· 3 DIFFERENTIATION RULES 3.1 Derivatives ofPolynomials and Exponential Functions 1. (a) 0is the...
Transcript of · 3 DIFFERENTIATION RULES 3.1 Derivatives ofPolynomials and Exponential Functions 1. (a) 0is the...
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
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 1ON DERIVATIVES OF POLYNOMIALS AND EXPS O
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
SECTION 3 1ON DERIVATIVES OF POLYNOMIALS AND EXPS O
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�� = (!
NOT FOR SALE SECTION 3 1ON DERIVATIVES OF POLYNOMIALS AND EXPS O
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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]
NOT FOR SALE SECTION 3 1ON DERIVATIVES OF POLYNOMIALS AND EXPS O
INSTRUCTOR USE ONLY [conti[con
© Cengage Learning. All Rights Reserved.
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
.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE APPLIED PROJECTPR BUILDING A BETT
INSTRUCTOR USE ONLY SS RR( )) 22 ( )( )
TT RR� ��
YYSTRUCTOR USE ONLYSTRUCTOR USE ONLY© Cengage Learning. All Rights Reserved.
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):
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE SECTION 3 2EC THE PRODUCT
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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..
© Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.)
NOT FOR SALE SECTION 3 2EC THE PRODUCT
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE SECTION 3 2EC THE PRODUCT
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 �).
NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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)
NOT FOR SALE SECTION 3 33 3 DERIVATIVES OF TRIGON
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY p qp q (( ))
33, g
© Cengage Learning. All Rights Reserved.
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).
NOT FOR SALE SECTION 3 33 3 DERIVATIVES OF TRIGON
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�]
NOT FOR SALE SECTION 3 33 3 DERIVATIVES OF TRIGON
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE SECTION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 = ��+ �.
NOT FOR SALE SECTION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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).
NOT FOR SALE SECTION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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).
NOT FOR SALE SECTION
INSTRUCTOR USE ONLY TOR USETOR USERR UUUUUUUUSUUSSSSSS© Cengage Learning. All Rights Reserved.
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)
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 �
�.
NOT FOR SALE SECTION
INSTRUCTOR USE ONLY ��180180
������ ����
�180180
�18018 180180 180180
© Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.)
NOT FOR SALE LABORATORY PROJ
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE SECTION 3 5 IMPL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 .
NOT FOR SALE SECTION 3 5 IMPL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE SECTION 3 5 IMPL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
�.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
SECTION 3 IMPL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 5 IMPL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 .
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 .
NOT FOR SALE SECTION 3 6N 3 INVERSE TRIGONOMETRIC FUNCTIONS ANGO
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 = �.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�)
NOT FOR SALE SECTION 3 6N 3 INVERSE TRIGONOMETRIC FUNCTIONS ANGO
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 6N 3 INVERSE TRIGONOMETRIC FUNCTIONS ANGO
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�|
© Cengage Learning. All Rights Reserved.
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
NOT FOR SALE SECTION 3N DERIVATIVES OF LOG
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 7ON DERIVATIVES OF LOG
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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� �
�.
NOT FOR SALE DISCOVERY PROJECTD HY
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESUL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 8TI RATES OF CHANGE IN THE NATURAL AF C
INSTRUCTOR USE ONLY (c) The particle is at rest when(c) The particle is at rest when ))((��) = 0) = 0. 00������22���� 11OO22
��+ 1+ 1��= 0= �� �� = 2= 2 s.s
© Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESUL
INSTRUCTOR USE ONLY from decreasing to increasing. Thus, atfrom decreasing to increasing. Thus, at � 1�5 s, the velocity has its minimum value.s, the velocity has its minimum value.
© Cengage Learning. All Rights Reserved.
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�.
NOT FOR SALE SECTION 3 8TI RATES OF CHANGE IN THE NATURAL AF C
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESUL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�
NOT FOR SALE SECTION 3 8TI RATES OF CHANGE IN THE NATURAL AF C
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESUL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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).
NOT FOR SALE SECTION 3 8TI RATES OF CHANGE IN THE NATURAL AF C
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESUL
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 99 LINEAR APPROXIMATION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE SECTION 3 99 LINEAR APPROXIMATION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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%.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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) = �.
NOT FOR SALE SECTION 3 99 LINEAR APPROXIMATION
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE LABORATORY PROJECT
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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(�)(�� �).
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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 � �
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.]
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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� �
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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).
NOT FOR SALE
INSTRUCTOR USE ONLY isis �� 11TT44 == 11TT44 ((��+ ln 4)+ ln 4) oror == 11CC44��++11TT44 (ln 4 + 1)(ln 4 + 1 ..
© Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE NTIATION RULESLE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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
.
NOT FOR SALE ON PROBLEM SOLVINGM S
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
CHAPTER 3 FOCUS O
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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�).
NOT FOR SALE ON PROBLEM SOLVINGM S
INSTRUCTOR USE ONLY ��(( ))
© Cengage Learning. All Rights Reserved.
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]
NOT FOR SALE CHAPTER 3 FOCUS O
INSTRUCTOR USE ONLY The result isThe result is �� == ++..[conti[con
© Cengage Learning. All Rights Reserved.
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�].
NOT FOR SALE ON PROBLEM SOLVINGM S
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE CHAPTER 3 FOCUS O
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
ON PROBLEM SOLVINGM S
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
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.
NOT FOR SALE CHAPTER 3 FOCUS O
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
NOT FOR SALE
INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.