Section 2.2.

29
Section 2.2. 4 6 f(x) x x 1. Find the derivative of the following function. Use the product rule. 4 6 4 5 6 3 9 9 9 d (x x) x 6x x 4x 6x 4x 10x dx

description

Section 2.2. 1. Find the derivative of the following function. Use the product rule. 2. Find the derivative of the following function. f (x) = x 2 (x 3 + 3). Use the product rule. 3. Find the derivative of the following function. f (x) = √ x (6x + 2). Use the product rule. - PowerPoint PPT Presentation

Transcript of Section 2.2.

Page 1: Section 2.2.

Section 2.2.

4 6f (x) x x 1. Find the derivative of the following function.

Use the product rule.

4 6 4 5 6 3 9 9 9d(x x ) x 6x x 4x 6x 4x 10x

dx

Page 2: Section 2.2.

2. Find the derivative of the following function. f (x) = x 2 (x 3 + 3)

Use the product rule.

2 2 3 4 4 4f '(x) x 3x 2x(x 3) 3x 2x 6x 5x 6x

Page 3: Section 2.2.

3. Find the derivative of the following function. f (x) = √x (6x + 2)

Use the product rule.

1 2f (x) x (6x 2)

1 2 1 2 1 2 1 2 1 2 1 2 1 21f '(x) x 6 x (6x 2) 6x 3x x 9x x

2

Page 4: Section 2.2.

4. Find the derivative of the following function. f (x) = (x 2 + x) (3x + 1)

Use the product rule.

2f '(x) (x x) 3 (2x 1) (3x 1)

2 2 23x 3x 6x 5x 1 9x 8x 1

Page 5: Section 2.2.

5. Find the derivative of the following function. f (x) = (2x 2 + 1) (1 - x)

Use the product rule.

2f '(x) (2x 1) ( 1) (1 x) (4x)

2 2 22x 1 4x 4x 6x 4x 1

Page 6: Section 2.2.

f (t) 6t4

3 (3t2

3 1)6. Find the derivative of the following function.

Use the product rule.

4 1 1 2 1

3 3 3 3 3f '(t) 6t 2t 8t (3t 1) 12t 24t 8t

1

336t 8t

Page 7: Section 2.2.

Use the product rule.

7. Find the derivative of the following function. f (x) = (x 4 + x 2 + 1) (x 3 - 3)

4 2 2 3 3f '(x) (x x 1) 3x (x 3) (4x 2x)

6 4 2 6 4 33x 3x 3x 4x 2x 12x 6x

6 4 3 27x 5x 12x 3x 6x

Page 8: Section 2.2.

8

2

xy

x8. Find the derivative of the following function.

Use the quotient rule.

8 2 7 8 9 9 95

2 2 2 4 4

d x x 8x x 2x 8x 2x 6x6x

dx x ((x )) x x

Page 9: Section 2.2.

4

3

x 1f (x)

x

Use the quotient rule.

9. Find the derivative of the following function.

3 3 2 4 6 6 2

3 2 6

x 4x 3x (x 1) 4x 3x 3xf '(x)

((x )) x

6 2 4

6 4

x 3x x 3

x x

Page 10: Section 2.2.

f (x) 3x 1

2 x10. Find the derivative of the following function.

Use the quotient rule.

Page 11: Section 2.2.

f (s) s3 1

s111. Find the derivative of the following function.

Use the quotient rule.

Page 12: Section 2.2.

f (x) x 4 x 2 1

x 2 112. Find the derivative of the following function.

Use the quotient rule.

Page 13: Section 2.2.

13. Economics: Marginal Average Revenue Use the Quotient Rule to find a generalexpression for the marginal average revenue. That is calculate

and simplify your answer.

d

dx[R(x)

x]

Page 14: Section 2.2.

14. Environmental Science: Water Purification If the cost (in cents) of purifying a gallon of water to a purify of x percent is for ( 50 x 100)

a.Find the instantaneous rate of change of the cost with respect to purity.b.Evaluate this rate of change for a purity of 95% and interpret your answer.c.Evaluate this rate of change for a purity of 98% and interpret your answer

C(x) 100

100 x

2 2

0 (100 x) (100) ( 1) 100a. C'(x)

(100 x) (100 x)

2 2

100 100b. C'(95) 4

(100 95) (5)

It will cost 4 cents per gallon to increase the purity from 95% to 96% pure.

2 2

100 100c. C'(99) 25

(100 98) (2)

It will cost 25 cents per gallon to increase the purity from 98% to 99% pure.

Page 15: Section 2.2.

15. Environmental Science: Water Purification (14 continued) a.Use a graphing calculator to graph the cost function C(x) from exercise14 on the window [50,100] by [0,20]. TRACE along the curve to see how rapidlycosts increase for purity (x-coordinate) increasing from 50 to near 100.

b. To check your answer to 14, use the “dy/dx” or SLOPE feature of your calculatorto find the slope of the cost curve at x = 95 and x = 98, The resulting rates of changeof the cost should agree with your answer to Exercise 14(b) and (c). Note thatfurther purification becomes increasingly expensive at higher purity levels.

Page 16: Section 2.2.

16. Business: Marginal Average Cost A company can produce LCD digital alarm clocks at a cost of $6 each while fixed costs are $45. Therefore, the company’s cost function C(x) = 6x+45.

a.Find the average cost function .b.Find the marginal average cost function.c.Evaluate marginal average cost function at x =30 and interpret your answer.

If you produce one more alarm clock, the 31st, the average cost will decrease by 5 cents.

Page 17: Section 2.2.

17. General: Body Temperature If a person;s temperature after x hours of strenuousexercise is T (x) = x 3 (4 – x 2) + 98.6 degrees Fahrenheit for (0 x 2), find the rate of change of the temperature after 1 hour.

Page 18: Section 2.2.

18. General: Body Temperature (17 continued)

a.Graph the temperature function T(x) given in 17, on the window [0,2] by [90, 110].TRACE along the temperature curve to see how the temperature rises and falls as time increases

b. To check you answer to 17, use the “dy/dx” or SLOPE feature of your calculator to find the slope (rate of change) of the curve at x =1. Your answer should agree with youranswer in 17.

c. Find the the maximum temperature.

Page 19: Section 2.2.

19. Find the first four derivatives of f (x) = x 4 - 2x 3 – 3x2 + 5x - 7

Page 20: Section 2.2.

20. Find the first four derivatives of 5x)x(f

Page 21: Section 2.2.

21. Find the first and second derivatives of

and evaluate the second derivative at x = 3.

f (x) x 1

x

Page 22: Section 2.2.

f (x) x 1

2x22. Find the first and second derivatives of

and evaluate the second derivative at x = 3.

Page 23: Section 2.2.

f (x) (x 2 2)(x 2 3)23. Find the first and second derivatives of

Page 24: Section 2.2.

xf (x)

x 1

24. Find the first and second derivatives of

Page 25: Section 2.2.

25. Find the first and second derivatives of f (x) = r 2.

Page 26: Section 2.2.

26. After t hours a freight train is s (t) = 18t 2 – 2t 3 miles due north of its starting point (for 0 ≤ t ≤ 9).a. Find its velocity at time t = 3 hours.b. Find its velocity at time t = 7 hours.c. Find its acceleration at time t = 1 hour.

a. Velocity = s’ (t) = 36 t – 6 t 2 . And s‘ (3) = 36 (3) – 6 (9) = 54 miles per hour.

b. Velocity = s’ (t) = 36 t – 6 t 2 . And s‘ (7) = 36 (7) – 6 (49) = - 42 miles per hour.

2.5 #33

c. acceleration = s” (t) = 36 – 12 t . And s” (1) = 36 – 12 = 24 miles per hour per hour.

Page 27: Section 2.2.

27. If a steel ball is dropped from the top of the Taipei 101, the tallest building in the world, its height above the ground t seconds after it is dropped will be

s (t) = 1667 – 16t 2 feet.a. How long will it take to reach the ground?b. Use your answer in part a to find the velocity at which it will strike

the ground.c. Find the acceleration at any time t.

a. To find when the steel ball will reach the ground, we need to determine what value of t produces s(t)=0. Thus, set s(t)=0 and solve the equation by find the x-intercepts (zeros) using our calculator. The steel ball will reach the ground after 10.2 seconds.

b. The velocity is the derivative of the distance function in part a. Use your calculator to find the derivative at x = 10.2 from part a. The velocity will be 326.4 feet per second. About 222.5 mph.

c. The acceleration is the second derivative of the given distance formula or s’ (t) = -32t and s” (t) = - 32 feet per second per second.

Page 28: Section 2.2.

28. ECONOMICS: National Debt The national debt of a South American country t years from now is predicted to be D (t) = 65 + 9 t 4/3 billion dollars.

Find D’ (8) and D” (8) and interpret your answers.

D’ (t) = 12 t 1/3 and D’ (8) = 24 billion dollars per year. This is the amount the national debt is expected to increase between the 8th and 9th year.

D” (t) = 4 t - 2/3 and D” (8) = 1 billion dollars per year. The rate of growth in the national debt is expected to increase by 1 billion dollars per year after the 8 th year.

Page 29: Section 2.2.

29. GENERAL: Wind-chill Index the wind-chill index for a temperature of 32 degrees Fahrenheit and a wind speed of x miles per hour is

W (x) = 55.628 – 22.07 x 0.16.a. Graph the wind-chill and find the wind-chill index for wind speeds of x = 15

and x = 30 miles per hour..b. Notice from your graph that the wind-chill index has a first derivative that is

negative and a second derivative that is positive. What does this mean about how successive 1-mph increases in wind speed will affect the wind-chill index?

a. W (15) = 21.6W (30) = 17.6

b. Each 1-mph increase in wind speed lowers the wind-chill index. As wind speed increases, the rate with which the wind-chill index decreases slows.