4 0 + 32 0 � 6� 0 = 32 � 5�4 � 4�3 �

(�4 + 32 � 6�) 0 = 32 � 5�4 � 4�3 � 0 =32 � 5�4 � 4�3�4 + 32 � 6�

8. �

��(5 + �23) =

��

�1 + 0�

2�

� 54 0 + (�2 · 32 0 + 3 · 2�) = 0 + · 0�2 · 2�+ 0�2 · 0 �

0�54 + 3�22 � 0�

2�= 2�0�

2 � 2�3 � 0 =2�

�0�

2 � 2�

54 + 3�22 � 0�2

9. �

��(�22 + � sin ) =

��(4) � �2 · 2 0 + 2 · 2�+ � cos · 0 + sin · 1 = 0 �

2�2 0 + � cos · 0 = �2�2 � sin � (2�2 + � cos )0 = �2�2 � sin � 0 =�2�2 � sin 2�2 + � cos

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210 ¤ CHAPTER 3 DIFFERENTIATION RULES

10. �

��(1 + �) =

��

sin(�2)

� 1 = [cos(�2)](� · 2 0 + 2 · 1) � 1 = 2� cos(�2) 0 + 2 cos(�2) �

1� 2 cos(�2) = 2� cos(�2) 0 � 0 =1� 2 cos(�2)

2� cos(�2)

11. �

��(4 cos� sin ) =

��(1) � 4 [cos� · cos · 0 + sin · (� sin�)] = 0 �

0(4 cos� cos ) = 4 sin� sin � 0 =4 sin� sin

4 cos� cos = tan� tan

12. �

��

sin(�2)

=

��

� sin(2)

� cos(�2) · 2�+ sin(�2) · 0 = � cos(2) · 2 0 + sin(2) · 1 �

0sin(�2)� 2� cos(2) = sin(2)� 2� cos(�2) � 0 =

sin(2)� 2� cos(�2)sin(�2)� 2� cos(2)

13. �

��(0���) =

��(� � ) � 0��� · �

��

��

�= 1 � 0 �

0��� · · 1� � · 02

= 1� 0 � 0��� · 1� �0���

2· 0 = 1� 0 � 0 � �0���

2· 0 = 1� 0���

0�1� �0���

2

�=

� 0���

� 0 =

� 0���

2 � �0���

2

=( � 0���)

2 � �0���

14. tan(�� ) =

1 + �2� (1 + �2) tan(�� ) = � (1 + �2) sec2(�� ) · (1� 0) + tan(�� ) · 2� = 0 �

(1 + �2) sec2(� � ) � (1 + �2) sec2(� � ) · 0 + 2� tan(�� ) = 0 �

(1 + �2) sec2(� � ) + 2� tan(� � ) =1 + (1 + �2) sec2(�� )

· 0 �

0 =(1 + �2) sec2(�� ) + 2� tan(�� )

1 + (1 + �2) sec2(�� )

15. �

��(0� cos�) =

��[1 + sin(�)] � 0�(� sin�) + cos� · 0� · 0 = cos(�) · (�0 + · 1) �

�0� sin�+ 0� cos� · 0 = � cos(�) · 0 + cos(�) � 0� cos� · 0 � � cos(�) · 0 = 0� sin�+ cos(�) �

[0� cos�� � cos(�)] 0 = 0� sin�+ cos(�) � 0 =0� sin�+ cos(�)

0� cos�� � cos(�)

16. sin�+ cos = sin� cos � cos�� sin · 0 = sin� (� sin · 0) + cos cos� �

(sin� sin � sin ) 0 = cos� cos � cos� � 0 =cos� (cos � 1)sin (sin�� 1)

17. �

��

�(�) + �2[�(�)]3

�=

��(10) � � 0(�) + �2 · 3[�(�)]2 · � 0(�) + [�(�)]3 · 2� = 0. If � = 1, we have

� 0(1) + 12 · 3[�(1)]2 · � 0(1) + [�(1)]3 · 2(1) = 0 � � 0(1) + 1 · 3 · 22 · � 0(1) + 23 · 2 = 0 �

� 0(1) + 12� 0(1) = �16 � 13� 0(1) = �16 � � 0(1) = �1613

.

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 211

18. �

��[�(�) + � sin �(�)] =

��(�2) � �0(�) + � cos �(�) · �0(�) + sin �(�) · 1 = 2�. If � = 0, we have

�0(0) + 0 + sin �(0) = 2(0) � �0(0) + sin 0 = 0 � �0(0) + 0 = 0 � �0(0) = 0.

19. �

�(�42 � �3 + 2�3) =

�(0) � �4 · 2 + 2 · 4�3 �0 � (�3 · 1 + · 3�2 �0) + 2(� · 32 + 3 · �0) = 0 �

4�32 �0 � 3�2 �0 + 23 �0 = �2�4 + �3 � 6�2 � (4�32 � 3�2 + 23)�0 = �2�4 + �3 � 6�2 �

�0 =��

�=�2�4 + �3 � 6�24�32 � 3�2 + 23

20. �

�( sec�) =

�(� tan ) � · sec� tan� · �0 + sec� · 1 = � · sec2 + tan · �0 �

sec� tan� · �0 � tan · �0 = � sec2 � sec� � ( sec� tan�� tan )�0 = � sec2 � sec� �

�0 =��

�=

� sec2 � sec� sec� tan�� tan

21. sin 2� = � cos 2 � · cos 2� · 2 + sin 2� · 0 = �(� sin 2 · 20) + cos(2) · 1 �sin 2� · 0 + 2� sin 2 · 0 = �2 cos 2�+ cos 2 �

0(sin 2�+ 2� sin 2) = �2 cos 2�+ cos 2 � 0 =�2 cos 2�+ cos 2sin 2�+ 2� sin 2

. When � = �2

and = �4

, we have

0 =(���2)(�1) + 0

0 + � · 1 =��2

�=1

2, so an equation of the tangent line is � �

4= 1

2(�� �

2), or = 1

2�.

22. sin(�+ ) = 2�� 2 � cos(�+ ) · (1 + 0) = 2� 20 � cos(�+ ) · 0 + 20 = 2� cos(�+ ) �

0[cos(�+ ) + 2] = 2� cos(�+ ) � 0 =2� cos(�+ )

cos(�+ ) + 2. When � = � and = �, we have 0 = 2� 1

1 + 2=1

3, so

an equation of the tangent line is � � = 13(�� �), or = 1

3�+ 2�

3.

23. �2 + � + 2 = 3 � 2�+ � 0 + · 1 + 20 = 0 � � 0 + 2 0 = �2�� � 0(�+ 2) = �2�� �

0 =�2��

�+ 2. When � = 1 and = 1, we have 0 = �2� 1

1 + 2 · 1 =�33= �1, so an equation of the tangent line is

� 1 = �1(�� 1) or = ��+ 2.

24. �2 + 2� � 2 + � = 2 � 2�+ 2(� 0 + · 1)� 2 0 + 1 = 0 � 2� 0 � 2 0 = �2�� 2 � 1 �

0(2�� 2) = �2�� 2 � 1 � 0 =�2�� 2 � 12�� 2 . When � = 1 and = 2, we have

0 =�2� 4� 12� 4 =

�7�2 =

7

2, so an equation of the tangent line is � 2 = 7

2(�� 1) or = 7

2�� 3

2.

25. �2 + 2 = (2�2 + 22 � �)2 � 2�+ 2 0 = 2(2�2 + 22 � �)(4�+ 4 0 � 1). When � = 0 and = 12

, we have

0 + 0 = 2( 12 )(20 � 1) � 0 = 20 � 1 � 0 = 1, so an equation of the tangent line is � 1

2 = 1(�� 0)

or = �+ 12 .

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212 ¤ CHAPTER 3 DIFFERENTIATION RULES

26. �2�3 + 2�3 = 4 � 23��1�3 + 2

3�1�30 = 0 � 1

3��+

0

3�= 0 � 0 = �

3�

3��

. When � = �3�3

and = 1, we have 0 = � 1��3�3 �1�3 = ���3�3 �2�3�3�3 =

3

3�3=

1�3

, so an equation of the tangent line is

� 1 = 1�3

��+ 3

�3�

or = 1�3�+ 4.

27. 2(�2 + 2)2 = 25(�2 � 2) � 4(�2 + 2)(2� + 2 0) = 25(2�� 2 0) �4(�+ 0)(�2 + 2) = 25(�� 0) � 4 0(�2 + 2) + 250 = 25�� 4�(�2 + 2) �

0 =25�� 4�(�2 + 2)

25 + 4(�2 + 2). When � = 3 and = 1, we have 0 = 75� 120

25+40= � 45

65= � 9

13,

so an equation of the tangent line is � 1 = � 913(�� 3) or = � 9

13�+ 40

13.

28. 2(2 � 4) = �2(�2 � 5) � 4 � 42 = �4 � 5�2 � 43 0 � 8 0 = 4�3 � 10�.

When � = 0 and = �2, we have�320 + 160 = 0 � �160 = 0 � 0 = 0, so an equation of the tangent line is

+ 2 = 0(�� 0) or = �2.

29. (a) 2 = 5�4 � �2 � 2 0 = 5(4�3)� 2� � 0 =10�3 � �

.

So at the point (1� 2) we have 0 = 10(1)3 � 12

=9

2, and an equation

of the tangent line is � 2 = 92(�� 1) or = 9

2�� 5

2.

(b)

30. (a) 2 = �3 + 3�2 � 2 0 = 3�2 + 3(2�) � 0 =3�2 + 6�

2. So at the point (1��2) we have

0 =3(1)2 + 6(1)

2(�2) = �94

, and an equation of the tangent line is + 2 = � 94(�� 1) or = � 9

4�+ 1

4.

(b) The curve has a horizontal tangent where 0 = 0 �3�2 + 6� = 0 � 3�(�+ 2) = 0 � � = 0 or � = �2.

But note that at � = 0, = 0 also, so the derivative does not exist.

At � = �2, 2 = (�2)3 + 3(�2)2 = �8 + 12 = 4, so = ±2.

So the two points at which the curve has a horizontal tangent are

(�2��2) and (�2� 2).

(c)

31. 9�2 + 2 = 9 � 18�+ 2 0 = 0 � 2 0 = �18� � 0 = �9�� �

00 = �9� · 1� � · 0

2

�= �9

� � �(�9��)

2

�= �9 ·

2 + 9�2

3= �9 · 9

3[since x and y must satisfy the

original equation, 9�2 + 2 = 9]. Thus, 00 = �81�3.

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 213

32.��+

� = 1 � 1

2��+

0

2�= 0 � 0 = �

���

00 = �

��

&1

2�

'0 �

1

2��

!�

= �

��

�1�

������

���

�1��

�2�

=

1 +

���

2�

=

��+

2���

=1

2���

since � and must satisfy the original equation,��+

� = 1.

33. �3 + 3 = 1 � 3�2 + 32 0 = 0 � 0 = ��2

2�

00 = �2(2�)� �2 · 2 0

(2)2= �2�

2 � 2�2(��2�2)4

= �2�4 + 2�4

6= �2�(

3 + �3)

6= �2�

5,

since � and must satisfy the original equation, �3 + 3 = 1.

34. �4 + 4 = �4 � 4�3 + 43 0 = 0 � 43 0 = �4�3 � 0 = ��3�3 �

00 = ��3 · 3�2 � �3 · 32 0

(3)2

�= �3�22 · � �(��3�3)

6= �3�2 ·

4 + �4

43= �3�2 · �

4

7=�3�4�27

35. If � = 0 in � + 0� = 0, then we get 0 + 0� = 0, so = 1 and the point where � = 0 is (0� 1). Differentiating implicitly

with respect to � gives us �0 + · 1 + 0�0 = 0. Substituting 0 for � and 1 for gives us

0 + 1 + 00 = 0 � 00 = �1 � 0 = �1�0. Differentiating �0 + + 0�0 = 0 implicitly with respect to � gives

us �00 + 0 · 1 + 0 + 0�00 + 0 · 0�0 = 0. Now substitute 0 for �, 1 for , and �1�0 for 0.

0 +

��10

�+

��10

�+ 000 +

��10

�(0)

��10

�= 0 � �2

0+ 000 +

1

0= 0 � 000 =

1

0� 00 =

1

02.

36. If � = 1 in �2 + �+ 3 = 1, then we get 1+ + 3 = 1 � 3 + = 0 � (2 +1) � = 0, so the point

where � = 1 is (1� 0). Differentiating implicitly with respect to � gives us 2�+ �0 + · 1 + 32 · 0 = 0. Substituting 1 for

� and 0 for gives us 2+ 0 +0+ 0 = 0 � 0 = �2. Differentiating 2�+ �0 + +320 = 0 implicitly with respect

to � gives us 2 + �00 + 0 · 1 + 0 + 3(200 + 0 · 20) = 0. Now substitute 1 for �, 0 for , and �2 for 0.

2 + 00 + (�2) + (�2) + 3(0 + 0) = 0 � 00 = 2. Differentiating 2 + �00 + 20 + 3200 + 6(0)2 = 0 implicitly

with respect to � gives us �000 + 00 · 1+ 200 +3(2000 + 00 · 20) + 6[ · 2000 + (0)20] = 0. Now substitute 1 for �, 0

for , �2 for 0, and 2 for 00. 000 + 2 + 4 + 3(0 + 0) + 6[0 + (�8)] = 0 � 000 = �2� 4 + 48 = 42.

37. (a) There are eight points with horizontal tangents: four at � 1�57735 and

four at � 0�42265.

(b) 0 = 3�2 � 6�+ 22(23 � 32 � + 1)

� 0 = �1 at (0� 1) and 0 = 13

at (0� 2).

Equations of the tangent lines are = ��+ 1 and = 13�+ 2.

(c) 0 = 0 � 3�2 � 6�+ 2 = 0 � � = 1± 13

�3

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214 ¤ CHAPTER 3 DIFFERENTIATION RULES

(d) By multiplying the right side of the equation by �� 3, we obtain the �rst

graph. By modifying the equation in other ways, we can generate the other

graphs.

(2 � 1)( � 2)= �(�� 1)(�� 2)(�� 3)

(2 � 4)( � 2)= �(�� 1)(�� 2)

( + 1)(2 � 1)( � 2)= �(�� 1)(�� 2)

( + 1)(2 � 1)( � 2)= (�� 1)(�� 2)

�( + 1)(2 � 1)( � 2)= (�� 1)(�� 2)

(2 + 1)( � 2)= �(�2 � 1)(�� 2)

( + 1)(2 � 2)= �(�� 1)(�2 � 2)

38. (a) (b) There are 9 points with horizontal tangents: 3 at � = 0, 3 at � = 12

,

and 3 at � = 1. The three horizontal tangents along the top of the

wagon are hard to �nd, but by limiting the -range of the graph

(to [1�6� 1�7], for example) they are distinguishable.

39. From Exercise 27, a tangent to the lemniscate will be horizontal if 0 = 0 � 25�� 4�(�2 + 2) = 0 ��[25� 4(�2 + 2)] = 0 � �2 + 2 = 25

4 (1). (Note that when � is 0, is also 0, and there is no horizontal tangent

at the origin.) Substituting 254

for �2 + 2 in the equation of the lemniscate, 2(�2 + 2)2 = 25(�2 � 2), we get

�2 � 2 = 258

(2). Solving (1) and (2), we have �2 = 7516

and 2 = 2516

, so the four points are�± 5

�3

4�± 5

4

�.

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 215

40. �2

�2+2

2= 1 � 2�

�2+20

2= 0 � 0 = � 2�

�2� an equation of the tangent line at (�0� 0) is

� 0 =�2�0�20

(�� �0). Multiplying both sides by 02

gives 02

� 202= ��0�

�2+�20�2

. Since (�0� 0) lies on the ellipse,

we have �0�

�2+0

2=

�20�2+202= 1.

41. �2 + 2 = �2 is a circle with center 2 and ��+ = 0 is a line through 2 [assume �

and are not both zero]. �2 + 2 = �2 � 2�+ 20 = 0 � 0 = ���, so the

slope of the tangent line at �0 (�0� 0) is ��0�0. The slope of the line 2�0 is 0��0,

which is the negative reciprocal of ��0�0. Hence, the curves are orthogonal, and the

families of curves are orthogonal trajectories of each other.

42. The circles �2 + 2 = �� and �2 + 2 = intersect at the origin where the tangents are vertical and horizontal [assume �

and are both nonzero]. If (�0� 0) is the other point of intersection, then �20 + 20 = ��0 (1) and �20 + 20 = 0 (2).

Now �2 + 2 = �� � 2�+ 20 = � � 0 =�� 2�2

and �2 + 2 = �

2�+ 20 = 0 � 0 =2�

� 2 . Thus, the curves are orthogonal at (�0� 0) �

�� 2�020

= �� 202�0

� 2��0 � 4�20 = 420 � 20 � ��0 + 0 = 2(�20 + 20),

which is true by (1) and (2).

43. = +�2 � 0 = 2+� and �2 + 22 = � [assume � 0] � 2�+ 40 = 0 �

20 = �� � 0 = � �

2()= � �

2(+�2)= � 1

2+�, so the curves are orthogonal if

+ 6= 0. If + = 0, then the horizontal line = +�2 = 0 intersects �2 + 22 = � orthogonally

at�±��� 0

�, since the ellipse �2 + 22 = � has vertical tangents at those two points.

44. = ��3 � 0 = 3��2 and �2 + 32 = [assume 0] � 2�+ 60 = 0 �

30 = �� � 0 = � �

3()= � �

3(��3)= � 1

3��2, so the curves are orthogonal if

� 6= 0. If � = 0, then the horizontal line = ��3 = 0 intesects �2 + 32 = orthogonally

at�±�� 0

�, since the ellipse �2 + 32 = has vertical tangents at those two points.

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216 ¤ CHAPTER 3 DIFFERENTIATION RULES

45. Since �2 � �2, we are assured that there are four points of intersection. .

(1) �2

�2+2

2= 1 � 2�

�2+20

2= 0 � 0

2= � �

�2� 0 = �1 = ��2

�2.

(2) �2

�2� 2

�2= 1 � 2�

�2� 20

�2= 0 � 0

�2=

�2� 0 = �2 =

��2

�2.

Now �1�2 = ��2

�2· ��

2

�2= � 2�2

�2�2· �

2

2(3). Subtracting equations, (1) � (2), gives us �

2

�2+2

2� �2

�2+

2

�2= 0 �

2

2+

2

�2=

�2

�2� �2

�2� 2�2 + 22

2�2=

�2�2 � �2�2

�2�2� 2(2 +�2)

2�2=

�2(�2 ��2)

�2�2(4). Since

�2 � 2 = �2 +�2, we have �2 ��2 = 2 +�2. Thus, equation (4) becomes 2

2�2=

�2

�2�2� �2

2=

�2�2

2�2, and

substituting for �2

2in equation (3) gives us �1�2 = � 2�2

�2�2· �

2�2

2�2= �1. Hence, the ellipse and hyperbola are orthogonal

trajectories.

46. = (�+ +)�1 � 0 = �(�+ +)�2 and = �(�+ �)1�3 � 0 = 13�(�+ �)�2�3, so the curves are othogonal if the

product of the slopes is �1, that is, �1(�+ +)2

· �

3(�+ �)2�3= �1 � � = 3(�+ +)2(�+ �)2�3 �

� = 3

�1

�2 ��

�2[since 2 = (�+ +)�2 and 2 = �2(�+ �)2�3] � � = 3

�1

�2

�� �3 = 3 � � = 3

�3.

47. (a)�� +

(2�

2

�( � () = (.� � � � �(+

(2�

� (3�

2= (.� �

��(� � �(+ (2� �1 � (3� �2) =

��((.� ) �

� 0 + · 1� (� (2� �2 · 0 + 2(3� �3 · 0 = 0 � 0(� � (2� �2 + 2(3� �3) = (� �

0 =(�

� � (2� �2 + 2(3� �3 or � ��

= 3((� )

� 3 � (2� + 2(3�

(b) Using the last expression for � ��� from part (a), we get

��=

(10 L)3[(1 mole)(0�04267 L�mole)� 10 L],-(2�5 atm)(10 L)3 � (1 mole)2(3�592 L2- atm� mole2)(10 L)

+ 2(1 mole)3(3�592 L2- atm� mole2)(0�04267 L� mole)

./

=�9957�33 L4

2464�386541 L3- atm �4�04 L� atm�

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 217

48. (a) �2 + � + 2 + 1 = 0 � 2�+ �0 + · 1 + 20 + 0 = 0 � 0(�+ 2) = �2�� � 0 =�2��

�+ 2

(b) Plotting the curve in part (a) gives us an empty graph, that is, there are no points that satisfy the equation. If there were any

points that satis�ed the equation, then � and must have opposite signs; otherwise, all the terms are positive and their sum

can not equal 0. �2 + �+ 2 +1 = 0 � �2 +2�+ 2 � �+1 = 0 � (�+ )2 = �� 1. The left side

of the last equation is nonnegative, but the right side is at most �1, so that proves there are no points that satisfy the

equation.

Another solution: �2 + � + 2 + 1 = 12�2 + � + 1

22 + 1

2�2 + 1

22 + 1 = 1

2(�2 + 2� + 2) + 1

2(�2 + 2) + 1

= 12(�+ )2 + 1

2(�2 + 2) + 1 � 1

Another solution: Regarding �2 + � + 2 + 1 = 0 as a quadratic in �, the discriminant is 2 � 4(2 + 1) = �32 � 4.

This is negative, so there are no real solutions.

(c) The expression for 0 in part (a) is meaningless; that is, since the equation in part (a) has no solution, it does not implicitly

de�ne a function of �, and therefore it is meaningless to consider 0.

49. If the circle has radius �, its equation is �2 + 2 = �2 � 2�+ 20 = 0 � 0 = ��

, so the slope of the tangent line

at � (�0� 0) is��00

. The negative reciprocal of that slope is �1��0�0 =

0�0

, which is the slope of 2� , so the tangent line at

� is perpendicular to the radius 2� .

50.��+

� =

�+ � 1

2��+

0

2�= 0 � 0 = �

���

� an equation of the tangent line at (�0� 0)

is � 0 = ��0��0(�� �0). Now � = 0 � = 0 �

�0��0(��0) = 0 +

��0

�0, so the -intercept is

0 +��0

�0. And = 0 � �0 = �

�0��0(�� �0) � �� �0 =

0��0�0

� = �0 +��0

�0, so the �-intercept is �0 +

��0

�0. The sum of the intercepts is�

0 +��0

�0

�+��0 +

��0

�0

�= �0 + 2

��0

�0 + 0 =

���0 +

�0

�2=��

+�2= +.

51. To �nd the points at which the ellipse �2 � � + 2 = 3 crosses the �-axis, let = 0 and solve for �.

= 0 � �2 � �(0) + 02 = 3 � � = ±�3. So the graph of the ellipse crosses the �-axis at the points�±�3� 0�.

Using implicit differentiation to �nd 0, we get 2�� �0 � + 20 = 0 � 0(2 � �) = � 2� � 0 = � 2�2 � �

.

So 0 at��3� 0

�is 0� 2�32(0)��3 = 2 and 0 at

���3� 0� is 0 + 2�3

2(0) +�3= 2. Thus, the tangent lines at these points are parallel.

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218 ¤ CHAPTER 3 DIFFERENTIATION RULES

52. (a) We use implicit differentiation to �nd 0 = � 2�2 � �

as in Exercise 51. The

slope of the tangent line at (�1� 1) is � =1� 2(�1)2(1)� (�1) =

3

3= 1, so the slope

of the normal line is� 1�= �1, and its equation is � 1 = �1(�+ 1) �

= ��. Substituting�� for in the equation of the ellipse, we get

�2 � �(��) + (��)2 = 3 � 3�2 = 3 � � = ±1. So the normal line

must intersect the ellipse again at � = 1, and since the equation of the line is

= ��, the other point of intersection must be (1��1).

(b)

53. �22 + � = 2 � �2 · 20 + 2 · 2�+ � · 0 + · 1 = 0 � 0(2�2 + �) = �2�2 � �

0 = �2�2 +

2�2 + �. So �2�

2 +

2�2 + �= �1 � 2�2 + = 2�2 + � � (2� + 1) = �(2� + 1) �

(2� + 1)� �(2� + 1) = 0 � (2� + 1)( � �) = 0 � � = �12

or = �. But � = � 12

�22 + � = 14� 1

26= 2, so we must have � = . Then �22 + � = 2 � �4 + �2 = 2 � �4 + �2 � 2 = 0 �

(�2 + 2)(�2 � 1) = 0. So �2 = �2, which is impossible, or �2 = 1 � � = ±1. Since � = , the points on the curve

where the tangent line has a slope of�1 are (�1��1) and (1� 1).

54. �2 + 42 = 36 � 2�+ 80 = 0 � 0 = � �

4. Let (�� ) be a point on �2 + 42 = 36 whose tangent line passes

through (12� 3). The tangent line is then � 3 = � �

4(�� 12), so � 3 = � �

4(�� 12). Multiplying both sides by 4

gives 42 � 12 = ��2 + 12�, so 42 + �2 = 12(�+ ). But 42 + �2 = 36, so 36 = 12(�+ ) � �+ = 3 � = 3� �. Substituting 3� � for into �2 + 42 = 36 gives �2 + 4(3� �)2 = 36 � �2 + 36� 24�+ 4�2 = 36 �5�2 � 24� = 0 � �(5�� 24) = 0, so � = 0 or � = 24

5. If � = 0, = 3� 0 = 3, and if � = 24

5, = 3� 24

5= � 9

5.

So the two points on the ellipse are (0� 3) and�245�� 9

5

�. Using

� 3 = � �

4(�� 12) with (�� ) = (0� 3) gives us the tangent line

� 3 = 0 or = 3. With (�� ) =�245 �� 9

5

�, we have

� 3 = � 24�54(�9�5) (�� 12) � � 3 = 2

3(�� 12) � = 2

3�� 5.

A graph of the ellipse and the tangent lines con�rms our results.

55. (a) = ?(�) and �00 + 0 + � = 0 � �? 00(�) + ? 0(�) + �?(�) = 0. If � = 0, we have 0 + ? 0(0) + 0 = 0,

so ? 0(0) = 0.

(b) Differentiating �00 + 0 + � = 0 implicitly, we get �000 + 00 · 1 + 00 + �0 + · 1 = 0 ��000 + 200 + �0 + = 0, so �? 000(�) + 2? 00(�) + �? 0(�) + ?(�) = 0. If � = 0, we have

0 + 2? 00(0) + 0 + 1 [?(0) = 1 is given] = 0 � 2? 00(0) = �1 � ? 00(0) = � 12 .

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SECTION 3.6 INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES ¤ 219

56. �2 + 42 = 5 � 2�+ 4(20) = 0 � 0 = � �

4. Now let � be the height of the lamp, and let (�� ) be the point of

tangency of the line passing through the points (3� �) and (�5� 0). This line has slope (�� 0)�[3� (�5)] = 18�. But the

slope of the tangent line through the point (�� ) can be expressed as 0 = � �

4, or as � 0

�� (�5) =

�+ 5[since the line

passes through (�5� 0) and (�� )], so � �

4=

�+ 5� 42 = ��2 � 5� � �2 + 42 = �5�. But �2 + 42 = 5

[since (�� ) is on the ellipse], so 5 = �5� � � = �1. Then 42 = ��2 � 5� = �1� 5(�1) = 4 � = 1, since the

point is on the top half of the ellipse. So �

8=

�+ 5=

1

�1 + 5 =1

4� � = 2. So the lamp is located 2 units above the

�-axis.

3.6 Inverse Trigonometric Functions and Their Derivatives

1. (a) sin�1��

32

�= �

3since sin �

3=

�32

and �3

is in��

2� �2

.

(b) cos�1(�1) = � since cos� = �1 and � is in [0� �].

2. (a) tan�1�

1�3

�= �

6since tan �

6= 1�

3and �

6is in

���2� �2

�.

(b) sec�1 2 = �3

since sec �3= 2 and �

3is in [0� �

2) [�� 3�

2).

3. (a) arctan 1 = �4

since tan �4= 1 and �

4is in

���2� �2

�.

(b) sin�1 1�2= �

4since sin �

4= 1�

2and �

4is in

��2� �2

.

4. (a) tan�1�tan

3�

4

�= tan�1(�1) = ��

4

(b) cos�arcsin

1

2

�= cos

��6

�=

�3

2

5. Let � = sin�1�23

�.

Then tan�sin�1

�23

��= tan � =

2�5

.

6. Let � = arccos 35

.

Then csc�arccos

�35

��= csc � = 5

4 .

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220 ¤ CHAPTER 3 DIFFERENTIATION RULES

7. Let � = tan�1�2. Then

sin�2 tan�1

�2�= sin(2�) = 2 sin = 2

��2�3

��1�3

�=2�2

3

8. Let � = tan�1 2 and = tan�1 3. Thencos

�tan�1 2 + tan�1 3

�= cos(�+ ) = cos� cos � sin� sin

=1�5

1�10� 2�

5

3�10

=�5�50=

�55�2=�1�2

9. Let = sin�1 �. Then ��2� � �

2� cos � 0, so cos(sin�1 �) = cos =

�1� sin2 = �1� �2.

10. Let = sin�1 �. Then sin = �, so from the triangle we see that

tan(sin�1 �) = tan =��1� �2

.

11. Let = tan�1 �. Then tan = �, so from the triangle we see that

sin(tan�1 �) = sin =��1 + �2

.

12. Let = tan�1 �. Then tan = �, so from the triangle we see that

cos(2 tan�1 �) = cos 2 = cos2 � sin2

=

�1�1 + �2

�2��

��1 + �2

�2=1� �2

1 + �2

13. The graph of sin�1 � is the re�ection of the graph of

sin� about the line = �.

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SECTION 3.6 INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES ¤ 221

14. The graph of tan�1 � is the re�ection of the graph of

tan� about the line = �.

15. Let = cos�1 �. Then cos = � and 0 � � � � � sin ���

= 1 �

��= � 1

sin = � 1�

1� cos2 = �1�1� �2

. [Note that sin � 0 for 0 � � �.]

16. (a) Let � = sin�1 � and = cos�1 �. Then cos � =�1� sin2 � = �1� �2 since cos � � 0 for ��

2 � � � �2 . Similarly,

sin =�1� �2. So

sin(sin�1 �+ cos�1 �) = sin(�+ ) = sin � cos + cos � sin = � · �+�1� �2�1� �2

= �2 + (1� �2) = 1

But ��2� sin�1 �+ cos�1 � � 3�

2, and so sin�1 �+ cos�1 � = �

2.

(b) We differentiate sin�1 �+ cos�1 � = �2

with respect to �, and get

1�1� �2

+�

��(cos�1 �) = 0 � �

��(cos�1 �) = � 1�

1� �2.

17. = (tan�1 �)2 � 0 = 2(tan�1 �)1 · ���(tan�1 �) = 2 tan�1 � · 1

1 + �2=2 tan�1 �1 + �2

18. = tan�1(�2) � 0 =1

1 + (�2)2· ���(�2) =

1

1 + �4· 2� = 2�

1 + �4

19. = sin�1(2� + 1) �

0 =1�

1� (2�+ 1)2 ·�

��(2�+ 1) =

1�1� (4�2 + 4�+ 1) · 2 =

2��4�2 � 4� =1���2 � �

20. ! (�) = arcsin�sin � = arcsin(sin �)1�2 �

! 0(�) =1�

1���sin �

�2 · ��� (sin �)1�2 = 1�1� sin � ·

1

2(sin �)�1�2 · cos � = cos �

2�1� sin ��sin �

21. #(�) =�1� �2 arccos� � #0(�) =

�1� �2 · �1�

1� �2+ arccos� · 1

2(1� �2)�1�2(�2�) = �1� � arccos��

1� �2

22. �(�) = � ln(arctan�) � � 0(�) = � · 1

arctan�· 1

1 + �2+ ln(arctan�) · 1 = �

(1 + �2) arctan�+ ln(arctan�)

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222 ¤ CHAPTER 3 DIFFERENTIATION RULES

23. = cos�1(02�) � 0 = � 1�1� (02�)2 ·

��(02�) = � 202��

1� 04�

24. = tan�1�����2 + 1 � �

0 =1

1 +�����2 + 1 �2

�1� ��

�2 + 1

�=

1

1 + �2 � 2���2 + 1 + �2 + 1

���2 + 1� ���2 + 1

=

��2 + 1� �

2�1 + �2 � �

��2 + 1

���2 + 1

=

��2 + 1� �

2�

�2 + 1 (1 + �2)� �(�2 + 1) = �

�2 + 1� �

2(1 + �2)

���2 + 1� �

�=

1

2(1 + �2)

25. = arctan(cos �) � 0 =1

1 + (cos �)2(� sin �) = � sin �

1 + cos2 �

26. = cos�1(sin�1 �) � 0 = � 1�1� (sin�1 �)2 ·

��sin�1 � = � 1�

1� (sin�1 �)2 ·1�1� �2

27. = � sin�1 �+�1� �2 �

0 = � · 1�1� �2

+ (sin�1 �)(1) +1

2(1� �2)�1�2(�2�) = ��

1� �2+ sin�1 �� ��

1� �2= sin�1 �

28. = arctan�1� �

1 + �= arctan

�1� �

1 + �

�1�2�

0 =1

1 +

��1� �

1 + �

�2 · ����1� �

1 + �

�1�2=

1

1 +1� �

1 + �

· 12

�1� �

1 + �

��1�2· (1 + �)(�1)� (1� �)(1)

(1 + �)2

=1

1 + �

1 + �+1� �

1 + �

· 12

�1 + �

1� �

�1�2· �2(1 + �)2

=1 + �

2· 12· (1 + �)1�2

(1� �)1�2· �2(1 + �)2

=�1

2(1� �)1�2(1 + �)1�2=

�12�1� �2

29. = arccos�+ � cos�

�+ cos�

��

0 = � 1%1�

�+ � cos�

�+ cos�

�2 (�+ cos�)(�� sin�)� (+ � cos�)(� sin�)(�+ cos�)2

=1�

�2 + 2 cos2 �� 2 � �2 cos2 �

(�2 � 2) sin�

|�+ cos�|

=1�

�2 � 2�1� cos2 �

(�2 � 2) sin�

|�+ cos�| =

��2 � 2

|�+ cos�|sin�

|sin�|

But 0 � � � �, so |sin�| = sin�. Also � 0 � cos� � � ��, so �+ cos� 0. Thus 0 =��2 � 2

�+ cos�.

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SECTION 3.6 INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES ¤ 223

30. �(�) = arcsin(0�) � � 0(�) =1�

1� (0�)2 · 0� =

0��1� 02�

.

Domain(�) = {� | �1 � 0� � 1} = {� | 0 � 0� � 1} = (��� 0].

Domain(� 0) = � | 1� 02� 0

�= � | 02� � 1� = {� | 2� � 0} = (��� 0).

31. �(�) = cos�1(3 � 2�) � �0(�) = � 1�1� (3� 2�)2

(�2) = 2�1� (3� 2�)2 .

Domain(�) = {� | �1 � 3� 2� � 1} = {� | �4 � �2� � �2} = {� | 2 � � � 1} = [1� 2].Domain(�0) =

� | 1� (3� 2�)2 0� =

� | (3� 2�)2 � 1� = {� | |3� 2�| � 1}= {� | �1 � 3� 2� � 1} = {� | �4 � �2� � �2} = {� | 2 � 1} = (1� 2)

32. tan�1(�) = 1 + �2 � 1

1 + �22(�0 + · 1) = 0 + �20 + 2� �

0�

1 + �22� �2

�= 2� �

1 + �22�

0 =2� �

1 + �22�

1 + �22� �2

=2�(1 + �22)�

�� �2(1 + �22)=

(�1� 2�� 2�32)�(1� �� �32)

.

33. �(�) = � sin�1��4

�+�16� �2 � �0(�) = sin�1

��4

�+

4�1� (��4)2 �

��16� �2

= sin�1��4

��

�0(2) = sin�1�12

�= �

6

34. = 3arccos �2

� 0 = 3

&� 1�

1� (��2)2

'�1

2

�, so at (1� �), 0 = � 3

2�1� 1

4

= ��3. An equation of the tangent

line is � � = ��3 (�� 1), or = ��3�+ � +�3.

35. �(�) =�1� �2 arcsin� � � 0(�) =

�1� �2 · 1�

1� �2+ arcsin� · 1

2

�1� �2

��1�2(�2�) = 1� � arcsin��

1� �2

Note that � 0 = 0 where the graph of � has a horizontal tangent. Also note

that � 0 is negative when � is decreasing and � 0 is positive when � is

increasing.

36. �(�) = arctan(�2 � �) � � 0(�) =1

1 + (�2 � �)2· ���(�2 � �) =

2�� 11 + (�2 � �)2

Note that � 0 = 0 where the graph of � has a horizontal tangent. Also note

that � 0 is negative when � is decreasing and � 0 is positive when � is

increasing.

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224 ¤ CHAPTER 3 DIFFERENTIATION RULES

37. lim���1+

sin�1 � = sin�1(�1) = ��2

38. Let � = 1 + �2

1 + 2�2� As � �, � = 1 + �2

1 + 2�2=1��2 + 1

1��2 + 2 1

2.

lim���

arccos

�1 + �2

1 + 2�2

�= lim

��1�2arccos � = arccos 1

2= �

3.

39. Let � = 0�. As � �, � �. lim���

arctan(0�) = lim���

arctan � = �2

by (3).

40. Let � = ln�. As � 0+, � ��. lim��0+

tan�1(ln�) = lim����

tan�1 � = ��2

by (3).

41. (a) If = ��1(�), then �() = �. Differentiating implicitly with respect to � and remembering that is a function of �,

we get � 0() ���

= 1, so �

��=

1

� 0()� �

��1�0(�) =

1

� 0(��1(�)).

(b) �(4) = 5 � ��1(5) = 4. By part (a),���1

�0(5) =

1

� 0(��1(5))=

1

� 0(4)= 1

��23

�= 3

2.

42. (a) �(�) = 2�+ cos� � � 0(�) = 2� sin� 0 for all �. Thus, � is increasing for all � and is therefore one-to-one.

(b) Since � is one-to-one, ��1(1) = � � �(�) = 1. By inspection, we see that �(0) = 2(0) + cos 0 = 1,

so � = ��1(1) = 0.

(c)���1

�0(1) =

1

� 0(��1(1))=

1

� 0(0)=

1

2� sin 0 =1

2

43. (a) Let �(�) = sin�, so ��1(�) = sin�1 � and � 0(�) = cos�.

(��1)0(�) =1

� 0(��1(�))�

��(sin�1 �) =

1

cos(sin�1 �)=

1�1� �2�1

=1�1� �2

, which is Formula 1.

(Note that we could have used the result of Exercise 9.)

(b) Let �(�) = tan�, so ��1(�) = tan�1 � and � 0(�) = sec2 �.

(��1)0(�) =1

� 0(��1(�))�

��(tan�1 �) =

1

sec2(tan�1 �)=

1

[sec(tan�1 �)]2=

1

(�1 + �2 )2

=1

1 + �2,

which is Formula 4.

44. (a) �(�) = sin(sin�1 �) (b) �(�) = sin�1(sin�)

(c) �0(�) = �

��sin�1(sin�) =

1�1� sin2 �

cos� =cos��cos2 �

=cos�

|cos�|

NOT FOR SALE NTIATION RULESLE

INSTRUCTOR USE ONLY )) ��0((��) =) = �

SS���� sinsin��11(sin(sin��) =) =11

RUCRUC��UCUC11� sinsin22 ��coscos�� ==

cosos��

TOTO��OOcosco 22 ��==cosco �

RR||coscos�|

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SECTION 3.7 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 225

(d) �(�) = cos�1(sin�), so

�0(�) = � cos��1� sin2 �

= � cos�

|cos�| .

Notice that �(�) = �2� �(�) because

sin�1 �+ cos�1 � = �2

for all �.

3.7 Derivatives of Logarithmic Functions

1. The differentiation formula for logarithmic functions, �

��(log� �) =

1

� ln �, is simplest when � = 0 because ln 0 = 1.

2. �(�) = � ln�� � � � 0(�) = � · 1�+ (ln�) · 1� 1 = 1 + ln�� 1 = ln�

3. �(�) = sin(ln�) � � 0(�) = cos(ln�) · ���ln� = cos(ln�) · 1

�=cos(ln�)

4. �(�) = ln(sin2 �) = ln(sin�)2 = 2 ln |sin�| � � 0(�) = 2 · 1

sin�· cos� = 2 cot�

5. �(�) = log2(1� 3�) � � 0(�) =1

(1� 3�) ln 2�

��(1� 3�) = �3

(1� 3�) ln 2 or 3

(3�� 1) ln 2

6. �(�) = log5(�0�) � � 0(�) =

1

�0� ln 5

��(�0�) =

1

�0� ln 5(�0� + 0� · 1) = 0�(�+ 1)

�0� ln 5=

�+ 1

� ln 5

Another solution:We can change the form of the function by �rst using logarithm properties.

�(�) = log5(�0�) = log5 �+ log5 0

� � � 0(�) =1

� ln 5+

1

0� ln 5· 0� = 1

� ln 5+

1

ln 5or 1 + �

� ln 5

7. �(�) = 5�ln� = (ln�)1�5 � � 0(�) = 1

5(ln�)�4�5

��(ln�) =

1

5(ln�)4�5· 1�=

1

5� 5�(ln�)4

8. �(�) = ln 5�� = ln�1�5 = 1

5ln� � � 0(�) =

1

5· 1�=1

5�

9. �(�) = sin� ln(5�) � � 0(�) = sin� · 15�· ���(5�)+ ln(5�) · cos� = sin� · 5

5�+cos� ln(5�) =

sin�

�+cos� ln(5�)

10. �(�) = 1 + ln �

1� ln � � � 0(�) =(1� ln �)(1��)� (1 + ln �)(�1��)

(1� ln �)2 =(1��)[(1� ln �) + (1 + ln �)]

(1� ln �)2 =2

�(1� ln �)2

11. ! (�) = ln (2�+ 1)3

(3�� 1)4 = ln(2�+ 1)3 � ln(3� � 1)4 = 3 ln(2�+ 1)� 4 ln(3�� 1) �

! 0(�) = 3 · 1

2�+ 1· 2� 4 · 1

3�� 1 · 3 =6

2�+ 1� 12

3�� 1 , or combined, �6(�+ 3)(2�+ 1)(3�� 1) .

12. �(�) = ln��+

��2 � 1 � � �0(�) =

1

�+��2 � 1

�1 +

���2 � 1

�=

1

�+��2 � 1 ·

��2 � 1 + ���2 � 1 =

1��2 � 1

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226 ¤ CHAPTER 3 DIFFERENTIATION RULES

13. �(�) = ln����2 � 1 � = ln� + ln(�2 � 1)1�2 = ln� + 1

2ln(�2 � 1) �

�0(�) =1

�+1

2· 1

�2 � 1 · 2� =1

�+

�2 � 1 =�2 � 1 + � · ��(�2 � 1) =

2�2 � 1�(�2 � 1)

14. ! () = ln(1 + 0�) � ! 0() = · 1

1 + 0�· 0� + ln(1 + 0�) · 1 = 0�

1 + 0�+ ln(1 + 0�)

15. = ln��2� �� 5�2�� � 0 =

1

2� �� 5�2 · (�1� 10�) =�10�� 12� �� 5�2 or 10�+ 1

5�2 + �� 2

16. "(8) = ln��2 � 82

�2 + 82= ln

��2 � 82

�2 + 82

�1�2=1

2ln

��2 � 82

�2 + 82

�= 1

2 ln(�2 � 82)� 1

2 ln(�2 + 82) �

"0(8) =1

2· 1

�2 � 82· (�28)� 1

2· 1

�2 + 82· (28) = 8

82 � �2� 8

82 + �2=

8(82 + �2)� 8(82 � �2)

(82 � �2)(82 + �2)

=83 + 8�2 � 83 + 8�2

(82 � �2)(82 + �2)=

2�28

84 � �4

17. = ln(0�� + �0��) = ln(0��(1 + �)) = ln(0��) + ln(1 + �) = ��+ ln(1 + �) �

0 = �1 + 1

1 + �=�1� �+ 1

1 + �= � �

1 + �

18. = [ln(1 + 0�)]2 � 0 = 2[ln(1 + 0�)] · 1

1 + 0�· 0� = 20� ln(1 + 0�)

1 + 0�

19. = 2� log10�� = 2� log10 �

1�2 = 2� · 12 log10 � = � log10 � � 0 = � · 1

� ln 10+ log10 � · 1 =

1

ln 10+ log10 �

Note: 1

ln 10=ln 0

ln 10= log10 0, so the answer could be written as 1

ln 10+ log10 � = log10 0+ log10 � = log10 0�.

20. = log2(0�� cos��) = log2 0

�� + log2 cos�� = �� log2 0 + log2 cos�� �

0 = � log2 0+1

cos�� (ln 2)

��(cos��) = � log2 0+

�� sin��cos�� (ln 2)

= � log2 0��

ln 2tan��

Note: 1

ln 2=ln 0

ln 2= log2 0, so the answer could be written as � log2 0� � log2 0 tan�� = (� log2 0) (1 + � tan��).

21. = �2 ln(2�) � 0 = �2 · 12�· 2 + ln(2�) · (2�) = � + 2� ln(2�) �

00 = 1 + 2� · 12�· 2 + ln(2�) · 2 = 1 + 2 + 2 ln(2�) = 3 + 2 ln(2�)

22. = ln�

�2� 0 =

�2(1��)� (ln�)(2�)(�2)2

=�(1� 2 ln�)

�4=1� 2 ln�

�3�

00 =�3(�2��)� (1� 2 ln�)(3�2)

(�3)2=

�2(�2� 3 + 6 ln�)�6

=6 ln�� 5

�4

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SECTION 3.7 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 227

23. �(�) = �

1� ln(�� 1) �

� 0(�) =[1� ln(�� 1)] · 1� � · �1

�� 1[1� ln(�� 1)]2 =

(�� 1)[1� ln(�� 1)] + �

�� 1[1� ln(�� 1)]2 =

�� 1� (�� 1) ln(�� 1) + �

(�� 1)[1� ln(�� 1)]2

=2�� 1� (�� 1) ln(�� 1)(�� 1)[1� ln(�� 1)]2

Dom(�) = {� | �� 1 0 and 1� ln(�� 1) 6= 0} = {� | � 1 and ln(�� 1) 6= 1}= � | � 1 and �� 1 6= 01

�= {� | � 1 and � 6= 1 + 0} = (1� 1 + 0) (1 + 0��)

24. �(�) = ln ln ln� � � 0(�) =1

ln ln�· 1

ln�· 1�

.

Dom(�) = {� | ln ln� 0} = {� | ln� 1} = {� | � 0} = (0��).

25. = ln(�2 � 3�+ 1) � 0 =1

�2 � 3�+ 1 · (2�� 3) � 0(3) = 11· 3 = 3, so an equation of a tangent line at

(3� 0) is � 0 = 3(�� 3), or = 3�� 9.

26. = ln(�3 � 7) � 0 =1

�3 � 7 · 3�2 � 0(2) =

12

8� 7 = 12, so an equation of a tangent line at (2� 0) is

� 0 = 12(�� 2) or = 12�� 24.

27. = ln��0�

2�= ln�+ ln 0�

2

= ln�+ �2 � 0 =1

�+ 2�. At (1� 1), the slope of the tangent line is

0(1) = 1 + 2 = 3, and an equation of the tangent line is � 1 = 3(�� 1), or = 3�� 2.

28. = ln�

�� 0 =

�(1��)� ln��2

=1� ln��2

.

0(1) =1� 012

= 1 and 0(0) = 1� 102

= 0 � equations of tangent

lines are � 0 = 1(�� 1) or = �� 1 and � 1�0 = 0(�� 0)

or = 1�0.

29. (a) The domain of �(�) = � ln� is (0��). � 0(�) = � (1��) + (ln�) · 1 = 1 + ln�. So � 0(�) � 0 when

1 + ln� � 0 � ln� � �1 � � � 0�1. Therefore, � is decreasing on (0� 1�0).

(b) � 0(�) = 1 + ln� � � 00(�) = 1�� 0 for � 0. So the curve is concave upward on (0��).

30. �(�) = sin�+ ln� � � 0(�) = cos�+ 1��.

This is reasonable, because the graph shows that � increases when � 0 is

positive, and � 0(�) = 0 when � has a horizontal tangent.

SECTION 3 7ON DERIVATIVES OF LOG

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228 ¤ CHAPTER 3 DIFFERENTIATION RULES

31. �(�) = +� + ln(cos�) � � 0(�) = + +1

cos�· (� sin�) = + � tan�.

� 0(�4) = 6 � +� tan �

4= 6 � +� 1 = 6 � + = 7.

32. �(�) = log�(3�2 � 2) � � 0(�) =

1

(3�2 � 2) ln� · 6�.

� 0(1) = 3 � 1

ln �· 6 = 3 � 2 = ln � � � = 02.

33. = (2�+ 1)5(�4 � 3)6 � ln = ln�(2�+ 1)5(�4 � 3)6� � ln = 5 ln(2�+ 1) + 6 ln(�4 � 3) �

1

0 = 5 · 1

2�+ 1· 2 + 6 · 1

�4 � 3 · 4�3 �

0 =

�10

2�+ 1+

24�3

�4 � 3�= (2�+ 1)5(�4 � 3)6

�10

2�+ 1+

24�3

�4 � 3�

.

[The answer could be simpli�ed to 0 = 2(2�+ 1)4(�4 � 3)5(29�4 + 12�3 � 15), but this is unnecessary.]

34. =�� 0�

2��2 + 1

�10 � ln = ln��+ ln 0�

2+ ln(�2 + 1)10 � ln = 1

2ln�+ �2 + 10 ln(�2 + 1) �

1

0 =

1

2· 1�+ 2�+ 10 · 1

�2 + 1· 2� � 0 =

��0�

2(�2 + 1)10

�1

2�+ 2�+

20�

�2 + 1

35. = sin2 � tan4 �

(�2 + 1)2� ln = ln(sin2 � tan4 �) � ln(�2 + 1)2 �

ln = ln(sin�)2 + ln(tan�)4 � ln(�2 + 1)2 � ln = 2 ln |sin�|+ 4 ln |tan�| � 2 ln(�2 + 1) �

1

0 = 2 · 1

sin�· cos�+ 4 · 1

tan�· sec2 �� 2 · 1

�2 + 1· 2� � 0 =

sin2 � tan4 �

(�2 + 1)2

�2 cot�+

4 sec2 �

tan�� 4�

�2 + 1

36. = 4

��2 + 1

�2 � 1 � ln = 14 ln(�

2 + 1)� 14 ln(�

2 � 1) � 1

0 =

1

4· 1

�2 + 1· 2�� 1

4· 1

�2 � 1 · 2� �

0 = 4

��2 + 1

�2 � 1 ·1

2

��

�2 + 1� �

�2 � 1�=1

24

��2 + 1

�2 � 1� �2��4 � 1

�=

1� �44

��2 + 1

�2 � 1

37. = �� � ln = ln�� � ln = � ln� � 0� = �(1��) + (ln�) · 1 � 0 = (1 + ln�) �0 = ��(1 + ln�)

38. = �cos � � ln = ln�cos � � ln = cos� ln� � 1

0 = cos� · 1

�+ ln� · (� sin�) �

0 = �cos�

�� ln� sin�

�� 0 = �cos �

�cos��

� ln� sin��

39. = (cos�)� � ln = ln(cos�)� � ln = � ln cos� � 1

0 = � · 1

cos�· (� sin�) + ln cos� · 1 �

0 =

�ln cos�� � sin�

cos�

�� 0 = (cos�)�(ln cos�� � tan�)

NTIATION RULESLE

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SECTION 3.7 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 229

40. =�� � � ln = ln

�� � � ln = � ln�1�2 � ln = 1

2� ln� � 1

0 =

1

2� · 1

�+ ln� · 1

2�

0 = �12+ 1

2ln�

� � 0 = 12

�� �(1 + ln�)

41. = (tan�)1�� � ln = ln(tan�)1�� � ln =1

�ln tan� �

1

0 =

1

�· 1

tan�· sec2 �+ ln tan� ·

�� 1

�2

�� 0 =

�sec2 �

� tan�� ln tan�

�2

��

0 = (tan�)1���sec2 �

� tan�� ln tan�

�2

�or 0 = (tan�)1�� · 1

�csc� sec�� ln tan�

42. = (sin�)ln � � ln = ln(sin�)ln� � ln = ln� · ln sin� � 1

0 = ln� · 1

sin�· cos�+ ln sin� · 1

��

0 =

�ln� · cos�

sin�+ln sin�

�� 0 = (sin�)ln�

�ln� cot�+

ln sin�

43. = ln(�2 + 2) � 0 =1

�2 + 2�

��(�2 + 2) � 0 =

2�+ 20

�2 + 2� �20 + 20 = 2�+ 20 �

�20 + 20 � 20 = 2� � (�2 + 2 � 2)0 = 2� � 0 =2�

�2 + 2 � 2

44. �� = � � ln� = � ln � · 1�+ (ln�) · 0 = � · 1

· 0 + ln � 0 ln�� �

0 = ln �

��

0 =ln � ��

ln�� ��

45. �(�) = ln(�� 1) � � 0(�) =1

(�� 1) = (�� 1)�1 � � 00(�) = �(�� 1)�2 � � 000 (�) = 2(�� 1)�3 �

� (4)(�) = �2 · 3(�� 1)�4 � · · · � � ()(�) = (�1)�1 · 2 · 3 · 4 · · · · · ((� 1)(�� 1)� = (�1)�1 ((� 1)!(�� 1)

46. = �8 ln�, so,9 = ,80 = ,8(8�7 ln�+ �7). But the eighth derivative of �7 is 0, so we now have

,8(8�7 ln�) = ,7(8 · 7�6 ln�+ 8�6) = ,7(8 · 7�6 ln�) = ,6(8 · 7 · 6�5 ln�) = · · · = ,(8!�0 ln�) = 8!���

47. If �(�) = ln (1 + �), then � 0(�) = 1

1 + �, so � 0(0) = 1.

Thus, lim��0

ln(1 + �)

�= lim

��0

�(�)

�= lim

��0

�(�)� �(0)

�� 0 = � 0(0) = 1.

48. Let � = (��. Then ( = ��, and as ( �, � �.

Therefore, lim��

�1 +

(

�= lim

���

�1 +

1

���

=

lim

���

�1 +

1

��!�= 0� by Equation 6.

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230 ¤ CHAPTER 3 DIFFERENTIATION RULES

DISCOVERY PROJECT Hyperbolic Functions

1. (a) (b) The domain of = cosh� is R.

The range is � 1.

2. The value of � is the -intercept. As � increases, the graph �attens out.

3. From the graphs, it appears that cosh is even and that sinh and tanh are odd.

�(�) = cosh� =0� + 0��

2�

�(��) = 0�� + 0�(��)

2=

0�� + 0�

2= �(�), so cosh is even.

�(�) = sinh� =0� � 0��

2� �(��) = 0�� � 0�(��)

2=

0�� � 0�

2= �0

� � 0��

2= ��(�), so sinh is odd.

�(�) = tanh� =sinh�

cosh�� �(��) = sinh(��)

cosh(��) =� sinh�cosh�

= ��(�), so tanh is odd.

4. cosh2 �� sinh2 � =�0� + 0��

2

�2��0� � 0��

2

�2=

02� + 2+ 0�2�

4� 02� � 2 + 0�2�

4=4

4= 1.

5. The curve looks like one branch of a hyperbola. We can con�rm this by

using the identity proved in Problem 4; �2 � 2 = cosh2 �� sinh2 � = 1.

So the curve is the right branch of the hyperbola �2 � 2 = 1.

6. sinh� cosh + cosh� sinh =12(0� � 0��)

12(0� + 0��)

+12(0� + 0��)

12(0� � 0��)

= 1

4[(0�+� + 0��� � 0��+� � 0����) + (0�+� � 0��� + 0��+� � 0����)]

= 14(20�+� � 20����) = 1

2[0�+� � 0�(�+�)] = sinh(�+ )

7. Substituting � for in Problem 6 and using the results from Problem 3 gives us

sinh(�� ) = sinh� cosh � cosh� sinh . By analogy with the trigonometric identity sin 2� = 2 sin� cos�, we might

guess that sinh 2� = 2 sinh� cosh� and we can prove it by putting = � in the identity in Problem 6. In view of the

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DISCOVERY PROJECT HYPERBOLIC FUNCTIONS ¤ 231

trigonometric identity cos(�+ ) = cos� cos � sin� sin , we expect that there is a similar hyperbolic identity.

By a calculation similar to that in Problem 6, we can prove that cosh(�+ ) = cosh� cosh + sinh� sinh .

Then, by putting = �, we get cosh 2� = cosh2 �+ sinh2 �. By substituting � for gives us

cosh(�� ) = cosh� cosh � sinh� sinh . Also, using the identities for sinh(�± ) and cosh(�± ), we get

tanh(�± ) =sinh(�± )

cosh(�± )=(sinh� cosh ± cosh� sinh )�(cosh� cosh )(cosh� cosh ± sinh� sinh )�(cosh� cosh ) =

tanh�± tanh 1± tanh� tanh �

8. (a) �

��(sinh�) =

��

�0� � 0��

2

�=

0� + 0��

2= cosh�

(b) �

��(cosh�) =

��

�0� + 0��

2

�=

0� � 0��

2= sinh�

��(tanh�) =

��

�sinh�

cosh�

�=cosh� cosh�� sinh� sinh�

(cosh�)2=cosh2 �� sinh2 �

cosh2 �=

1

cosh2 �= sech2 �

9. (a) sinh is an increasing function on its domain, so it passes the Horizontal Line Test, and is one-to-one.

(b) Let = sinh�1 �. Then sinh = �. If we differentiate this equation implicitly with respect to �, we get cosh ���

= 1.

Since cosh2 � sinh2 = 1 and cosh � 0, we have cosh =�1 + sinh2 , so

��=

1

cosh =

1�1 + sinh2

=1�1 + �2

.

(c) Let = sinh�1 �. Then � = sinh = 0� � 0��

2, so 0� � 2�� 0�� = 0 or, multiplying by 0� , 02� � 2�0� � 1 = 0.

This is really a quadratic equation in 0�: (0�)2 � 2�(0�)� 1 = 0. Solving by the quadratic formula, we get

0� =2�±�4�2 + 4

2= �±��2 + 1. Note that 0� 0, but ����2 + 1 � 0

because � ���2 + 1

. Thus,

the minus sign is inadmissible and we have 0� = �+��2 + 1. Therefore, = ln(0�) = ln

��+

��2 + 1

�.

(d) �

��

ln��+

��2 + 1

�=

1

�+��2 + 1

�1 + 1

2(�2 + 1)�1�2(2�)

�=

1

�+��2 + 1

�1 +

���2 + 1

=1

�+��2 + 1

·��2 + 1 + ���2 + 1

=1�

�2 + 1

10. (a) From the graph in Problem 3, tanh is an increasing function on its domain, so it passes the Horizontal Line Test, and is

one-to-one.

(b) Let = tanh�1 �. Then tanh = � � sech2 · 0 = 1 � 0 =1

sech2 =

1

1� tanh2 =1

1� �2.

To show that sech2 = 1� tanh2 , divide the equation cosh2 � sinh2 = 1 by cosh2 .

(c) Let = tanh�1 �. Then � = tanh = sinh

cosh =

0� � 0��

0� + 0��· 0

0�=

02� � 102� + 1

� �02� + � = 02� � 1 �

1 + � = 02� � �02� � 02� =1 + �

1� �� 2 = ln

�1 + �

1� �

�� =

1

2ln

�1 + �

1� �

�.

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232 ¤ CHAPTER 3 DIFFERENTIATION RULES

(d) �

��

1

2ln

�1 + �

1� �

�!=

��

12ln(1 + �)� 1

2ln(1� �)

=1

2

1

1 + �� 1

2

1

1� �(�1)

=(1� �) + (1 + �)

2(1 + �)(1� �)=

2

2(1� �2)=

1

1� �2

11. The tangent to = cosh� has slope 1 when 0 = sinh� = 1 � � = sinh�1 1 = ln�1 +

�2�, by Problem 9(c).

Since sinh� = 1 and = cosh� =�1 + sinh2 �, we have cosh� =

�2. The point is

�ln�1 +

�2���2�.

3.8 Rates of Change in the Natural and Social Sciences

1. (a) � = �(�) = �3 � 12�2 + 36� � )(�) = � 0(�) = 3�2 � 24�+ 36

(b) )(3) = 27� 72 + 36 = �9 ft�s

(c) The particle is at rest when )(�) = 0. 3�2 � 24�+ 36 = 0 � 3(�� 2)(�� 6) = 0 � � = 2 s or 6 s.

(d) The particle is moving in the positive direction when )(�) 0. 3(�� 2)(�� 6) 0 � 0 � � � 2 or � 6.

(e) Since the particle is moving in the positive direction and in the

negative direction, we need to calculate the distance traveled in the

intervals [0� 2], [2� 6], and [6� 8] separately.

|�(2)� �(0)| = |32� 0| = 32.

|�(6)� �(2)| = |0� 32| = 32.

|�(8)� �(6)| = |32� 0| = 32.

The total distance is 32 + 32 + 32 = 96 ft.

(f )

(g) )(�) = 3�2 � 24�+ 36 ��(�) = )0(�) = 6�� 24.

�(3) = 6(3)� 24 = �6 (ft�s)�s or ft�s2.

(h )

(i) The particle is speeding up when ) and � have the same sign. This occurs when 2 � � � 4 [) and � are both negative] and

when � 6 [) and � are both positive]. It is slowing down when ) and � have opposite signs; that is, when 0 � � � 2 and

when 4 � � � 6.

2. (a) � = �(�) = 0�01�4 � 0�04�3 � )(�) = � 0(�) = 0�04�3 � 0�12�2

(b) )(3) = 0�04(3)3 � 0�12(3)2 = 0 ft�s

(c) The particle is at rest when )(�) = 0. 0�04�3 � 0�12�2 = 0 � 0�04�2(�� 3) = 0 � � = 0 s or 3 s.

(d) The particle is moving in the positive direction when )(�) 0. 0�04�2(�� 3) 0 � � 3.

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SECTION 3.8 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 233

(e) See Exercise 1(e).

|�(3)� �(0)| = |�0�27� 0| = 0�27.

|�(8)� �(3)| = |20�48� (�0�27)| = 20�75.

The total distance is 0�27 + 20�75 = 21�02 ft.

(f )

(g) )(�) = 0�04�3 � 0�12�2 � �(�) = )0(�) = 0�12�2 � 0�24�. �(3) = 0�12(3)2 � 0�24(3) = 0�36 (ft�s)�s or ft�s2.

(h) Here we show the graph of �, ), and �

for 0 � � � 4 and 4 � � � 8.

(i) The particle is speeding up when ) and � have the same sign. This occurs when 0 � � � 2 [) and � are both negative]

and when � 3 [) and � are both positive]. It is slowing down when ) and � have opposite signs; that is,

when 2 � � � 3.

3. (a) � = �(�) = cos(���4) � )(�) = � 0(�) = � sin(���4) · (��4)

(b) )(3) = ��4sin 3�

4= ��

4·�22= ��

�2

8ft�s [ �0�56]

(c) The particle is at rest when )(�) = 0. ��4sin ��

4= 0 � sin ��

4= 0 � ��

4= �( � � = 0, 4, 8 s.

(d) The particle is moving in the positive direction when )(�) 0. ��4sin ��

4 0 � sin ��

4� 0 � 4 � � � 8.

(e) From part (c), )(�) = 0 for � = 0� 4� 8. As in Exercise 1, we’ll

�nd the distance traveled in the intervals [0� 4] and [4� 8].

|�(4)� �(0)|= |�1� 1| = 2|�(8)� �(4)|= |1� (�1)| = 2.

The total distance is 2 + 2 = 4 ft.

(f )

(g) )(�) = ��4sin

��

4�

�(�) = )0(�) = ��4cos

��

4· �4= ��

2

16cos

��

4.

�(3) = ��2

16cos

3�

4= ��

2

16

���2

2

�=

�2�2

32(ft�s)�s or ft�s2.

(h)

(i) The particle is speeding up when ) and � have the same sign. This occurs when 0 � � � 2 or 8 � � � 10 [) and � are

both negative] and when 4 � � � 6 [) and � are both positive]. It is slowing down when ) and � have opposite signs;

that is, when 2 � � � 4 and when 6 � � � 8.

4. (a) � = �(�) = �0���2 � )(�) = � 0(�) = �0���2�� 1

2

�+ 0���2 · 1 = 0���2

�� 12�+ 1

�(b) )(3) = 0�3�2

�� 32 + 1

�= � 1

20�3�2 ft�s [ �0�11]

(c) The particle is at rest when )(�) = 0. 0���2�� 1

2�+ 1

�= 0 � � = 2 s.

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��+ 1+ 1��= 0= �� �� = 2= 2 s.s

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234 ¤ CHAPTER 3 DIFFERENTIATION RULES

(d) The particle is moving in the positive direction when )(�) 0. 0���2�� 1

2�+ 1

� 0 � � 1

2�+ 1 0 �

� 12� �1 � 0 � � � 2.

(e) See Exercise 1(a).

|�(2)� �(0)| = ��20�1 � 0�� = 20�1.

|�(8)� �(2)| = ��80�4 � 20�1�� = 20�1 � 80�4.

The total distance is 20�1 +�20�1 � 80�4� = 40�1 � 80�4 ft. [ 1�32]

(f )

(g) )(�) = 0���2�� 1

2�+ 1

� �

�(�) = )0(�) = 0���2��1

2

�+�� 1

2�+ 1

�0���2

�� 12

�= � 1

20���2

1 +

�� 12�+ 1

�= � 1

20���2

�� 12�+ 2

�= 1

40���2(�� 4)

�(3) = � 140�3�2 ft�s2

(h )

(i) The particle is speeding up when ) and � have the same sign. This occurs when 2 � � � 4 [) and � are both negative].

It is slowing down when ) and � have opposite signs; that is, when 0 � � � 2 and when � 4.

5. (a) From the �gure, the velocity ) is positive on the interval (0� 2) and negative on the interval (2� 3). The acceleration � is

positive (negative) when the slope of the tangent line is positive (negative), so the acceleration is positive on the interval

(0� 1), and negative on the interval (1� 3). The particle is speeding up when ) and � have the same sign, that is, on the

interval (0� 1) when ) 0 and � 0, and on the interval (2� 3) when ) � 0 and � � 0. The particle is slowing down

when ) and � have opposite signs, that is, on the interval (1� 2) when ) 0 and � � 0.

(b) ) 0 on (0� 3) and ) � 0 on (3� 4). � 0 on (1� 2) and � � 0 on (0� 1) and (2� 4). The particle is speeding up on (1� 2)

[) 0, � 0] and on (3� 4) [) � 0, � � 0]. The particle is slowing down on (0� 1) and (2� 3) [) 0, � � 0].

6. (a) The velocity ) is positive when � is increasing, that is, on the intervals (0� 1) and (3� 4); and it is negative when � is

decreasing, that is, on the interval (1� 3). The acceleration � is positive when the graph of � is concave upward () is

increasing), that is, on the interval (2� 4); and it is negative when the graph of � is concave downward () is decreasing), that

is, on the interval (0� 2). The particle is speeding up on the interval (1� 2) [) � 0, � � 0] and on (3� 4) [) 0, � 0].

The particle is slowing down on the interval (0� 1) [) 0, � � 0] and on (2� 3) [) � 0, � 0].

(b) The velocity ) is positive on (3� 4) and negative on (0� 3). The acceleration � is positive on (0� 1) and (2� 4) and negative

on (1� 2). The particle is speeding up on the interval (1� 2) [) � 0, � � 0] and on (3� 4) [) 0, � 0]. The particle is

slowing down on the interval (0� 1) [) � 0, � 0] and on (2� 3) [) � 0, � 0].

7. (a) �(�) = �3 � 4�5�2 � 7� � )(�) = �0(�) = 3�2 � 9�� 7 = 5 � 3�2 � 9�� 12 = 0 �

3(�� 4)(�+ 1) = 0 � � = 4 or �1. Since � � 0, the particle reaches a velocity of 5 m�s at � = 4 s.

(b) �(�) = )0(�) = 6�� 9 = 0 � � = 1�5. The acceleration changes from negative to positive, so the velocity changes

from decreasing to increasing. Thus, at � = 1�5 s, the velocity has its minimum value.

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SECTION 3.8 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 235

8. (a) � = 5�+ 3�2 � )(�) =��

��= 5 + 6�, so )(2) = 5 + 6(2) = 17 m�s.

(b) )(�) = 35 � 5 + 6� = 35 � 6� = 30 � � = 5 s.

9. (a) � = 10�� 0�83�2 � )(�) =��

��= 10� 1�66�, so )(3) = 10� 1�66(3) = 5�02 m�s.

(b) � = 25 � 10�� 0�83�2 = 25 � 0�83�2 � 10�+ 25 = 0 � � = 10±�17

166 3�54 or 8�51.

The value �1 = 10��17

166 corresponds to the time it takes for the stone to rise 25 m and �2 = 10+�17

166 corresponds to the

time when the stone is 25 m high on the way down. Thus, )(�1) = 10� 1�66�10��

17166

�=�17 4�12 m�s.

10. (a) At maximum height the velocity of the ball is 0 ft�s. )(�) = �0(�) = 80� 32� = 0 � 32� = 80 � � = 52

.

So the maximum height is ��52

�= 80

�52

�� 16� 52

�2= 200� 100 = 100 ft.

(b) �(�) = 80�� 16�2 = 96 � 16�2 � 80�+ 96 = 0 � 16(�2 � 5�+ 6) = 0 � 16(�� 3)(�� 2) = 0.

So the ball has a height of 96 ft on the way up at � = 2 and on the way down at � = 3. At these times the velocities are

)(2) = 80� 32(2) = 16 ft�s and )(3) = 80� 32(3) = �16 ft�s, respectively.

11. (a) �(�) = �2 � �0(�) = 2�. �0(15) = 30 mm2�mm is the rate at which

the area is increasing with respect to the side length as � reaches 15 mm.

(b) The perimeter is � (�) = 4�, so �0(�) = 2� = 12 (4�) =

12� (�). The

�gure suggests that if�� is small, then the change in the area of the square

is approximately half of its perimeter (2 of the 4 sides) times��. From the

�gure,�� = 2� (��) + (��)2. If�� is small, then�� 2� (��) and

so����� 2�.

12. (a) (�) = �3 � �

��= 3�2. �

��

�����=3

= 3(3)2 = 27 mm3�mm is the

rate at which the volume is increasing as � increases past 3 mm.

(b) The surface area is &(�) = 6�2, so 0(�) = 3�2 = 12(6�2) = 1

2&(�).

The �gure suggests that if�� is small, then the change in the volume of the

cube is approximately half of its surface area (the area of 3 of the 6 faces)

times��. From the �gure,� = 3�2(��) + 3�(��)2 + (��)3.

If�� is small, then� 3�2(��) and so� ��� 3�2.

13. (a) Using �(�) = ��2, we �nd that the average rate of change is:

(i) �(3)��(2)

3� 2 =9� � 4�1

= 5� (ii) �(2�5)��(2)

2�5� 2 =6�25� � 4�

0�5= 4�5�

(iii) �(2�1)��(2)

2�1� 2 =4�41� � 4�

0�1= 4�1�

(b) �(�) = ��2 � �0(�) = 2��, so �0(2) = 4�.

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236 ¤ CHAPTER 3 DIFFERENTIATION RULES

(c) The circumference is �(�) = 2�� = �0(�). The �gure suggests that if�� is small,

then the change in the area of the circle (a ring around the outside) is approximately

equal to its circumference times��. Straightening out this ring gives us a shape that

is approximately rectangular with length 2�� and width��, so�� 2��(��).Algebraically,�� = �(�+��)��(�) = �(�+��)2 � ��2 = 2��(��) + �(��)2.

So we see that if�� is small, then �� 2��(��) and therefore,����� 2��.

14. After � seconds the radius is � = 60�, so the area is �(�) = �(60�)2 = 3600��2 � �0(�) = 7200�� �(a) �0(1) = 7200� cm2�s (b) �0(3) = 21� 600� cm2�s (c) �0(5) = 36,000� cm2�s

As time goes by, the area grows at an increasing rate. In fact, the rate of change is linear with respect to time.

15. &(�) = 4��2 � &0(�) = 8�� �(a) &0(1) = 8� ft2�ft (b) &0(2) = 16� ft2�ft (c) &0(3) = 24� ft2�ft

As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the

radius.

16. (a) Using (�) = 43��3, we �nd that the average rate of change is:

(i) (8)� (5)

8� 5 =43�(512)� 4

3�(125)

3= 172� :m3�:m

(ii) (6)� (5)

6� 5 =43�(216)� 4

3�(125)

1= 121�3� :m3�:m

(iii) (5�1)� (5)

5�1� 5 =43�(5�1)

3 � 43�(5)

3

0�1= 102�013� :m3�:m

(b) 0(�) = 4��2, so 0(5) = 100� :m3�:m.

(c) (�) = 43��3 � 0(�) = 4��2 = &(�). By analogy with Exercise 13(c), we can say that the change in the volume

of the spherical shell,� , is approximately equal to its thickness, ��, times the surface area of the inner sphere. Thus,

� 4��2(��) and so� ��� 4��2.

17. The mass is �(�) = 3�2, so the linear density at � is @(�) = � 0(�) = 6�.

(a) @(1) = 6 kg�m (b) @(2) = 12 kg�m (c) @(3) = 18 kg�m

Since @ is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end.

18. (�) = 5000�1� 1

40 ��2 � 0(�) = 5000 · 2�1� 1

40 ���� 1

40

�= �250�1� 1

40 ��

(a) 0(5) = �250�1� 540

�= �218�75 gal�min (b) 0(10) = �250�1� 10

40

�= �187�5 gal�min

(c) 0(20) = �250�1� 2040

�= �125 gal�min (d) 0(40) = �250�1� 40

40

�= 0 gal�min

The water is �owing out the fastest at the beginning — when � = 0, 0(�) = �250 gal�min. The water is �owing out the

slowest at the end — when � = 40, 0(�) = 0. As the tank empties, the water �ows out more slowly.

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SECTION 3.8 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 237

19. The quantity of charge is /(�) = �3 � 2�2 + 6�+ 2, so the current is /0(�) = 3�2 � 4�+ 6.

(a) /0(0�5) = 3(0�5)2 � 4(0�5) + 6 = 4�75 A (b) /0(1) = 3(1)2 � 4(1) + 6 = 5 A

The current is lowest when /0 has a minimum. /00(�) = 6�� 4 � 0 when � � 23

. So the current decreases when � � 23

and

increases when � 23 . Thus, the current is lowest at � = 2

3 s.

20. (a) ! = #�-

�2= (#�-)��2 � �!

��= �2(#�-)��3 = �2#�-

�3, which is the rate of change of the force with

respect to the distance between the bodies. The minus sign indicates that as the distance � between the bodies increases,

the magnitude of the force ! exerted by the body of mass � on the body of mass - is decreasing.

(b) Given ! 0(20,000) = �2, �nd ! 0(10,000). �2 = �2#�-20,0003

� #�- = 20,0003.

! 0(10� 000) = �2(20,0003)10,0003

= �2 · 23 = �16 N�km

21. (a) To �nd the rate of change of volume with respect to pressure, we �rst solve for in terms of � .

� = � � =�

�� �

��= � �

� 2.

(b) From the formula for � ��� in part (a), we see that as � increases, the absolute value of � ��� decreases.

Thus, the volume is decreasing more rapidly at the beginning.

(c) ; = � 1

��= � 1

�� �

� 2

�=

(� )�=

��=1

22. (a) [C] = �2��

���+ 1� rate of reaction = �[C]

��=(���+ 1)(�2�)� (�2��)(��)

(���+ 1)2=

�2�(���+ 1� ���)

(���+ 1)2=

�2�

(���+ 1)2

(b) If � = [C], then �� � = �� �2��

���+ 1=

�2��+ �� �2��

���+ 1=

���+ 1.

So �(�� �)2 = �

��

���+ 1

�2=

�2�

(���+ 1)2=

�[C]��

[from part (a)] =��

��.

(c) As � �, [C] = �2��

���+ 1=

(�2��)��

(���+ 1)��=

�2�

�� + (1��) �2�

��= � moles�L.

(d) As � �, �[C]��

=�2�

(���+ 1)2 0.

(e) As � increases, nearly all of the reactants A and B are converted into product C. In practical terms, the reaction virtually

stops.

23. In Example 6, the population function was ( = 2� (0. Since we are tripling instead of doubling and the initial population is

400, the population function is ((�) = 400 · 3�. The rate of growth is (0(�) = 400 · 3� · ln 3, so the rate of growth after

2�5 hours is (0(2�5) = 400 · 325 · ln 3 6850 bacteria�hour�

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238 ¤ CHAPTER 3 DIFFERENTIATION RULES

24. ( = �(�) =�

1 + 0�07�� (0 = �� · 0

�07�(�0�7)(1 + 0�07�)2

[Reciprocal Rule]. When � = 0, ( = 20 and (0 = 12.

�(0) = 20 � 20 =�

1 + � � = 20(1 + ). � 0(0) = 12 � 12 =

0�7�

(1 + )2� 12 =

0�7(20)(1 + )

(1 + )2�

12

14=

1 + � 6(1 + ) = 7 � 6 + 6 = 7 � = 6 and � = 20(1 + 6) = 140. For the long run, we let �

increase without bound. lim���

�(�) = lim���

140

1 + 60�07�=

140

1 + 6 · 0 = 140, indicating that the yeast population stabilizes

at 140 cells.

25. (a) 1920: �1 =1860 � 1750

1920 � 1910=110

10= 11, �2 =

2070 � 1860

1930 � 1920=210

10= 21,

(�1 +�2)/ 2 = (11 + 21)�2 = 16 million�year

1980: �1 =4450 � 3710

1980 � 1970=740

10= 74, �2 =

5280 � 4450

1990 � 1980=830

10= 83,

(�1 +�2)/ 2 = (74 + 83)�2 = 78�5 million�year

(b) � (�) = ��3 + �2 + +�+ � (in millions of people), where � 0�0012937063, �7�061421911, + 12,822�97902,

and � �7,743,770�396.

(c) � (�) = ��3 + �2 + +�+ � � � 0(�) = 3��2 + 2�+ + (in millions of people per year)

(d) � 0(1920) = 3(0�0012937063)(1920)2 + 2(�7�061421911)(1920) + 12,822�97902

14.48 million�year [smaller than the answer in part (a), but close to it]

� 0(1980) 75.29 million�year (smaller, but close)

(e) � 0(1985) 81.62 million�year, so the rate of growth in 1985 was about 81�62 million�year.

26. (a) �(�) = ��4 + �3 + +�2 + ��+ 0, where � �3�076923× 10�6,

0�0243620824, + �72�33147086, � 95442�50365, and

0 �47,224,986�6.

(d)

(b) �(�) = ��4 + �3 + +�2 + ��+ 0 �

�0(�) = 4��3 + 3�2 + 2+�+ �.

(c) Part (b) gives �0(1990) 0�11 years of age per year.

27. (a) Using ) = �

4A9(.2 � �2) with . = 0�01, 9 = 3, � = 3000, and A = 0�027, we have ) as a function of �:

)(�) =3000

4(0�027)3(0�012 � �2). )(0) = 0�925 cm�s, )(0�005) = 0�694 cm�s, )(0�01) = 0.

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SECTION 3.8 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 239

(b) )(�) = �

4A9(.2 � �2) � )0(�) =

4A9(�2�) = ���

2A9. When 9 = 3, � = 3000, and A = 0�027, we have

)0(�) = � 3000�

2(0�027)3. )0(0) = 0, )0(0�005) = �92�592 (cm�s)�cm, and )0(0�01) = �185�185 (cm�s)�cm.

(c) The velocity is greatest where � = 0 (at the center) and the velocity is changing most where � = . = 0�01 cm

(at the edge).

28. (a) (i) � = 1

2$

��

@=

�1

2

��

@

�$�1 � ��

�$= �

�1

2

��

@

�$�2 = � 1

2$2

��

@

(ii) � = 1

2$

��

@=

�1

2$�@

�� 1�2 � ��

��=1

2

�1

2$�@

���1�2 =

1

4$��@

(iii) � = 1

2$

��

@=

���

2$

�@�1�2 � ��

�@= �1

2

���

2$

�@�3�2 = �

��

4$@3�2

(b) Note: Illustrating tangent lines on the generic �gures may help to explain the results.

(i) ��

�$� 0 and $ is decreasing � � is increasing � higher note

(ii) ��

�� 0 and � is increasing � � is increasing � higher note

(iii) ��

�@� 0 and @ is increasing � � is decreasing � lower note

29. (a) �(�) = 1200 + 12�� 0�1�2 + 0�0005�3 � �0(�) = 12� 0�2�+ 0�0015�2 $�yard, which is the marginal cost

function.

(b) �0(200) = 12� 0�2(200) + 0�0015(200)2 = $32�yard, and this is the rate at which costs are increasing with respect to

the production level when � = 200. �0(200) predicts the cost of producing the 201st yard�

(c) The cost of manufacturing the 201st yard of fabric is �(201)� �(200) = 3632�2005� 3600 $32�20, which is

approximately �0(200).

30. (a) �(�) = 339 + 25�� 0�09�2 + 0�0004�3 � �0(�) = 25� 0�18�+ 0�0012�2. �0(100) = $19�item, and this is

the rate at which costs are increasing with respect to the production level when � = 100.

(b) The cost of producing the 101st item is �(101)� �(100) = 2358�0304� 2339 = $19�03, which is approximately

�0(100).

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240 ¤ CHAPTER 3 DIFFERENTIATION RULES

31. (a) �(�) = �(�)

�� �0(�) =

��0(�)� �(�) · 1�2

=��0(�)� �(�)

�2.

�0(�) 0 � �(�) is increasing; that is, the average productivity increases as the size of the workforce increases.

(b) �0(�) is greater than the average productivity � �0(�) �(�) � �0(�) �(�)

�� ��0(�) �(�) �

��0(�)� �(�) 0 � ��0(�)� �(�)

�2 0 � �0(�) 0.

32. (a) & = �.

��=(1 + 4�04)(9�6��06)� (40 + 24�04)(1�6��06)

(1 + 4�04)2

=9�6��06 + 38�4��02 � 64��06 � 38�4��02

(1 + 4�04)2= � 54�4��06

(1 + 4�04)2

(b) At low levels of brightness, . is quite large [.(0) = 40] and is quickly

decreasing, that is, & is negative with large absolute value. This is to be

expected: at low levels of brightness, the eye is more sensitive to slight

changes than it is at higher levels of brightness.

33. � = (.� � � =�

(.=

(10)(0�0821)=

1

0�821(� ). Using the Product Rule, we have

��

��=

1

0�821[� (�) 0(�) + (�)� 0(�)] =

1

0�821[(8)(�0�15) + (10)(0�10)] �0�2436 K�min.

34. (a) If ����� = 0, the population is stable (it is constant).

(b) ����

= 0 � ;� = �0

�1� �

��

�� � ;

�0= 1� �

��� �

��= 1� ;

�0� � = ��

�1� ;

�0

�.

If �� = 10,000, �0 = 5% = 0�05, and ; = 4% = 0�04, then � = 10,000�1� 4

5

�= 2000.

(c) If ; = 0�05, then � = 10,000�1� 5

5

�= 0. There is no stable population.

35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, ����

= 0 and �%

��= 0.

(b) “The caribou go extinct” means that the population is zero, or mathematically, � = 0.

(c) We have the equations ����

= �� � �% and �%

��= �+% + ��% . Let ����� = �%��� = 0, � = 0�05, = 0�001,

+ = 0�05, and � = 0�0001 to obtain 0�05� � 0�001�% = 0 (1) and �0�05% + 0�0001�% = 0 (2). Adding 10 times

(2) to (1) eliminates the �% -terms and gives us 0�05� � 0�5% = 0 � � = 10% . Substituting � = 10% into (1)

results in 0�05(10% )� 0�001(10% )% = 0 � 0�5% � 0�01% 2 = 0 � 50% �% 2 = 0 �

% (50�% ) = 0 � % = 0 or 50. Since � = 10% , � = 0 or 500. Thus, the population pairs (��% ) that lead to

stable populations are (0� 0) and (500� 50). So it is possible for the two species to live in harmony.

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SECTION 3.9 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 241

3.9 Linear Approximations and Differentials

1. As in Example 1, � (0) = 185, � (10) = 172, � (20) = 160, and

� 0(20) � (10)� � (20)

10� 20 =172� 160�10 = �1�2�F�min.

� (30) � (20) + � 0(20)(30� 20) 160� 1�2(10) = 148�F.

We would expect the temperature of the turkey to get closer to 75�F as time

increases. Since the temperature decreased 13�F in the �rst 10 minutes and 12�F

in the second 10 minutes, we can assume that the slopes of the tangent line are

increasing through negative values: �1�3��1�2� � � �. Hence, the tangent lines are

under the curve and 148�F is an underestimate. From the �gure, we estimate the slope of the tangent line at � = 20 to be184� 1470� 30

= � 3730

. Then the linear approximation becomes � (30) � (20) + � 0(20) · 10 160� 3730(10) = 147 2

3 147�7.

2. � 0(2) � (1)� � (2)

1� 2 =87�1� 74�9

�1 = �12�2 kilopascals�km.

� (3) � (2) + � 0(2)(3� 2) 74�9� 12�2(1) = 62�7 kPa.

From the �gure, we estimate the slope of the tangent line at � = 2

to be 98 � 63

0 � 3= �35

3. Then the linear approximation becomes

� (3) � (2) + � 0(2) · 1 74�9� 353 63�23 kPa.

3. Extend the tangent line at the point (2030� 21) to the �-axis. Answers

will vary based on this approximation—we’ll use � = 1900 as our

�-intercept. The linearization is then

� (�) � (2030) + � 0(2030)(�� 2030) 21 + 21130(�� 2030)

� (2040) = 21 + 21130(2040� 2030) 22�6%

� (2050) = 21 + 21130 (2050� 2030) 24�2%

These predictions are probably too high since the tangent line lies above the graph at � = 2030�

4. Let � = �(1985)��(1990)

1985� 1990 =17�04� 19�33

�5 = 0�458 and � =�(1995)��(1990)

1995� 1990 =

21�91� 19�335

= 0�516 . Then � 0(1990) = lim��1990

�(�)��(1990)

�� 1990 �+�

2= 0�487 million�year.

So �(1989) �(1990) +� 0(1990)(1989� 1990) 19�33 + 0�487(�1) = 18�843 million.

� 0(2005) �(2000)��(2005)

2000� 2005 =24�70� 27�68

�5 = 0�596 million�year.

�(2010) �(2005) +� 0(2005)(2010� 2005) 27�68 + 0�596(5) = 30�66 million.

5. �(�) = �4 + 3�2 � � 0(�) = 4�3 + 6�, so �(�1) = 4 and � 0(�1) = �10.

Thus, $(�) = �(�1) + � 0(�1)(�� (�1)) = 4 + (�10)(�+ 1) = �10�� 6.

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242 ¤ CHAPTER 3 DIFFERENTIATION RULES

6. �(�) = ln� � � 0(�) = 1��, so �(1) = 0 and � 0(1) = 1. Thus, $(�) = �(1) + � 0(1) (�� 1) = 0 + 1(�� 1) = �� 1.

7. �(�) = cos� � � 0(�) = � sin�, so ���2

�= 0 and � 0

��2

�= �1.

Thus, $(�) = ���2

�+ � 0

��2

���� �

2

�= 0� 1��� �

2

�= ��+ �

2.

8. �(�) = �3�4 � � 0(�) = 34�

�1�4, so �(16) = 8 and � 0(16) = 38 .

Thus, $(�) = �(16) + � 0(16)(�� 16) = 8 + 38(�� 16) = 3

8�+ 2.

9. �(�) =�1� � � � 0(�) =

�12�1� �

, so �(0) = 1 and � 0(0) = � 12

.

Therefore,�1� � = �(�) �(0) + � 0(0)(�� 0) = 1 + �� 1

2

�(�� 0) = 1� 1

2�.

So�0�9 =

�1� 0�1 1� 1

2(0�1) = 0�95

and�0�99 =

�1� 0�01 1� 1

2 (0�01) = 0�995.

10. �(�) = 3�1 + � = (1 + �)1�3 � �0(�) = 1

3 (1 + �)�2�3, so �(0) = 1 and

�0(0) = 13

. Therefore, 3�1 + � = �(�) �(0) + �0(0)(�� 0) = 1 + 1

3�.

So 3�0�95 = 3

�1 + (�0�05) 1 + 1

3(�0�05) = 0�983,

and 3�1�1 = 3

�1 + 0�1 1 + 1

3(0�1) = 1�03.

11. �(�) = 3�1� � = (1� �)1�3 � � 0(�) = � 1

3(1� �)�2�3, so �(0) = 1

and � 0(0) = �13 . Thus, �(�) �(0) + � 0(0)(�� 0) = 1� 1

3�. We need

3�1� �� 0�1 � 1� 1

3� � 3

�1� �+ 0�1, which is true when

�1�204 � � � 0�706.

12. �(�) = tan� � � 0 (�) = sec2 �, so �(0) = 0 and � 0(0) = 1.

Thus, �(�) �(0) + � 0(0)(�� 0) = 0 + 1(�� 0) = �.

We need tan�� 0�1 � � � tan�+ 0�1, which is true when

�0�63 � � � 0�63.

13. �(�) = 1

(1 + 2�)4= (1 + 2�)�4 �

� 0(�) = �4(1 + 2�)�5(2) = �8(1 + 2�)5

, so �(0) = 1 and � 0(0) = �8.

Thus, �(�) �(0) + � 0(0)(�� 0) = 1 + (�8)(�� 0) = 1� 8�.

We need 1

(1 + 2�)4� 0�1 � 1� 8� � 1

(1 + 2�)4+ 0�1, which is true

when � 0�045 � � � 0�055.

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SECTION 3.9 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 243

14. �(�) = 0� � � 0(�) = 0�, so �(0) = 1 and � 0(0) = 1.

Thus, �(�) �(0) + � 0(0)(�� 0) = 1 + 1(�� 0) = 1 + �.

We need 0� � 0�1 � 1 + � � 0� + 0�1, which is true when �0�483 � � � 0�416.

15. To estimate (2�001)5, we’ll �nd the linearization of �(�) = �5 at � = 2. Since � 0(�) = 5�4, �(2) = 32, and � 0(2) = 80,

we have $(�) = 32 + 80(�� 2) = 80�� 128. Thus, �5 80�� 128 when � is near 2 , so

(2�001)5 80(2�001)� 128 = 160�08� 128 = 32�08.

16. To estimate 0�0015, we’ll �nd the linearization of �(�) = 0� at � = 0. Since � 0(�) = 0�, �(0) = 1, and � 0(0) = 1, we have

$(�) = 1 + 1(�� 0) = �+ 1. Thus, 0� �+ 1 when � is near 0, so 0�0015 �0�015 + 1 = 0�985.

17. To estimate (8�06)2�3, we’ll �nd the linearization of �(�) = �2�3 at � = 8. Since � 0(�) = 23��1�3 = 2�

�3

3���

,

�(8) = 4, and � 0(8) = 13 , we have $(�) = 4 + 1

3 (�� 8) = 13�+

43 . Thus, �2�3 1

3�+43 when � is near 8, so

(8�06)2�3 13 (8�06) +

43 =

12063 = 4�02.

18. To estimate 1�1002, we’ll �nd the linearization of �(�) = 1�� at � = 1000. Since � 0(�) = �1��2, �(1000) = 0�001,

and � 0(1000) = �0�000001, we have $(�) = 0�001� 0�000001(�� 1000) = �0�000001�+ 0�002. Thus,

1�� �0�000001�+ 0�002 when � is near 1000, so 1�1002 �0�000001(1002) + 0�002 = 0�000998.

19. = �(�) = sec� � � 0(�) = sec� tan�, so �(0) = 1 and � 0(0) = 1 · 0 = 0. The linear approximation of � at 0 is

�(0) + � 0(0)(�� 0) = 1 + 0(�) = 1. Since 0�08 is close to 0, approximating sec 0�08 with 1 is reasonable.

20. If = �6, 0 = 6�5 and the tangent line approximation at (1� 1) has slope 6. If the change in � is 0�01, the change in on the

tangent line is 0�06, and approximating (1�01)6 with 1�06 is reasonable.

21. = �(�) = ln� � � 0(�) = 1��, so �(1) = 0 and � 0(1) = 1. The linear approximation of � at 1 is

�(1) + � 0(1)(�� 1) = 0 + 1(�� 1) = �� 1. Now �(1�05) = ln 1�05 1�05� 1 = 0�05, so the approximation

is reasonable.

22. (a) �(�) = (� � 1)2 � � 0(�) = 2(� � 1), so �(0) = 1 and � 0(0) = �2.

Thus, �(�) $ (�) = �(0) + � 0(0)(�� 0) = 1� 2�.

�(�) = 0�2� � �0(�) = �20�2�, so �(0) = 1 and �0(0) = �2.

Thus, �(�) $�(�) = �(0) + �0(0)(�� 0) = 1� 2�.

�(�) = 1 + ln(1 � 2�) � �0(�) =�2

1� 2� , so �(0) = 1 and �0(0) = �2.

Thus, �(�) $�(�) = �(0) + �0(0)(�� 0) = 1� 2�.

Notice that $ = $� = $�. This happens because � , �, and � have the same function values and the same derivative

values at � = 0.

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244 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b) The linear approximation appears to be the best for the function � since it is

closer to � for a larger domain than it is to � and �. The approximation

looks worst for � since � moves away from $ faster than � and � do.

23. (a) For = �(�) =�+ 1

�� 1 , � 0(�) = (�� 1)(1)� (�+ 1)(1)(�� 1)2 =

�2(�� 1)2 , so � = �2

(�� 1)2 ��.

(b) For = �(�) = (1 + �3)�2, � 0(�) = �2(1 + �3)�3(3�2) =�6�2

(1 + �3)3, so � = �6�2

(1 + �3)3��.

24. (a) For = �(�) = 0tan��, � 0(�) = 0tan�� · sec2(��) · �, so � = � sec2(��)0tan�� ��.

(b) For = �(8) =�1 + ln 8, � 0(8) = 1

2 (1 + ln 8)�1�2 · 1

8, so � = 1

28�1 + ln 8

�8.

25. (a) = 0��10 � � = 0��10 · 110�� = 1

100��10��

(b) � = 0 and �� = 0�1 � � = 11000�10(0�1) = 0�01.

� = � (�+��)� � (�) = 00+01 � 00 = 001 � 1 0�0101.

26. (a) =�� � � = 1

2��1�2�� =

1

2����

(b) � = 1 and �� = 1 � � = 12(1)

(1) = 12

. � = � (�+��)� � (�) =�1 + 1��1 = �2� 1 0�414.

(c) Remember,� represents the amount that the curve = �(�) rises

or falls when � changes by an amount ��, whereas � represents

the amount that the tangent line rises or falls (the change in the

linearization).

27. (a) If � is the edge length, then = �3 � � = 3�2 ��. When � = 30 and �� = 0�1, � = 3(30)2(0�1) = 270, so the

maximum possible error in computing the volume of the cube is about 270 cm3. The relative error is calculated by dividing

the change in ,� , by . We approximate� with � .

Relative error= �

=3�2 ��

�3= 3

��

�= 3

�0�1

30

�= 0�01.

Percentage error = relative error× 100% = 0�01× 100% = 1%.

(b) & = 6�2 � �& = 12���. When � = 30 and �� = 0�1, �& = 12(30)(0�1) = 36, so the maximum possible error in

computing the surface area of the cube is about 36 cm2.

Relative error= �&

& �&

&=12���

6�2= 2

��

�= 2

�0�1

30

�= 0�006.

Percentage error = relative error× 100% = 0�006× 100% = 0�6%.

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SECTION 3.9 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 245

28. (a) � = ��2 � �� = 2�� ��. When � = 24 and �� = 0�2, �� = 2�(24)(0�2) = 9�6�, so the maximum possible error

in the calculated area of the disk is about 9�6� 30 cm2.

(b) Relative error = ��

� ��

�=2�� ��

��2=2 ��

�=2(0�2)

24=0�2

12=1

60= 0�016.

Percentage error = relative error×100% = 0�016× 100% = 1�6%.

29. (a) For a sphere of radius �, the circumference is � = 2�� and the surface area is & = 4��2, so

� =�

2�� & = 4�

��

2�

�2=

�2

�� �& =

2

�� ��. When � = 84 and �� = 0�5, �& = 2

�(84)(0�5) =

84

�,

so the maximum error is about 84� 27 cm2. Relative error �&

&=84��

842��=1

84 0�012

(b) =4

3��3 =

4

3�

��

2�

�3=

�3

6�2� � =

1

2�2�2 ��. When � = 84 and �� = 0�5,

� =1

2�2(84)2(0�5) =

1764

�2, so the maximum error is about 1764

�2 179 cm3.

The relative error is approximately �

=

1764��2

(84)3�(6�2)=1

56 0�018.

30. For a hemispherical dome, = 23��3 � � = 2��2 ��. When � = 1

2(50) = 25 m and �� = 0�05 cm = 0�0005 m,

� = 2�(25)2(0�0005) = 5�8 , so the amount of paint needed is about 5�8 2 m3.

31. (a) = ��2� � � � = 2����� = 2�����

(b) The error is

� � � = [�(� +��)2�� ��2�]� 2����� = ��2�+ 2����� + �(��)2�� ��2�� 2����� = �(��)2�.

32. (a) sin � = 20

�� � = 20 csc � �

��= 20(� csc � cot �) �� = �20 csc 30� cot 30� (±1�)

= �20(2)��3 ��± �

180

�= ±2

�3

9�

So the maximum error is about ± 29

�3� ±1�21 cm.

(b) The relative error is ��� ��

�=± 29

�3�

20(2)= ±

�3

180� ±0�03, so the percentage error is approximately±3%.

33. ! = �.4 � �! = 4�.3 �. � �!

!=4�.3 �.

�.4= 4

��.

.

�. Thus, the relative change in ! is about 4 times the

relative change in .. So a 5% increase in the radius corresponds to a 20% increase in blood �ow.

34. (a) �(�) = sin� � � 0(�) = cos�, so �(0) = 0 and � 0(0) = 1. Thus, �(�) �(0)+ � 0(0)(�� 0) = 0+1(�� 0) = �.

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246 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b)

We want to know the values of � for which = � approximates = sin� with less than a 2% difference; that is, the

values of � for which������ sin�sin�

���� � 0�02 � �0�02 � �� sin�sin�

� 0�02 �

��0�02 sin� � �� sin� � 0�02 sin� if sin� 0

�0�02 sin� �� sin� 0�02 sin� if sin� � 0�

�0�98 sin� � � � 1�02 sin� if sin� 0

1�02 sin� � � � 0�98 sin� if sin� � 0

In the �rst �gure, we see that the graphs are very close to each other near � = 0. Changing the viewing rectangle

and using an intersect feature (see the second �gure) we �nd that = � intersects = 1�02 sin� at � 0�344.

By symmetry, they also intersect at � �0�344 (see the third �gure). Converting 0�344 radians to degrees, we get

0�344�180��

� 19�7� 20�, which veri�es the statement.

35. (a) The graph shows that � 0(1) = 2, so $(�) = �(1) + � 0(1)(�� 1) = 5 + 2(�� 1) = 2�+ 3.

�(0�9) $(0�9) = 4�8 and �(1�1) $(1�1) = 5�2.

(b) From the graph, we see that � 0(�) is positive and decreasing. This means that the slopes of the tangent lines are positive,

but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too

large.

36. (a) �0(�) =��2 + 5 � �0(2) =

�9 = 3. �(1�95) �(2) + �0(2)(1�95� 2) = �4 + 3(�0�05) = �4�15.

�(2�05) �(2) + �0(2)(2�05� 2) = �4 + 3(0�05) = �3�85.

(b) The formula �0(�) =��2 + 5 shows that �0(�) is positive and increasing. This means that the slopes of the tangent lines

are positive and the tangents are getting steeper. So the tangent lines lie below the graph of �. Hence, the estimates in

part (a) are too small.

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LABORATORY PROJECT TAYLOR POLYNOMIALS ¤ 247

LABORATORY PROJECT Taylor Polynomials

1. We �rst write the functions described in conditions (i), (ii), and (iii):

� (�) = �+��+ ��2 �(�) = cos�

� 0(�) = � + 2�� � 0(�) = � sin�� 00(�) = 2� � 00(�) = � cos�

So, taking � = 0, our three conditions become

� (0) = �(0): � = cos 0 = 1

� 0(0) = � 0(0): � = � sin 0 = 0� 00(0) = � 00(0): 2� = � cos 0 = �1 � � = �1

2

The desired quadratic function is � (�) = 1� 12�2, so the quadratic approximation is cos� 1� 1

2�2.

The �gure shows a graph of the cosine function together with its linear

approximation $(�) = 1 and quadratic approximation � (�) = 1� 12�2

near 0. You can see that the quadratic approximation is much better than the

linear one.

2. Accuracy to within 0�1 means that��cos�� �

1� 12�2��� � 0�1 � �0�1 � cos�� �

1� 12�2�� 0�1 �

0�1 �1� 1

2�2�� cos� �0.1 � cos�+0�1 1� 1

2�2 cos�� 0�1 � cos�� 0�1 � 1� 1

2�2 � cos�+0�1.

From the �gure we see that this is true between � and �. Zooming in or

using an intersect feature, we �nd that the �-coordinates of � and � are

about ±1�26. Thus, the approximation cos� 1� 12�2 is accurate to

within 0�1 when �1�26 � � � 1�26.

3. If � (�) = �+�(�� �) +�(�� �)2, then � 0(�) = � + 2�(�� �) and � 00(�) = 2�. Applying the conditions (i), (ii),

and (iii), we get

� (�) = �(�): � = �(�)

� 0(�) = � 0(�): � = � 0(�)

� 00(�) = � 00(�): 2� = � 00(�) � � = 12� 00(�)

Thus, � (�) = �+�(�� �) +�(�� �)2 can be written in the form � (�) = �(�) + � 0(�)(�� �) + 12� 00(�)(�� �)2.

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248 ¤ CHAPTER 3 DIFFERENTIATION RULES

4. From Example 2 in Section 3.9, we have �(1) = 2, � 0(1) = 14

, and

� 0(�) = 12 (�+ 3)

�1�2. So � 00(�) = � 14 (�+ 3)

�3�2 � � 00(1) = � 132 .

From Problem 3, the quadratic approximation � (�) is��+ 3 �(1)+ � 0(1)(�� 1)+ 1

2� 00(1)(�� 1)2 = 2+ 1

4(�� 1)� 1

64(�� 1)2.

The �gure shows the function �(�) =��+ 3 together with its linear

approximation $(�) = 14�+ 7

4and its quadratic approximation � (�). You can see that � (�) is a better approximation than

$(�) and this is borne out by the numerical values in the following chart.

from $(�) actual value from � (�)�3�98 1�9950 1�99499373 � � � 1�99499375�4�05 2�0125 2�01246118 � � � 2�01246094�4�2 2�0500 2�04939015 � � � 2�04937500

5. �(�) = +0 + +1(�� �) + +2(�� �)2 + +3(�� �)3 + · · ·+ +(�� �). If we put � = � in this equation,

then all terms after the �rst are 0 and we get �(�) = +0. Now we differentiate �(�) and obtain

� 0(�) = +1 + 2+2(�� �) + 3+3(�� �)2 + 4+4(�� �)3 + · · ·+ (+(�� �)�1. Substituting � = � gives � 0(�) = +1.

Differentiating again, we have � 00 (�) = 2+2 + 2 · 3+3(�� �) + 3 · 4+4(�� �2) + · · ·+ ((� 1)(+(�� �)�2 and so

� 00 (�) = 2+2. Continuing in this manner, we get � 000 (�) = 2 · 3+3+2 · 3 · 4+4(�� �) + · · ·+((� 2)((� 1)(+(�� �)�3

and � 000 (�) = 2 · 3+3. By now we see the pattern. If we continue to differentiate and substitute � = �, we obtain

�(4) (�) = 2 · 3 · 4+4 and in general, for any integer � between 1 and (, � (�) (�) = 2 · 3 · 4 · 5 · · · · · �+� = �! +� �

+� =�(�) (�)

�!. Because we want � and � to have the same derivatives at �, we require that +� =

� (�)(�)

�!for

� = 1� 2� � � � � (.

6. �(�) = �(�) + � 0(�)(�� �) +� 00(�)2!

(�� �)2 + · · ·+ � ()(�)

(!(�� �). To compute the coef�cients in this equation we

need to calculate the derivatives of � at 0:

�(�) = cos� �(0) = cos 0 = 1

� 0(�) = � sin� � 0(0) = � sin 0 = 0� 00(�) = � cos� � 00(0) = �1� 000(�) = sin� � 000(0) = 0

� (4)(�) = cos� � (4)(0) = 1

We see that the derivatives repeat in a cycle of length 4, so � (5)(0) = 0, � (6)(0) = �1, � (7)(0) = 0, and � (8)(0) = 1.

From the original expression for �(�), with ( = 8 and � = 0, we have

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CHAPTER 3 REVIEW ¤ 249

�8(�) = �(0) + � 0(0)(�� 0) + � 00(0)2!

(�� 0)2 + � 000(0)3!

(�� 0)3 + · · ·+ � (8)(0)

8!(�� 0)8

= 1 + 0 · �+ �12!�2 + 0 · �3 + 1

4!�4 + 0 · �5 + �1

6!�6 + 0 · �7 + 1

8!�8 = 1� �2

2!+�4

4!� �6

6!+�8

8!

and the desired approximation is cos� 1� �2

2!+�4

4!� �6

6!+�8

8!. The Taylor polynomials �2, �4, and �6 consist of the

initial terms of �8 up through degree 2, 4, and 6, respectively. Therefore, �2(�) = 1� �2

2!, �4(�) = 1� �2

2!+�4

4!, and

�6(�) = 1� �2

2!+�4

4!� �6

6!. We graph �2, �4, �6, �8, and � :

Notice that �2(�) is a good approximation to cos�

near 0, �4(�) is a good approximation on a larger

interval, �6(�) is a better approximation, and

�8(�) is better still. Each successive Taylor

polynomial is a good approximation on a larger

interval than the previous one.

3 Review

1. (a) The Power Rule: If ( is any real number, then �

��(�) = (��1. The derivative of a variable base raised to a constant

power is the power times the base raised to the power minus one.

(b) The Constant Multiple Rule: If + is a constant and � is a differentiable function, then �

��[+�(�)] = +

���(�).

The derivative of a constant times a function is the constant times the derivative of the function.

(c) The Sum Rule: If � and � are both differentiable, then �

��[�(�) + �(�)] =

���(�) +

���(�). The derivative of a sum

of functions is the sum of the derivatives.

(d) The Difference Rule: If � and � are both differentiable, then �

��[�(�)� �(�)] =

���(�)� �

���(�). The derivative of a

difference of functions is the difference of the derivatives.

(e) The Product Rule: If � and � are both differentiable, then �

��[�(�) �(�)] = �(�)

���(�) + �(�)

���(�). The

derivative of a product of two functions is the �rst function times the derivative of the second function plus the second

function times the derivative of the �rst function.

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250 ¤ CHAPTER 3 DIFFERENTIATION RULES

(f ) The Quotient Rule: If � and � are both differentiable, then �

��

�(�)

�(�)

!=

�(�)�

���(�)� �(�)

���(�)

[ �(�)]2.

The derivative of a quotient of functions is the denominator times the derivative of the numerator minus the numerator

times the derivative of the denominator, all divided by the square of the denominator.

(g) The Chain Rule: If � and � are both differentiable and ! = � � � is the composite function de�ned by ! (�) = �(�(�)),

then ! is differentiable and ! 0 is given by the product ! 0(�) = � 0(�(�)) �0(�). The derivative of a composite function is

the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

2. (a) = � � 0 = (��1 (b) = 0� � 0 = 0�

(c) = �� � 0 = �� ln � (d) = ln� � 0 = 1��

(e) = log� � � 0 = 1�(� ln �) (f ) = sin� � 0 = cos�

(g) = cos� � 0 = � sin� (h) = tan� � 0 = sec2 �

(i) = csc� � 0 = � csc� cot� ( j) = sec� � 0 = sec� tan�

(k) = cot� � 0 = � csc2 � (l) = sin�1 � � 0 = 1��1� �2

(m) = cos�1 � � 0 = �1��1� �2 (n) = tan�1 � � 0 = 1�(1 + �2)

3. (a) 0 is the number such that lim��0

0� � 1�

= 1.

(b) 0 = lim��0

(1 + �)1��

(c) The differentiation formula for = �� [0 = �� ln �] is simplest when � = 0 because ln 0 = 1.

(d) The differentiation formula for = log� � [0 = 1�(� ln �)] is simplest when � = 0 because ln 0 = 1.

4. (a) Implicit differentiation consists of differentiating both sides of an equation involving � and with respect to �, and then

solving the resulting equation for 0. Use implicit differentiation when it is dif�cult to solve an equation for in terms

of �.

(b) Logarithmic differentiation consists of taking natural logarithms of both sides of an equation = �(�), simplifying,

differentiating implicitly with respect to �, and then solving the resulting equation for 0. Use logarithmic differentiation

when the calculation of derivatives of complicated functions involving products, quotients, or powers can be simpli�ed by

taking logarithms.

5. You can �nd �

��as a function of � by calculating �

��=

����

�����=

�0(�)� 0(�)

[if ����� = � 0(�) 6= 0].

6. The linearization $ of � at � = � is $(�) = �(�) + � 0(�)(�� �).

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CHAPTER 3 REVIEW ¤ 251

1. True. This is the Sum Rule.

2. False. See the warning before the Product Rule.

3. True. This is the Chain Rule.

4. True by the Chain Rule.

5. False. �

���(�� ) =

� 0(�� )

2��

by the Chain Rule.

6. False. 02 is a constant, so 0 = 0.

7. False. �

��10� = 10� ln 10

8. False. ln 10 is a constant, so its derivative is 0.

9. True. �

��(tan2 �) = 2 tan� sec2 �, and �

��(sec2 �) = 2 sec� (sec� tan�) = 2 tan� sec2 �.

Or: �

��(sec2 �) =

��(1 + tan2 �) =

��(tan2 �).

10. False. �(�) =���2 + �

�� = �2 + � for � � 0 or � � �1 and���2 + �

�� = �(�2 + �) for �1 � � � 0.

So � 0(�) = 2�+ 1 for � 0 or � � �1 and � 0(�) = �(2�+ 1) for �1 � � � 0. But |2�+ 1| = 2�+ 1for � � � 1

2and |2�+ 1| = �2�� 1 for � � � 1

2.

11. True. �(�) = �5 � �0(�) = 5�4 � �0(2) = 5(2)4 = 80, and by the de�nition of the derivative,

lim��2

�(�)� �(2)

�� 2 = �0(2) = 80.

12. False. A tangent line to the parabola = �2 has slope ���� = 2�, so at (�2� 4) the slope of the tangent is 2(�2) = �4and an equation of the tangent line is � 4 = �4(�+ 2). [The given equation, � 4 = 2�(�+ 2), is not even

linear!]

1. = (�4 � 3�2 + 5)3 �

0 = 3(�4 � 3�2 + 5)2 �

��(�4 � 3�2 + 5) = 3(�4 � 3�2 + 5)2(4�3 � 6�) = 6�(�4 � 3�2 + 5)2(2�2 � 3)

2. = cos(tan�) � 0 = � sin(tan�) �

��(tan�) = � sin(tan�)(sec2 �)

3. =��+

13��4= �1�2 + ��4�3 � 0 = 1

2��1�2 � 4

3��7�3 =

1

2��� 4

33��7

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252 ¤ CHAPTER 3 DIFFERENTIATION RULES

4. = 3�� 2�2�+ 1

0 =

�2�+ 1 (3)� (3�� 2) 1

2(2�+ 1)�1�2(2)��

2�+ 1�2 · (2�+ 1)

1�2

(2�+ 1)1�2=3(2�+ 1)� (3�� 2)

(2�+ 1)3�2=

3�+ 5

(2�+ 1)3�2

5. = 2���2 + 1 �

0 = 2� · 12(�2 + 1)�1�2(2�) +

��2 + 1 (2) =

2�2��2 + 1

+ 2��2 + 1 =

2�2 + 2(�2 + 1)��2 + 1

=2(2�2 + 1)�

�2 + 1

6. = 0�

1 + �2� 0 =

(1 + �2)0� � 0�(2�)

(1 + �2)2=

0�(�2 � 2�+ 1)(1 + �2)2

=0�(�� 1)2(1 + �2)2

7. = 0sin 2� � 0 = 0sin 2��

��(sin 2�) = 0sin 2�(cos 2�)(2) = 2 cos 2� 0sin 2�

8. = 0��(�2 � 2�+ 2) �0 = 0��(2�� 2) + (�2 � 2�+ 2)(�0��) = 0��(2�� 2� �2 + 2�� 2) = 0��(��2 + 4�� 4)

9. = �

1� �2� 0 =

(1� �2)(1)� �(�2�)(1� �2)2

=1� �2 + 2�2

(1� �2)2=

�2 + 1

(1� �2)2

10. = 0�� cos(� �0 = 0��(cos(�)0 + cos(� (0��)0 = 0��(� sin(� · () + cos(� (0�� ·�) = 0��(� cos(�� ( sin(�)

11. = 01��

�2� 0 =

�2(01��)0 � 01����2�0

(�2)2=

�2(01��)(�1��2)� 01��(2�)

�4=�01��(1 + 2�)

�4

12. = (arcsin 2�)2 � 0 = 2(arcsin 2�) · (arcsin 2�)0 = 2arcsin 2� · 1�1� (2�)2 · 2 =

4 arcsin 2��1� 4�2

13. �

��(�4 + �2) =

��(�+ 3) � � · 430 + 4 · 1 + �2 · 0 + · 2� = 1 + 30 �

0(4�3 + �2 � 3) = 1� 4 � 2� � 0 =1� 4 � 2�4�3 + �2 � 3

14. = ln(csc 5�) � 0 =1

csc 5�(� csc 5� cot 5�)(5) = �5 cot 5�

15. = sec 2�

1 + tan 2��

0=(1 + tan 2�)(sec 2� tan 2� · 2)� (sec 2�)(sec2 2� · 2)

(1 + tan 2�)2=2 sec 2� [(1 + tan2�) tan 2� � sec2 2�]

(1 + tan 2�)2

=2 sec 2� (tan 2� + tan2 2� � sec2 2�)

(1 + tan 2�)2=2 sec 2� (tan 2� � 1)(1 + tan 2�)2

1 + tan2 � = sec2 �

16. �

��(�2 cos + sin 2) =

��(�) � �2(� sin · 0) + (cos )(2�) + cos 2 · 20 = � · 0 + · 1 �

0(��2 sin + 2cos 2 � �) = � 2� cos � 0 = � 2� cos

2 cos 2 � �2 sin � �

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CHAPTER 3 REVIEW ¤ 253

17. = 0��(+ sin�� cos�) �0 = 0��(+ cos�+ sin�) + +0��(+ sin�� cos�) = 0��(+2 sin�� + cos�+ + cos�+ sin�)

= 0��(+2 sin�+ sin�) = 0�� sin� (+2 + 1)

18. = ln(�20�) = ln�2 + ln 0� = 2 ln�+ � � 0 = 2��+ 1

19. = log5(1 + 2�) � 0 =1

(1 + 2�) ln 5

��(1 + 2�) =

2

(1 + 2�) ln 5

20. = (ln�)cos � � ln = cos� ln(ln�) � 0

= cos� · 1

ln�· 1�+ (ln ln�)(� sin�) �

0 = (ln�)cos �� cos�� ln�

� sin� ln ln��

21. sin(�) = �2 � � cos(�)(�0 + · 1) = 2�� 0 � � cos(�)0 + 0 = 2�� cos(�) �

0[� cos(�) + 1] = 2�� cos(�) � 0 =2�� cos(�)

� cos(�) + 1

22. =�� ln(�4) �

0 =1

2[� ln(�4)]�1�2

��[� ln(�4)] =

1

2�� ln(�4)

· 1 · ln(�4) + � · 1

�4· 4�3

!=

1

2�� ln(�4)

· [ln(�4) + 4] = ln(�4) + 4

2�� ln(�4)

Or: Since y is only de�ned for � 0, we can write =�� · 4 ln � = 2

�� ln �. Then

0 = 2 · 1

2�� ln �

·�1 · ln �+ � · 1

�=ln �+ 1�� ln �

. This agrees with our �rst answer since

ln(�4) + 4

2�� ln(�4)

=4 ln �+ 4

2�� · 4 ln � =

4(ln �+ 1)

2 · 2�� ln � =ln �+ 1�� ln �

.

23. = 3� ln � � 0 = 3� ln �(ln 3)�

��(� ln�) = 3� ln�(ln 3)

�� · 1

�+ ln� · 1

�= 3� ln �(ln 3)(1 + ln�)

24. �0� = � 1 � �0�0 + 0� = 0 � 0� = 0 � �0�0 � 0 = 0��(1� �0�)

25. = ln sin�� 12 sin

2 � � 0 =1

sin�· cos�� 1

2 · 2 sin� · cos� = cot�� sin� cos�

26. = (�2 + 1)4

(2�+ 1)3(3�� 1)5 �

ln = ln(�2 + 1)4

(2�+ 1)3(3�� 1)5 = ln(�2 + 1)4 � ln[(2�+ 1)3(3�� 1)5] = 4 ln(�2 + 1)� [ln(2�+ 1)3 + ln(3�� 1)5]

= 4 ln(�2 + 1)� 3 ln(2�+ 1)� 5 ln(3�� 1) �

0

= 4 · 1

�2 + 1· 2�� 3 · 1

2�+ 1· 2� 5 · 1

3�� 1 · 3 � 0 =(�2 + 1)4

(2�+ 1)3(3�� 1)5�

8�

�2 + 1� 6

2�+ 1� 15

3�� 1�

.

[The answer could be simpli�ed to 0 = � (�2 + 56�+ 9)(�2 + 1)3

(2�+ 1)4(3�� 1)6 , but this is unnecessary.]

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254 ¤ CHAPTER 3 DIFFERENTIATION RULES

27. = � tan�1(4�) � 0 = � · 1

1 + (4�)2· 4 + tan�1(4�) · 1 = 4�

1 + 16�2+ tan�1(4�)

28. = 0cos� + cos(0 �) � 0 = 0cos�(� sin�) + [� sin(0 �) · 0 �] = � sin�0cos � � 0 � sin(0 �)

29. = ln |sec 5�+ tan 5�| �

0 =1

sec 5�+ tan 5�(sec 5� tan 5� · 5 + sec2 5� · 5) = 5 sec 5� (tan 5�+ sec 5�)

sec 5�+ tan 5�= 5 sec 5�

30. = 10 tan�� � 0 = 10 tan�� · ln 10 · sec2 �� · � = �(ln 10)10 tan�� sec2 ��

31. = tan2(sin �) = [tan(sin �)]2 � 0 = 2[tan(sin �)] · sec2(sin �) · cos �

32. = ln�����2 � 42�+ 5

���� = ln ���2 � 4��� ln |2�+ 5| � 0 =2�

�2 � 4 �2

2�+ 5or 2(�+ 1)(�+ 4)

(�+ 2)(�� 2)(2�+ 5)

33. = sin�tan

�1 + �3

� � 0 = cos�tan

�1 + �3

��sec2

�1 + �3

�3�2

��2�1 + �3

�34. = arctan

�arcsin

���

� 0 =1

1 +�arcsin

���2 · 1�

1� �· 1

2��

35. = cos�0�tan 3�

��

0 = � sin�0�tan 3�

�·�0�tan 3�

�0= � sin

�0�tan 3�

�0�tan 3� · 12 (tan 3�)�1�2 · sec2(3�) · 3

=�3 sin

�0�tan 3�

�0�tan 3� sec2(3�)

2�tan 3�

36. = sin2�cos

�sin��

�=�sin

�cos

�sin��

��2 �

0 = 2�sin

�cos

�sin��

���sin

�cos

�sin��

��0= 2 sin

�cos

�sin��

�cos

�cos

�sin��

��cos

�sin��

�0= 2 sin

�cos

�sin��

�cos

�cos

�sin��

��� sin�sin��

���sin��

�0= �2 sin

�cos

�sin��

�cos

�cos

�sin��

�sin�sin�� · 1

2(sin��)�1�2(sin��)0

=� sin

�cos

�sin��

�cos

�cos

�sin��

�sin�sin��

�sin��

· cos�� · �

=�� sin

�cos

�sin��

�cos

�cos

�sin��

�sin�sin�� cos��

�sin��

37. �(�) =�4�+ 1 � � 0(�) = 1

2(4� + 1)�1�2 · 4 = 2(4� + 1)�1�2 �

� 00(�) = 2(� 12 )(4�+ 1)

�3�2 · 4 = �4�(4�+ 1)3�2, so � 00(2) = �4�93�2 = � 427 .

38. �(�) = � sin � � �0(�) = � cos � + sin � · 1 � �00(�) = �(� sin �) + cos � · 1 + cos � = 2 cos � � � sin �,

so �00(��6) = 2 cos(��6)� (��6) sin(��6) = 2 ��3�2�� (��6)(1�2) = �3� ��12.

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CHAPTER 3 REVIEW ¤ 255

39. �(�) = 2� � � 0(�) = 2� ln 2 � � 00(�) = (2� ln 2) ln 2 = 2�(ln 2)2 �� 000(�) = (2� ln 2)(ln 2)2 = 2�(ln 2)3 � · · · � � ()(�) = (2� ln 2)(ln 2)�1 = 2�(ln 2)

40. �6 + 6 = 1 � 6�5 + 650 = 0 � 0 = ��5�5 �

00 = �5(5�4)� �5(540)

(5)2= �5�

44 � �(��5�5)10

= �5�4(6 + �6)�5

6

= �5�4

11

41. = 4 sin2 � � 0 = 4 · 2 sin� cos�. At��6� 1�, 0 = 8 · 1

2·�32= 2

�3, so an equation of the tangent line

is � 1 = 2�3 ��� �6

�, or = 2

�3�+ 1� �

�3�3.

42. = �2 � 1�2 + 1

� 0 =(�2 + 1)(2�)� (�2 � 1)(2�)

(�2 + 1)2=

4�

(�2 + 1)2.

At (0��1), 0 = 0, so an equation of the tangent line is + 1 = 0(�� 0), or = �1.

43. � = ln �, = �2 + 1, (0� 2). If (�� ) = (0� 2), then � = ln � = 0 � � = 1.

��= 2�, ��

��=1

�, and �

��=

����

�����=2�

1��= 2�2, so �

��= 2 when � = 1.

An equation of the tangent to the curve at (0� 2) is � 2 = 2(�� 0), or = 2�+ 2.

44. � = �3 � 2�2 + �+ 1, = �2 + �, (1� 0). If (�� ) = (1� 0), then = �2 + � = 0 � �(�+ 1) = 0 �� = 0 or � = �1, but it is easy to see that � = 0 satis�es the equation � = �3 � 2�2 + �+ 1 = 1.

��= 2�+ 1, ��

��= 3�2 � 4�+ 1, and �

��=

����

�����=

2�+ 1

3�2 � 4�+ 1 , so �

��=

0 + 1

0� 0 + 1 = 1 when � = 0.

An equation of the tangent to the curve at (1� 0) is � 0 = 1(�� 1), or = �� 1.

45. = (2 + �)0�� � 0 = (2 + �)(�0��) + 0�� · 1 = 0��[�(2 + �) + 1] = 0��(��� 1).At (0� 2), 0 = 1(�1) = �1, so an equation of the tangent line is � 2 = �1(�� 0), or = ��+ 2.

The slope of the normal line is 1, so an equation of the normal line is � 2 = 1(�� 0), or = �+ 2.

46. �2 + 4� + 2 = 13 � 2�+ 4(�0 + · 1) + 20 = 0 � �+ 2�0 + 2 + 0 = 0 �

2�0 + 0 = ��� 2 � 0(2�+ ) = ��� 2 � 0 =��� 22�+

.

At (2� 1), 0 = �2� 24 + 1

= �45

, so an equation of the tangent line is � 1 = � 45 (�� 2), or = � 4

5�+135 .

The slope of the normal line is 54

, so an equation of the normal line is � 1 = 54(�� 2), or = 5

4�� 3

2.

47. (a) �(�) = ��5� � �

� 0(�) = �

1

2(5� �)�1�2(�1)

!+�5� � =

��2�5� �

+�5� � · 2

�5� �

2�5� �

=��

2�5� �

+2(5� �)

2�5� �

=��+ 10� 2�2�5� �

=10� 3�2�5� �

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256 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b) At (1� 2): � 0(1) = 74

.

So an equation of the tangent line is � 2 = 74 (�� 1) or = 7

4�+14 .

At (4� 4): � 0(4) = � 22= �1.

So an equation of the tangent line is � 4 = �1(�� 4) or = ��+ 8.

(c)

(d) The graphs look reasonable, since � 0 is positive where � has tangents with

positive slope, and � 0 is negative where � has tangents with negative slope.

48. (a) �(�) = 4�� tan� � � 0(�) = 4� sec2 � � � 00(�) = �2 sec� (sec� tan�) = �2 sec2 � tan�.

(b) We can see that our answers are reasonable, since the graph of � 0 is 0 where �

has a horizontal tangent, and the graph of � 0 is positive where � has tangents

with positive slope and negative where � has tangents with negative slope. The

same correspondence holds between the graphs of � 0 and � 00.

49. �(�) = �0sin � � � 0(�) = �[0sin�(cos�)] + 0sin�(1) = 0sin �(� cos�+ 1). As a check on our work, we notice from the

graphs that � 0(�) 0 when � is increasing. Also, we see in the larger viewing rectangle a certain similarity in the graphs of �

and � 0: the sizes of the oscillations of � and � 0 are linked.

50. (a) (b) The average rate of change is larger on [2� 3].

(c) The instantaneous rate of change (the slope of the tangent) is larger at � = 2.

(d) �(�) = �� 2 sin� � � 0(�) = 1� 2 cos�,

so � 0(2) = 1� 2 cos 2 1�8323 and � 0(5) = 1� 2 cos 5 0�4327.

So � 0(2) � 0(5), as predicted in part (c).

51. (a) �(�) = �(�) �(�) � �0(�) = �(�) �0(�) + �(�) � 0(�) ��0(2) = �(2) �0(2) + �(2) � 0(2) = (3)(4) + (5)(�2) = 12� 10 = 2

(b) ! (�) = �(�(�)) � ! 0(�) = � 0(�(�)) �0(�) � ! 0(2) = � 0(�(2)) �0(2) = � 0(5)(4) = 11 · 4 = 44

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CHAPTER 3 REVIEW ¤ 257

52. (a) � (�) = �(�) �(�) � � 0(�) = �(�) �0(�) + �(�) � 0(�) �

� 0(2) = �(2) �0(2) + �(2) � 0(2) = (1)�6�03�0

�+ (4)

�0�33�0

�= (1)(2) + (4)(�1) = 2� 4 = �2

(b) /(�) = �(�)

�(�)� /0(�) =

�(�) � 0(�)� �(�) �0(�)[�(�)]2

/0(2) =�(2) � 0(2)� �(2) �0(2)

[�(2)]2=(4)(�1)� (1)(2)

42=�616

= �38

(c) �(�) = �(�(�)) � �0(�) = � 0(�(�))�0(�) �

�0(2) = � 0(�(2))�0(2) = � 0(4)�0(2) =�6�05�3

�(2) = (3)(2) = 6

53. �(�) = �2�(�) � � 0(�) = �2�0(�) + �(�)(2�) = �[��0(�) + 2�(�)]

54. �(�) = �(�2) � � 0(�) = �0(�2)(2�) = 2��0(�2)

55. �(�) = [ �(�)]2 � � 0(�) = 2[ �(�)] · �0(�) = 2�(�) �0(�)

56. �(�) = �(�(�)) � � 0(�) = �0(�(�)) �0(�)

57. �(�) = �(0�) � � 0(�) = �0(0�) 0�

58. �(�) = 0�(�) � � 0(�) = 0�(�)�0(�)

59. �(�) = ln |�(�)| � � 0(�) =1

�(�)�0(�) =

�0(�)�(�)

60. �(�) = �(ln�) � � 0(�) = �0(ln�) · 1�=

�0(ln�)�

61. �(�) = �(�) �(�)

�(�) + �(�)�

�0(�) =[�(�) + �(�)] [�(�) �0(�) + �(�) � 0(�)]� �(�) �(�) [� 0(�) + �0(�)]

[�(�) + �(�)]2

=[�(�)]2 �0(�) + �(�) �(�) � 0(�) + �(�) �(�) �0(�) + [ �(�)]2 � 0(�)� �(�) �(�) � 0(�)� �(�) �(�) �0(�)

[�(�) + �(�)]2

=� 0(�) [ �(�)]2 + �0(�) [�(�)]2

[�(�) + �(�)]2

62. Using the Chain Rule repeatedly, �(�) = �(�(sin 4�)) �

�0(�) = � 0(�(sin 4�)) · ���(�(sin 4�)) = � 0(�(sin 4�)) · �0(sin 4�) · �

��(sin 4�) = � 0(�(sin 4�))�0(sin 4�)(cos 4�)(4).

63. = [ln(�+ 4)]2 � 0 = 2[ln(�+ 4)]1 · 1

�+ 4· 1 = 2 ln(�+ 4)

�+ 4and 0 = 0 � ln(�+ 4) = 0 �

�+ 4 = 00 � �+ 4 = 1 � � = �3, so the tangent is horizontal at the point (�3� 0).

64. (a) The line �� 4 = 1 has slope 14

. A tangent to = 0� has slope 14

when 0 = 0� = 14

� � = ln 14= � ln 4.

Since = 0�, the -coordinate is 14

and the point of tangency is�� ln 4� 1

4

�. Thus, an equation of the tangent line

is � 14 =

14 (�+ ln 4) or = 1

4�+14 (ln 4 + 1).

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258 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b) The slope of the tangent at the point (�� 0�) is �

��0������= �

= 0�. Thus, an equation of the tangent line is

� 0� = 0�(�� �). We substitute � = 0, = 0 into this equation, since we want the line to pass through the origin:

0� 0� = 0�(0� �) � �0� = 0�(��) � � = 1. So an equation of the tangent line at the point (�� 0�) = (1� 0)

is � 0 = 0(�� 1) or = 0�.

65. �2 + 22 = 1 � 2�+ 40 = 0 � 0 = ���(2) = 1 � � = �2. Since the points lie on the ellipse,

we have (�2)2 + 22 = 1 � 62 = 1 � = ± 1�6

. The points are�� 2�

6� 1�

6

�and

�2�6�� 1�

6

�.

66. (a) �(�) = ln�

�� � 0(�) =

�(1��)� (ln�) · 1�2

=1� ln��2

. � 0(�) 0 � 1� ln� 0 �ln� � 1 � � � 0. Since the domain of � is (0��), � is increasing on (0� 0).

(b) � 00(�) = �2(�1��)� (1� ln�)(2�)(�2)2

=�[�1� 2(1� ln�)]

�4=2 ln�� 3

�3.

� 00(�) 0 � 2 ln�� 3 0 � ln� 32

� � 03�2 4�48. � is concave upward on (03�2��).

67. = �(�) = ��2 + �+ + � � 0(�) = 2��+ . We know that � 0(�1) = 6 and � 0(5) = �2, so �2�+ = 6 and

10�+ = �2. Subtracting the �rst equation from the second gives 12� = �8 � � = � 23

. Substituting � 23

for � in the

�rst equation gives = 143

. Now �(1) = 4 � 4 = �+ + +, so + = 4 + 23� 14

3= 0 and hence, �(�) = �2

3�2 + 14

3�.

68. (a) = �3 � 12�+ 3 � )(�) = 0 = 3�2 � 12 � �(�) = )0(�) = 6�

(b) )(�) = 3(�2 � 4) 0 when � 2, so it moves upward when � 2 and downward when 0 � � � 2.

(c) Distance upward = (3)� (2) = �6� (�13) = 7,

Distance downward = (0)� (2) = 3� (�13) = 16. Total distance = 7 + 16 = 23.

(d) (e) The particle is speeding up when ) and � have the same sign, that is,

when � 2. The particle is slowing down when ) and � have opposite

signs; that is, when 0 � � � 2.

69. �(�) = �0��� cos(<�+ =) �)(�) = �0(�) = �{0��� [�< sin(<�+ =)] + cos(<�+ =)(�+0���)} = ��0��� [< sin(<�+ =) + + cos(<�+ =)] ��(�) = )0(�) = ��{0���[<2 cos(<�+ =)� +< sin(<�+ =)] + [< sin(<�+ =) + + cos(<�+ =)](�+0���)}

= ��0���[<2 cos(<�+ =)� +< sin(<�+ =)� +< sin(<�+ =)� +2 cos(<�+ =)]

= ��0���[(<2 � +2) cos(<�+ =)� 2+< sin(<�+ =)] = �0���[(+2 � <2) cos(<�+ =) + 2+< sin(<�+ =)]

70. (a) � =�2 + +2 �2 � )(�) = �0 =

1��2�2 + +2 �2

�2+2� = +2��

�2 + +2 �2 �

�(�) = )0(�) =+2�2 + +2 �2 � +2�

�+2��

�2 + +2 �2

�2 + +2 �2

=2+2

(2 + +2 �2)3�2

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CHAPTER 3 REVIEW ¤ 259

(b) )(�) 0 for � 0, so the particle always moves in the positive direction.

71. The linear density @ is the rate of change of mass � with respect to length �.

� = ��1 +

���= �+ �3�2 � @ = ����� = 1 + 3

2

��, so the linear density when � = 4 is 1 + 3

2

�4 = 4 kg�m.

72. (a) = 13��2� � � ��� = 1

3��2 [� constant]

(b) = 13��

2� � � ��� = 23��� [� constant]

73. (a) �(�) = 920 + 2�� 0�02�2 + 0�00007�3 � � 0(�) = 2� 0�04�+ 0�00021�2

(b) � 0(100) = 2� 4 + 2�1 = $0�10�unit. This value represents the rate at which costs are increasing as the hundredth unit is

produced, and is the approximate cost of producing the 101st unit.

(c) The cost of producing the 101st item is �(101)��(100) = 990�10107� 990 = $0�10107, slightly larger than � 0(100).

(d) � 00(�) = �0�04 + 0�00042� = 0 � � = 004000042

95�24 and � 00 changes from negative to positive at this value

of �. This is the value of � at which the marginal cost is minimized.

74. (a) lim���

�(�) = lim���

[B(0��� � 0� �)] = B lim���

(0��� � 0� �) = B(0� 0) = 0 because ��� �� and �� ��

as � �.

(b) �(�) = B(0��� � 0� �) � � 0(�) = B(0���(��)� 0� �(�)) = B(��0��� + 0� �)

(c) � 0(�) = 0 � 0� � = �0��� �

�= 0(��+ )� � ln

�= (� �)� � � =

ln(��)

� �

75. (a) �(�) = 3�1 + 3� = (1 + 3�)1�3 � � 0(�) = (1 + 3�)�2�3, so the linearization of � at � = 0 is

$(�) = �(0) + � 0(0)(�� 0) = 11�3 + 1�2�3� = 1 + �. Thus, 3�1 + 3� 1 + � �

3�1�03 = 3

�1 + 3(0�01) 1 + (0�01) = 1�01.

(b) The linear approximation is 3�1 + 3� 1 + �, so for the required accuracy

we want 3�1 + 3�� 0�1 � 1 + � � 3

�1 + 3�+ 0�1. From the graph,

it appears that this is true when �0�23 � � � 0�40.

76. � = �2 + 12��12��2=�1 + �

8

��2 � �� =

�2 + �

4

����. When � = 60

and �� = 0�1, �� =�2 + �

4

�60(0�1) = 12 + 3�

2, so the maximum error is

approximately 12 + 3�2 16�7 cm2.

77. lim����3

cos � � 0�5� � ��3

=

��cos �

!�=��3

= � sin �3= �

�3

2

78. �

��[�(2�)] = �2 � � 0(2�) · 2 = �2 � � 0(2�) = 1

2�2. Let � = 2�. Then � 0(�) = 1

2

�12��2= 1

8�2, so � 0(�) = 1

8�2.

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260 ¤ CHAPTER 3 DIFFERENTIATION RULES

79. lim��0

�1 + tan���1 + sin�

�3= lim

��0

��1 + tan���1 + sin� ���1 + tan�+�1 + sin� �

�3��1 + tan�+

�1 + sin�

�= lim

��0

(1 + tan�)� (1 + sin�)�3��1 + tan�+

�1 + sin�

� = lim��0

sin� (1� cos�� 1)�3��1 + tan�+

�1 + sin�

� · cos�cos�

= lim��0

sin� (1� cos�)�3��1 + tan�+

�1 + sin�

�cos�

· 1 + cos�1 + cos�

= lim��0

sin� · sin2 ��3��1 + tan�+

�1 + sin�

�cos� (1 + cos�)

=

�lim��0

sin�

�3lim��0

1��1 + tan�+

�1 + sin�

�cos� (1 + cos�)

= 13 · 1��1 +

�1� · 1 · (1 + 1) = 1

4

80. Let (� +) be on the curve, that is, 2�3 + +2�3 = �2�3. Now �2�3 + 2�3 = �2�3 � 23��1�3 + 2

3�1�3

��= 0, so

��= �1�3

�1�3= �

��

�1�3, so at (� +) the slope of the tangent line is �(+�)1�3 and an equation of the tangent line is

� + = �(+�)1�3(�� ) or = �(+�)1�3�+ (++ 2�3+1�3). Setting = 0, we �nd that the �-intercept is

1�3+2�3 + = 1�3(+2�3 + 2�3) = 1�3�2�3 and setting � = 0 we �nd that the -intercept is

++ 2�3+1�3 = +1�3(+2�3 + 2�3) = +1�3�2�3. So the length of the tangent line between these two points is

�(1�3�2�3)2 + (+1�3�2�3)2 =

�2�3�4�3 + +2�3�4�3 =

�(2�3 + +2�3)�4�3

=��2�3�4�3 =

��2 = � = constant

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FOCUS ON PROBLEM SOLVING1. We must �nd a value �0 such that the normal lines to the parabola = �2 at � = ±�0 intersect at a point one unit from the

points�±�0� �20�. The normals to = �2 at � = ±�0 have slopes � 1

±2�0 and pass through�±�0� �20� respectively, so the

normals have the equations � �20 = � 1

2�0(�� �0) and � �20 =

1

2�0(�+ �0). The common -intercept is �20 +

1

2.

We want to �nd the value of �0 for which the distance from�0� �20 +

12

�to��0� �

20

�equals 1. The square of the distance is

(�0 � 0)2 +�20 �

��20 +

12

�2= �20 +

14= 1 � �0 = ±

�32

. For these values of �0, the -intercept is �20 + 12= 5

4, so

the center of the circle is at�0� 5

4

�.

Another solution: Let the center of the circle be (0� �). Then the equation of the circle is �2 + ( � �)2 = 1.

Solving with the equation of the parabola, = �2, we get �2 + (�2 � �)2 = 1 � �2 + �4 � 2��2 + �2 = 1 ��4 + (1� 2�)�2 + �2 � 1 = 0. The parabola and the circle will be tangent to each other when this quadratic equation in �2

has equal roots; that is, when the discriminant is 0. Thus, (1� 2�)2 � 4(�2 � 1) = 0 �1� 4�+ 4�2 � 4�2 + 4 = 0 � 4� = 5, so � = 5

4. The center of the circle is

�0� 5

4

�.

2. = �3 � 3�+ 4 � 0 = 3�2 � 3, and = 3(�2 � �) � 0 = 6�� 3.

The slopes of the tangents of the two curves are equal when 3�2 � 3 = 6�� 3;

that is, when � = 0 or 2. At � = 0, both tangents have slope �3, but the curves do

not intersect. At � = 2, both tangents have slope 9 and the curves intersect at

(2� 6). So there is a common tangent line at (2� 6), = 9�� 12.

3. We must show that � (in the �gure) is halfway between � and �, that is,

� = (�+ �)�2. For the parabola = ��2 + �+ +, the slope of the tangent line is

given by 0 = 2��+ . An equation of the tangent line at � = � is

� (��2 + �+ +) = (2��+ )(�� �). Solving for gives us

= (2��+ )�� 2��2 � �+ (��2 + �+ +)

or = (2��+ )�+ +� ��2 (1)

Similarly, an equation of the tangent line at � = � is = (2�� + )�+ +� ��2 (2)

We can eliminate and solve for � by subtracting equation (1) from equation (2).

[(2�� + )� (2��+ )]�� ��2 + ��2 = 0

(2�� � 2��)�= ��2 � ��2

2�(� � �)�= �(�2 � �2)

�=�(� + �)(� � �)

2�(� � �)=

�+ �

2

Thus, the �-coordinate of the point of intersection of the two tangent lines, namely �, is (�+ �)�2.

261

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262 ¤ CHAPTER 3 FOCUS ON PROBLEM SOLVING

4. We could differentiate and then simplify or we can simplify and then differentiate. The latter seems to be the simpler method.

sin2 �

1 + cot�+

cos2 �

1 + tan�=

sin2 �

1 +cos�

sin�

· sin�sin�

+cos2 �

1 +sin�

cos�

· cos�cos�

=sin3 �

sin�+ cos�+

cos3 �

cos�+ sin�

=sin3 �+ cos3 �

sin�+ cos�[factor sum of cubes] =

(sin�+ cos�)(sin2 �� sin� cos�+ cos2 �)sin�+ cos�

= sin2 �� sin� cos�+ cos2 � = 1� sin� cos� = 1� 12 (2 sin� cos�) = 1� 1

2 sin 2�

Thus, �

��

�sin2 �

1 + cot�+

cos2 �

1 + tan�

�=

��

�1� 1

2sin 2�

�= � 1

2cos 2� · 2 = � cos 2�.

5. Using � 0(�) = lim���

�(�)� �(�)

�� �, we recognize the given expression, �(�) = lim

���

sec �� sec��� �

, as

�0(�) with �(�) = sec�. Now � 0(�4) = �00(�

4), so we will �nd �00(�).

�0(�) = sec� tan� � �00(�) = sec� sec2 �+ tan� sec� tan� = sec�(sec2 �+ tan2 �), so

�00(�4) =

�2(�22+ 12) =

�2(2 + 1) = 3

�2.

6. lim���

�(�)� �(�)���

��

= lim���

&�(�)� �(�)�

����·��+

���

�+��

'= lim

���

�(�)� �(�)

�� �·��

�+���!

= lim���

�(�)� �(�)

�� �· lim���

���+

���= � 0(�) ·

���+

���= 2

�� � 0(�)

7. We can assume without loss of generality that � = 0 at time � = 0, so that � = 12�� rad. [The angular velocity of the wheel

is 360 rpm = 360 · (2� rad)�(60 s) = 12� rad�s.] Then the position of � as a function of time is

� = (40 cos �� 40 sin �) = (40 cos 12��� 40 sin 12��), so sin1 =

1�2 m=40 sin �

120=sin �

3=1

3sin 12��.

(a) Differentiating the expression for sin1, we get cos1 · �1��=1

3· 12� · cos 12�� = 4� cos �. When � = �

3, we have

sin1 =1

3sin � =

�3

6, so cos1 =

%1�

��3

6

�2=

�11

12and �1

��=4� cos �

3

cos1=

2��11�12

=4��3�

11 6�56 rad�s.

(b) By the Law of Cosines, |�� |2 = |2�|2 + |2� |2 � 2 |2�| |2� | cos � �

1202 = 402 + |2� |2 � 2 · 40 |2� | cos � � |2� |2 � (80 cos �) |2� | � 12,800 = 0 �

|2� | = 12

�80 cos � ±�6400 cos2 � + 51,200

�= 40 cos � ± 40�cos2 � + 8 = 40�cos � +�8 + cos2 � � cm

[since |2� | 0]. As a check, note that |2� | = 160 cm when � = 0 and |2� | = 80�2 cm when � = �2 .

(c) By part (b), the �-coordinate of � is given by � = 40�cos � +

�8 + cos2 �

�, so

��

��=

��

��

��

��= 40

�� sin � � 2 cos � sin �

2�8 + cos2 �

�· 12� = �480� sin �

�1 +

cos ��8 + cos2 �

�cm�s.

In particular, ����� = 0 cm�s when � = 0 and ����� = �480� cm�s when � = �2

.

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CHAPTER 3 FOCUS ON PROBLEM SOLVING ¤ 263

8. The equation of �1 is � �21 = 2�1(�� �1) = 2�1�� 2�21 or = 2�1�� �21.

The equation of �2 is = 2�2�� �22. Solving for the point of intersection, we

get 2�(�1 � �2) = �21 � �22 � � = 12 (�1 + �2). Therefore, the coordinates

of � are�12 (�1 + �2)� �1�2

�. So if the point of contact of � is

��� �2

�, then

/1 is�12(�+ �1)� ��1

�and /2 is

�12(�+ �2)� ��2

�. Therefore,

|�/1|2 = 14(�� �2)

2 + �21(�� �2)2 = (�� �2)

2� 14+ �21

�and

|��1|2 = 14(�1 � �2)

2 + �21(�1 � �2)2 = (�1 � �2)

2� 14+ �21

�.

So |�/1|2|��1|2

=(�� �2)

2

(�1 � �2)2 , and similarly |�/2|2

|��2|2=(�1 � �)2

(�1 � �2)2 . Finally, |�/1|

|��1| +|�/2||��2| =

�� �2�1 � �2

+�1 � �

�1 � �2= 1.

9. Consider the statement that �

��(0�� sin �) = �0�� sin(� + (�). For ( = 1,

��(0�� sin �) = �0�� sin �+ 0�� cos �, and

�0�� sin(�+ �) = �0��[sin � cos � + cos � sin �] = �0����

�sin �+

�cos �

�= �0�� sin �+ 0�� cos �

since tan � =

�� sin � =

�and cos � = �

�. So the statement is true for ( = 1.

Assume it is true for ( = �. Then

��+1

���+1(0�� sin �) =

��

��0�� sin(�+ ��)

= ���0�� sin(�+ ��) + ��0�� cos(�+ ��)

= ��0��[� sin(�+ ��) + cos(�+ ��)]

But

sin[�+ (� + 1)�] = sin[(�+ ��) + �] = sin(�+ ��) cos � + sin � cos(�+ ��)

= �� sin(�+ ��) +

� cos(�+ ��)�

Hence, � sin(�+ ��) + cos(�+ ��) = � sin[�+ (� + 1)�]. So

��+1

���+1(0�� sin �) = ��0��[� sin(�+��)+ cos(�+��)] = ��0��[� sin(�+(�+1)�)] = ��+10��[sin(�+(�+1)�)].

Therefore, the statement is true for all ( by mathematical induction.

10. Using � 0(0) = lim��0

�(�)� �(0)

�� 0 , we see that for the given equation, lim��0

3���+ � 2

�=5

12, we have �(�) = 3

���+ ,

�(0) = 2, and � 0(0) = 512

. Now �(0) = 2 � 3� = 2 � = 8. Also � 0(�) = 1

3(��+ )�2�3 · �, so

� 0(0) = 512

� 13(8)�2�3 · � = 5

12� 1

3( 14)� = 5

12� � = 5.

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264 ¤ CHAPTER 3 FOCUS ON PROBLEM SOLVING

11. It seems from the �gure that as � approaches the point (0� 2) from the right, �! � and ! 2+. As � approaches the

point (3� 0) from the left, it appears that �! 3+ and ! �. So we guess that �! � (3��) and ! � (2��). It is

more dif�cult to estimate the range of values for �" and " . We might perhaps guess that �" � (0� 3),and " � (��� 0) or (�2� 0).

In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to �nd the equation of the

tangent line: �2

9+2

4= 1 � 2�

9+2

40 = 0, so 0 = �4

9

. So at the point (�0� 0) on the ellipse, an equation of the

tangent line is � 0 = �49

�00(�� �0) or 4�0�+ 90 = 4�20 + 920 . This can be written as �0�

9+0

4=

�209+204= 1,

because (�0� 0) lies on the ellipse. So an equation of the tangent line is �0�9+0

4= 1.

Therefore, the �-intercept �! for the tangent line is given by �0�!9

= 1 � �! =9

�0, and the -intercept ! is given

by 0!4

= 1 � ! =4

0.

So as �0 takes on all values in (0� 3), �! takes on all values in (3��), and as 0 takes on all values in (0� 2), ! takes on

all values in (2��). At the point (�0� 0) on the ellipse, the slope of the normal line is � 1

0(�0� 0)=9

4

0�0

, and its

equation is � 0 =9

4

0�0(�� �0). So the �-intercept �" for the normal line is given by 0� 0 =

9

4

0�0(�" � �0) �

�" = �4�09+ �0 =

5�09

, and the -intercept " is given by " � 0 =9

4

0�0(0� �0) � " = �90

4+ 0 = �50

4.

So as �0 takes on all values in (0� 3), �" takes on all values in�0� 5

3

�, and as 0 takes on all values in (0� 2), " takes on

all values in�� 5

2� 0�.

12. lim��0

�(�)

�(�)= lim

��0

�(�)� 0�(�)� 0 = lim

��0

�(�)� �(0)

�(�)� �(0)= lim

��0

�(�)� �(0)

�� 0�(�)� �(0)

�� 0=lim��0

�(�)� �(0)

�� 0lim��0

�(�)� �(0)

�� 0=

� 0(0)�0(0)

13. = ���2 � 1 �

2��2 � 1 arctan

sin�

�+��2 � 1 + cos� . Let � = �+

��2 � 1. Then

0 =1�

�2 � 1 �2�

�2 � 1 ·1

1 + sin2 ��(� + cos�)2· cos�(� + cos�) + sin

2 �

(� + cos�)2

=1�

�2 � 1 �2�

�2 � 1 ·� cos�+ cos2 �+ sin2 �

(� + cos�)2 + sin2 �=

1��2 � 1 �

2��2 � 1 ·

� cos�+ 1

�2 + 2� cos�+ 1

=�2 + 2� cos�+ 1� 2� cos�� 2�

�2 � 1 (�2 + 2� cos�+ 1) =�2 � 1�

�2 � 1 (�2 + 2� cos�+ 1)

But �2 = 2�2 + 2���2 � 1� 1 = 2���+��2 � 1 �� 1 = 2�� � 1, so �2 + 1 = 2��, and �2 � 1 = 2(�� � 1).

So 0 = 2(�� � 1)��2 � 1 (2�� + 2� cos�) =

�� � 1��2 � 1� (�+ cos�) . But �� � 1 = �2 + �

��2 � 1� 1 = �

��2 � 1,

so 0 = 1�(�+ cos�).

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CHAPTER 3 FOCUS ON PROBLEM SOLVING ¤ 265

14. We see that at � = 0, �(�) = �� = 1 + � = 1, so if = �� is to lie above = 1 + �,

the two curves must just touch at (0� 1), that is, we must have � 0(0) = 1. [To see this

analytically, note that �� � 1 + � � �� � 1 � � � �� � 1�

� 1 for � 0, so

� 0(0) = lim��0+

�� � 1�

� 1. Similarly, for � � 0, �� � 1 � � � �� � 1�

� 1, so

� 0(0) = lim��0�

�� � 1�

� 1.

Since 1 � � 0(0) � 1, we must have � 0(0) = 1.] But � 0(�) = �� ln � � � 0(0) = ln�, so we have ln � = 1 � � = 0.

Another method: The inequality certainly holds for � � �1, so consider � �1, � 6= 0. Then �� � 1 + � �� � (1 + �)1�� for � 0 � � � lim

��0+(1 + �)1�� = 0, by Equation 3.7.5. Also, �� � 1 + � � � � (1 + �)1��

for � � 0 � � � lim��0�

(1 + �)1�� = 0. So since 0 � � � 0, we must have � = 0.

15. Let �(�) = 02� and �(�) = ��� [� 0]. From the graphs of � and �,

we see that � will intersect � exactly once when � and � share a tangent

line. Thus, we must have � = � and � 0 = �0 at � = �.

�(�) = �(�) � 02� = ��� (3)

and � 0(�) = �0(�) � 202� =�

2��

� 02� =�

4��

.

So we must have ��� =

4��

���

��2=

4�� � = 1

4. From (3), 02(1�4) = �

�1�4 �

� = 201�2 = 2�0 3�297.

16. (a) �(�) = �(�� 2)(�� 6) = �3 � 8�2 + 12� �

� 0(�) = 3�2 � 16�+ 12. The average of the �rst pair of zeros is

(0 + 2)�2 = 1. At � = 1, the slope of the tangent line is � 0(1) = �1, so an

equation of the tangent line has the form = �1�+ . Since �(1) = 5, we

have 5 = �1 + � = 6 and the tangent has equation = ��+ 6.

Similarly, at � = 0 + 6

2= 3, = �9�+ 18; at � = 2 + 6

2= 4, = �4�. From the graph, we see that each tangent line

drawn at the average of two zeros intersects the graph of � at the third zero.

(b) A CAS gives � 0(�) = (�� )(�� +) + (�� �)(�� +) + (�� �)(�� ) or

� 0(�) = 3�2 � 2(�+ + +)�+ �+ �++ +. Using the Simplify command, we get

� 0��+

2

�= � (�� )2

4and �

��+

2

�= � (�� )2

8(�+ � 2+), so an equation of the tangent line at � = �+

2is

= � (�� )2

4

��� �+

2

�� (�� )2

8(�+ � 2+). To �nd the �-intercept, let = 0 and use the Solve command.

The result is � = +.[continued]

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266 ¤ CHAPTER 3 FOCUS ON PROBLEM SOLVING

Using Derive, we can begin by authoring the expression (�� �)(�� )(�� +). Now load the utility �le

DifferentiationApplications. Next we author tangent (#1� �� (�+ )�2)—this is the command to �nd an

equation of the tangent line of the function in #1 whose independent variable is � at the �-value (�+ )�2. We then

simplify that expression and obtain the equation = #4. The form in expression #4 makes it easy to see that the

�-intercept is the third zero, namely +. In a similar fashion we see that is the �-intercept for the tangent line at (�+ +)�2

and � is the �-intercept for the tangent line at (+ +)�2.

17. (a) If the two lines $1 and $2 have slopes �1 and �2 and angles of

inclination 61 and 62, then �1 = tan 61 and �2 = tan 62. The triangle

in the �gure shows that 61 + 1+ (180� � 62) = 180� and so

1 = 62 � 61. Therefore, using the identity for tan(�� ), we have

tan1 = tan(62 � 61) =tan62 � tan611 + tan62 tan61

and so tan 1 = �2 ��1

1 +�1�2.

(b) (i) The parabolas intersect when �2 = (�� 2)2 � � = 1. If = �2, then 0 = 2�, so the slope of the tangent

to = �2 at (1� 1) is �1 = 2(1) = 2. If = (�� 2)2, then 0 = 2(�� 2), so the slope of the tangent to

= (�� 2)2 at (1� 1) is �2 = 2(1� 2) = �2. Therefore, tan1 = �2 ��1

1 +�1�2=

�2� 21 + 2(�2) =

4

3and

so 1 = tan�1�43

� 53� [or 127�].

(ii) �2 � 2 = 3 and �2 � 4�+ 2 + 3 = 0 intersect when �2 � 4�+ (�2 � 3) + 3 = 0 � 2�(�� 2) = 0 �

� = 0 or 2, but 0 is extraneous. If � = 2, then = ±1. If �2 � 2 = 3 then 2�� 20 = 0 � 0 = �� and

�2 � 4�+ 2 + 3 = 0 � 2�� 4 + 20 = 0 � 0 =2� �

. At (2� 1) the slopes are �1 = 2 and

�2 = 0, so tan1 = 0� 21+ 2 · 0 = �2 � 1 117�. At (2��1) the slopes are �1 = �2 and �2 = 0�

so tan1 = 0� (�2)1 + (�2) (0) = 2 � 1 63� [or 117�].

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CHAPTER 3 FOCUS ON PROBLEM SOLVING ¤ 267

18. 2 = 4�� � 20 = 4� � 0 = 2�� � slope of tangent at � (�1� 1) is �1 = 2��1. The slope of !� is

�2 =1

�1 � �, so by the formula from Problem 17(a),

tan1 =

1�1 � �

� 2�

1

1 +

�2�

1

��1

�1 � �

� · 1 (�1 � �)

1 (�1 � �)=

21 � 2�(�1 � �)

1(�1 � �) + 2�1

=4��1 � 2��1 + 2�2�11 � �1 + 2�1

=2�(�+ �1)

1(�+ �1)=2�

1

= slope of tangent at � = tan;

Since 0 � 1, ; � �2

, this proves that 1 = ;.

19. Since �.2/ = �2/� = �, the triangle /2. is isosceles, so

|/.| = |.2| = �. By the Law of Cosines, �2 = �2 + �2 � 2�� cos �. Hence,

2�� cos � = �2, so � = �2

2� cos �=

2 cos �. Note that as 0+, � 0+ (since

sin � = ��), and hence � �

2 cos 0=

2. Thus, as � is taken closer and closer

to the �-axis, the point . approaches the midpoint of the radius �2.

20. Suppose that = ��+ + is a tangent line to the ellipse. Then it intersects the ellipse at only one point, so the discriminant

of the equation �2

�2+(��+ +)2

2= 1 � (2 + �2�2)�2 + 2�+�2�+ �2+2 � �22 = 0 must be 0; that is,

0 = (2�+�2)2 � 4(2 + �2�2)(�2+2 � �22) = 4�4+2�2 � 4�22+2 + 4�24 � 4�4+2�2 + 4�42�2

= 4�22(�2�2 + 2 � +2)

Therefore, �2�2 + 2 � +2 = 0.

Now if a point (1� ;) lies on the line = ��+ +, then + = ; ��1, so from above,

0 = �2�2 + 2 � (; ��1)2 = (�2 � 12)�2 + 21;�+ 2 � ;2 � �2 +21;

�2 � 12�+

2 � ;2

�2 � 12= 0.

(a) Suppose that the two tangent lines from the point (1� ;) to the ellipse

have slopes � and 1

�. Then � and 1

�are roots of the equation

82 +21;

�2 � 128 +

2 � ;2

�2 � 12= 0. This implies that (8 ��)

�8 � 1

�= 0 �

82 ���+

1

�8 +�

�1

�= 0, so equating the constant terms in the two

quadratic equations, we get 2 � ;2

�2 � 12= �

�1

�= 1, and hence 2 � ;2 = �2 � 12. So (1� ;) lies on the

hyperbola �2 � 2 = �2 � 2.

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268 ¤ CHAPTER 3 FOCUS ON PROBLEM SOLVING

(b) If the two tangent lines from the point (� �) to the ellipse have slopes�

and � 1

�, then� and � 1

�are roots of the quadratic equation, and so

(� ��)� +

1

= 0, and equating the constant terms as in part (a), we get

�2 � �2�2 � �2 = �1, and hence �

2 � �2 = �2 � �2. So the point (� �) lies on the

circle �2 + �2 = �2 + �2.

21. � = �4 � 2�2 � � �0 = 4�3 � 4�� 1. The equation of the tangent line at � = � is

� � (�4 � 2�2 � �) = (4�3 � 4�� 1)(�� �) or � = (4�3 � 4�� 1)�+ (�3�4 + 2�2) and similarly for � = �. So if at

� = � and � = � we have the same tangent line, then 4�3 � 4�� 1 = 4�3 � 4�� 1 and�3�4 +2�2 = �3�4 +2�2. The �rstequation gives �3 � �3 = �� � (�� �)(�2 + ��+ �2) = (�� �). Assuming � 6= �, we have 1 = �2 + ��+ �2.

The second equation gives 3(�4 � �4) = 2(�2 � �2) 3(�2 � �2)(�2 + �2) = 2(�2 � �2) which is true if � = ��.

Substituting into 1 = �2 + ��+ �2 gives 1 = �2 � �2 + �2 � = ±1 so that � = 1 and � = �1 or vice versa. Thus,the points (1�2) and (�1 0) have a common tangent line.As long as there are only two such points, we are done. So we show that these are in fact the only two such points.

Suppose that �2 � �2 6= 0. Then 3(�2 � �2)(�2 + �2) = 2(�2 � �2) gives 3(�2 + �2) = 2 or �2 + �2 = 23.

Thus, �� = (�2 + ��+ �2)� (�2 + �2) = 1� 2

3=1

3, so � = 1

3�. Hence, �2 + 1

9�2=2

3, so 9�4 + 1 = 6�2

0 = 9�4 � 6�2 + 1 = (3�2 � 1)2. So 3�2 � 1 = 0 �2 =1

3 �2 =

1

9�2=1

3= �2, contradicting our assumption

that �2 6= �2.

22. Suppose that the normal lines at the three points��1 �

21

�,��2 �

22

�, and

��3 �

23

�intersect at a common point. Now if one of

the �� is 0 (suppose �1 = 0) then by symmetry �2 = ��3, so �1 + �2 + �3 = 0. So we can assume that none of the �� is 0.

The slope of the tangent line at��� �

2�

�is 2��, so the slope of the normal line is � 1

2��and its equation is

� � �2� = � 1

2��(�� ��). We solve for the �-coordinate of the intersection of the normal lines from

��1 �

21

�and

��2 �

22

�:

� = �21 � 1

2�1(�� �1) = �22 � 1

2�2(�� �2) �

1

2�2� 1

2�1

= �22 � �21

�1 � �22�1�2

= (��1 � �2)(�1 + �2) � � = �2�1�2(�1 + �2) (1). Similarly, solving for the �-coordinate of the

intersections of the normal lines from��1 �

21

�and

��3 �

23

�gives � = �2�1�3(�1 + �3) (2).

Equating (1) and (2) gives �2(�1 + �2) = �3(�1 + �3) � �1(�2 � �3) = �23 � �22 = �(�2 + �3)(�2 � �3) �

�1 = �(�2 + �3) � �1 + �2 + �3 = 0.

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CHAPTER 3 FOCUS ON PROBLEM SOLVING ¤ 269

23. Because of the periodic nature of the lattice points, it suf�ces to consider the points in the 5× 2 grid shown. We can see that

the minimum value of � occurs when there is a line with slope 25

which touches the circle centered at (3� 1) and the circles

centered at (0� 0) and (5� 2).

To �nd � , the point at which the line is tangent to the circle at (0� 0), we simultaneously solve �2 + 2 = �2 and

= � 52� � �2 + 25

4�2 = �2 � �2 = 4

29�2 � � = 2�

29�, = � 5�

29�. To �nd /, we either use symmetry or

solve (�� 3)2 + (� 1)2 = �2 and � 1 = � 52(�� 3). As above, we get � = 3� 2�

29�, = 1+ 5�

29�. Now the slope of

the line �/ is 25

, so ��� =1 + 5�

29� �

�� 5�

29��

3� 2�29� � 2�

29�

=1 + 10�

29�

3� 4�29�=

�29 + 10�

3�29� 4� =

2

5�

5�29 + 50� = 6

�29� 8� � 58� =

�29 � � =

�2958

. So the minimum value of � for which any line with slope 25

intersects circles with radius � centered at the lattice points on the plane is � =�2958

0�093.

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