Chapter 3 Determining Change: Derivatives · Chapter 3 Determining Change: Derivatives Section 3.1 Drawing Rate-of-Change Graphs 1. The slopes are negative to the left of A and positive

Copyright © Houghton Mifflin Company. All rights reserved. 79

Chapter 3 Determining Change: Derivatives

Section 3.1 Drawing Rate-of-Change Graphs 1. The slopes are negative to the left of A and positive

to the right of A. The slope is zero at A.

2. The slope is constant and positive everywhere.

3. The slopes are positive everywhere, near zero to the

left of zero, and increasingly positive to the right of zero.

4. The slopes are positive to the left of the origin and

negative to the right of the origin. Moving from left to right, the slopes begin near zero, increase to a maximum at the first inflection point of the graph, decrease to zero at x = 0, continuing decreasing (becoming negative) until the point of most negative slope is reached (the second inflection point of the graph), then increasing toward zero while remaining negative.

5. The slope is zero everywhere.

80 Chapter 3: Determining Change: Derivatives Calculus Concepts

Copyright © Houghton Mifflin Company. All rights reserved.

6. The slopes are positive to the left of A, negative from A to B, positive from C to D, and negative to the right of C.

7. The slopes are negative everywhere. The magnitude is

large close to x = 0 and is near zero to the far right.

8. The slopes are positive everywhere. The slopes are

large close to x = 0 and approach zero to the far right.

9. The slopes are negative to the left and right of A. The

slope appears to be zero at A.

10. The slopes are negative everywhere. The slopes begin

near zero at the left side of the graph, decrease to a minimum at B, and then increase while approaching zero.

11. a.

b.

Calculus Concepts Section 3.1: Drawing Rate-of-Change Graphs 81


12. a,b.

Year Slope 1991 −6.6

1993 −6.2

1997 −2.5

1999 1.0 2000 5.4

13. a, b.

(Table values may vary.)

-8

-6

-4

-2

0

2

4

6

8

1985 1990 1995 2000 2005

Rat

e o

f ch

ang

e o

f av

erag

e b

ill (

do

llars

p

er y

ear)

14. a.

Year Slope

(thousands of people per year)

1990.5 17 1992 14 1994 10

1996 7

1998 3

(Table values may vary.)

Graph Slope graph

Graph

Slope graph



b.

15. a. (Table values may vary slightly.)

Year 1985.25 1990 1995 1997 2000

Slope 8 46.8 79.2 55.2 21.4

b. Graph may vary, but its basic shape should be concave down with a maximum between 1993 and 1995.

16. a.

(Table values may vary.) Year Estimated slope

(gallons/year2) Year Estimated slope

(gallons/year2) Year Estimated slope

(gallons/year2)

1971 10

4

− = − 2.5 1980 45

4

− = − 11.25 1990 5

4= 1.25

1975 50

4

− = − 12.5 1985 15

5

− = − 3 1995 30

3= 10



b.

17. a. The average rate of change during the year (found by estimating the slope of the secant line

drawn from September to May) is approximately 14 members per month. (One possible answer.)

b,c. By estimating the slopes of tangent lines we obtain the following. (One possible answer.)

Month Slope (members per month)

Sept 98 Nov −9 Feb 30 Apr 11

d. Membership was growing most rapidly around March. This point on the membership graph is an inflection point.

e. The average rate of change is not useful in sketching an instantaneous-rate-of-change graph.

18. a.

b. The number of calls is a minimum around 6 a.m. and a maximum around 10 p.m.

c. The graph appears to have additional zero slopes around 10 a.m. and 2 p.m.



Time Slope

8 a.m. 55

9.2 calls/hour6

≈

Noon 5

0.3 call/hour18

− ≈ −

6 p.m. 45

7.5 calls/hour6

=

d.

e.

19.

20.



21. a. $18,000

Slope $600 per car30 cars

≈ =

Profit is increasing on average by approximately $600 per car.

b. By sketching tangent lines we obtain the following estimates: (Answers will vary depending on points picked and estimates of slopes.)

Number of cars

Slope (dollars per car)

20 0 40 160 60 750 80 10

100 −1200

c.

d. The average monthly profit is increasing most rapidly for approximately 60 cars sold and is decreasing most rapidly when approximately 100 cars are sold. The corresponding points on the graph are inflection points.

e. Average rates of change are not useful when graphing an instantaneous-rate-of-change graph.

22.

1940: Slope 11 deaths

15 years≈

≈ 0.7 death per 100,000 males per year


20 years≈

≈ 1.8 deaths per 100,000 males per year


15 years≈

≈ 0.7 death per 100,000 males per year



Rate of change graph:

23. The derivative does not exist at x = 0, x = 3, and x = 4 because the graph is not continuous at those input values.

24. The derivative does not exist at x = 3 because the slopes from the right and left are different,

and it does not exist at x = 6 because the graph is not continuous at that input.

25. The derivative does not exist at x = 2 and x = 3 because the slopes from the left and right are

different at those inputs.



26. The derivative does not exist at x = 0, x = A, and x = B because the slope approaches infinity and the function is undefined at these

inputs.

27. One possible graph:





30. a.

b. i. q tt

( ) =1

ii. q t ht h

( )+ =+1

iii. q t h q t

t h tt h t

ht h t

h

t t h

t t h

t t h

ht t h

h

ht t h

( ) ( ) ( )

( )

( )

( ) ( )

+ −+ −

= +−

= +−

⋅ ++

= − ++

= −+

1 1 1 1

iv. lim( )

lim( )h h

h

ht t h t t h t→ →

−+

=−+

=−

0 0 2

1 1. Therefore, ′ =

−q t

t( )

12 . The graph of ′q is the

same as the graph in part a.



31. a.

b. i. p m m m( ) = + ii. p m h m h m h( )+ = + + +

iii. ( )( )

( )( )

( )( )

( ) ( )

( )

m h mp m h p m h m h m h m h m

m h m h h m h m

h m h m m h m h m h m h

h m h m h m h m

+ + + − + + − + + −= = + − + +

+ + + + − + + += =

+ + + +

iv. ( )

( )( )

0 0

1 2 1 1lim lim 1

2 2h h

h m h m h m h m m

m h m m mh m h m→ →

+ + + + + + += = = ++ ++ +

dp

dm m= +1

1

2. The graph of dp

dmis the same as the one in part a.

32. a.

b. i. k x xx

( ) = + 1

ii. k x h x hx h

( )+ = + ++1

iii.

1 1 1 1( ) ( ) ( )

( )

( ) ( ) ( ) 11

( ) ( ) ( )

x h x hk x h k x x x hx h x x h xx h x h h x x h

hx x h x x h hx x h h

hx x h hx x h x x h

+ + − + + − + − ++ += = ⋅+ − +

+ + − + + −= = = −+ + +

iv. lim( )

lim lim( )h h h

h

hx x h x x h x→ → →−

+

= −

+= −

0 0 0 21 1

11

1. Therefore, ′ = −k x

x( ) 1

12

.

The graph of ′k is the same as the graph in part a.



33. One possible answer: When sketching a rate-of-change graph, it is important to identify the following features on the graph of the original function: 1) input values for which the derivative does not exist, 2) input intervals over which the function is increasing, 3) input intervals over which the function is decreasing, 4) input values that correspond to a relative maximum or minimum of the function, 5) input values for which the function appears to be increasing or decreasing most rapidly, 6) input values that correspond to inflection points where the function has zero slope.

34. One possible answer: It is important to identify the intervals on which ( )f x is increasing,

decreasing or constant. These intervals correspond to intervals on '( )f x that are positive,

negative or zero. Assuming ( )f x is a smooth, continuous curve, maximum and minimum

points of ( )f x become the zeros of '( )f x . Inflections points of ( )f x become relative

maximum and minimum points on '( )f x . If ( )f x is not smooth and/or continuous, it is

important to identify the points where '( )f x does not exist. These points are identified as

sharp points, discontinuities or points where a vertical tangent line would be drawn on ( )f x . Section 3.2 Simple Rate-of-Change Formulas 1.

The function y x= −2 7 is a line with slope −7. Thus 7dydx

= − .

2.

The slope equation is dy

dxx x= − + = −−1 2 0 22 1( ) .

Calculus Concepts Section 3.2: Simple Rate-of-Change Formulas 91


3.

The slope formula is 4 1 34 4dydx

x x−= = .

4.

The slope formula is dy

dxx x= =−3 33 1 2.

5.

The slope of any horizontal line is 0.

6.

Rewrite the function as y x= −2 , then calculate the slope formula as

dy

dxx x

x= − = − = −− − −2 2

22 1 33

.



7. The power rule applies: 5 1 4'( ) 5 5f x x x−= =

8. The power rule applies: 4 1 3'( ) 4 4f x x x−= =

9. The power rule and the constant multiplier rule apply: ( )3 1 2'( ) 3 3 9f x x x−= =

10. The power rule and the constant multiplier rule apply: ( )2 1 1( ) 0.5 2 1f x x x x−′ = − = − = −

11. The linear function rule (or the power rule) apply: '( ) 5= −f x 12. The linear function rule (or the power rule) apply:'( ) 7f x = 13. The constant rule applies: '( ) 0f x = 14. The constant rule applies: '( ) 0f x =

15. The linear function rule (or the power rule and the constant multiplier rule) apply: 12dy

dx=

16. The sum rule, the constant multiplier rule, the power rule, and the constant rule apply:

( )2 1'( ) 7 2 9.4 0 14 9.4f x x x−= − + = −


( ) ( )3 1 2 1 25 3 3 2 2 15 6 2dy

x x x xdx

− −= + − = + −


( )3 1 23.2 3 6.1 9.5 6.1dy

x xdx

−= − + = − +

19. 3 3 1 43 4

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

x x− − − − −= = → = − = − =

20. 3 3 1 23

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

x−

−= = → = =

21. ( )2 2 1 32 3

9 18( ) 9 '( ) 9 2 18f x x f x x x

x x− − − −−= = − → = − − = =

22. 23

( ) 3 '( ) 3x

f x x f xx

= = → =

23. 2 2

1 1 1 22

3 1 3 1 1( ) 3 '( ) 3 1 3 3

x xf x x x f x x x

x x x x− − − −+= = + = + → = − = − = −



24.

( )

2 21

1 1 22

4 19 6 4 19 6( ) 4 19 6

6'( ) 4 0 6 1 4 6 4

x x x xj x x x

x x x x

j x x xx

−

− − −

+ += = + + = + +

= + + − = − = −

25. 1 1 12 2 2

11 1 1( ) '( )

2 2 2f x x x f x x x

x

− −= = → = = =

26. 1 1 12 2 2

11 4( ) 17 8 17 8 '( ) 0 8 4

2h x x x h x x x

x

− − − = − = − → = − = − =

27. a. ′ =A t( ) .01333 dollars per year gives the rate of change in the average ATM transaction

fee t years after 1990, 6 ≤ t ≤ 9. b. A( ) . ( ) . $1.10 01333 10 017 50= + ≈

c. ′ =A ( ) $0.9 1333 per year The transaction fee in 1999 was increasing by approximately $0.13 per year. 28. a. ′ =P t( ) .1548 thousand people/year gives the rate of change in the population of Hawaii

t years after 1950, 20 ≤ t ≤ 40.

b. (20) 15.48(2) 485.4

795 thousand people

P = +=

c. ′ =P ( ) .40 1548 thousand people/year Hawaii’s population was increasing by 15.48 thousand people per year in 1990.

29. a. ( ) 1.6 2 FT x t′ = − + � per hour is the rate of change of the temperature t hours after noon

(1.5) 1.6(1.5) 2 0.4 F per hourT ′ = − + = − �

( 5) 1.6( 5) 2 10T ′ − = − − + = � F/hour Steepness does not consider the sign of the slope, so we compare 0.4 and 10 and conclude

that the graph is steeper at 7 a.m. than it is at 1:30 p.m.

b. ( 5) 1.6( 5) 2 10 F per hourT ′ − = − − + = �

c. (0) 1.6(0) 2 2 F per hourT ′ = − + = �

d. (4) 1.6(4) 2 4.4 F per hourT ′ = − + = − �

Because this value is negative, the temperature is falling at a rate of 4.4 F� per hour. 30. a. 2( 5) 0.03(5) 0.315( 5) 34.23 33.405N − = + − + = million senior citizens in 1995

2(29) 0.03(29) 0.315(29) 34.23 68.595N = + + = million senior citizens in 2029

2(30) 0.03(30) 0.315(30) 34.23 70.68N = + + = million senior citizens in 2030



b. ( ) 0.03(2 ) 0.315 0 0.06 0.315N x x x′ = + + = + million senior citizens per year gives the rate

of change of the projected number of Americans 65 or older x years after 2000, -5 ≤ x ≤ 30. ( 4) 0.06( 4) 0.315 0.075N ′ − = − + = million senior citizens per year in 1996

(29) 0.06(29) 0.315 2.055N ′ = + = million senior citizens per year in 2029

c. (29) 2.055

100% 100% 2.996%(29) 68.595

N

N

′⋅ = ⋅ ≈ per year in 2029

d. Solve 70.68 = 0.201P for P to obtain P = 70 68

0201

.

. ≈ 351.6 million senior citizens in 2030.

31. a,b. 684.942.318055.00684.94)2(6.15)3(2685.0)(' 22 +−=++−= xxxxxB births per year is the rate of change in the number of live births to U.S. women aged 45 and over, where x is the number of years since 1950, 0 ≤ x ≤ 50.

12.207684.94)20(2.31)20(8055.0)20(' 2 −=+−=B births per year in 1970

82.321684.94)45(2.31)45(8055.0)45(' 2 =+−=B births per year in 1995 In 1970, the number of live births to U.S. women over 45 and over was falling by

approximately 207 births per year. In 1995, the number of live births to this group was rising by approximately 322 births per year.

32. 3 1 2( ) 0.044(3 ) 0.918(2 ) 0 0.132 1.836G t t t t t−′ = − + + = − + points per hour is the rate of change

of the test grade after t hours of study.

a. ′ = − + =G ( ) . ( ) . ( ) .1 0132 1 1836 1 17042 points per hour

b. 2(4) 0.132(4) 1.836(4) 5.232G′ = − + = points per hour

c. 2(15) 0.132(15) 1.836(15) 2.16 2G′ = − + = − ≈ points per hour 33. a. m w w( ) . .= +6 930 682188 kilocalories per day is the metabolic rate of a typical 18- to

30-year old male who weighs w pounds, 88 ≤ w ≤ 200. b. ′ =m w( ) .6 930 kilocalories per day per pound is the rate of change in the metabolic rate of

a typical 18- to 30-year old male who weighs w pounds, 88 ≤ w ≤ 200. c. Regardless of the man’s weight, if he gains weight, his metabolic rate will increase by

6.930 kilocalories per day per pound.

34. a. R b b b( ) . . .= − + −20 982 417 857 1433042 dollars gives the weekly revenue when b benches

are sold, 1 ≤ b ≤ 13.

3 2( ) 0.069 1.113 16.002 43.327C b b b b= − + + dollars gives the weekly cost when b benches

are produced, 1 ≤ b ≤ 13. 3 2( ) ( ) ( ) 0.069 19.869 401.855 186.631P b R b C b b b b= − = − − + − dollars gives the weekly

profit when b benches are produced and sold each week, 1 ≤ b ≤ 13.

b. 2( ) 0.208 39.738 401.855P b b b′ = − − + dollars per bench is the rate of change in the weekly

profit when b benches are produced and sold each week, 1 ≤ b ≤ 13.



c. (6) $155.93P′ ≈ per bench ; When 6 benches are sold each week, the profit is increasing by approximately $156 per bench.

(9) $27.35P′ ≈ dollars per bench; When 9 benches are sold each week, the profit is increasing by approximately $27 per bench.

(10) $16.34P′ ≈ − dollars per bunch; When 10 benches are sold each week, the profit is decreasing by approximately $16 per week.

d. Because ′P changes from positive for b = 9 to negative for b = 10, we know that the maximum of the profit function occurs between b = 9 and b = 10. Because the artisan cannot sell a fraction of a bench, the maximum profit occurs at either 9 or 10 benches. By evaluating the profit function at b = 9 and b = 10, we determine that profit will be greatest when she sells 10 benches.

35. a. 98.166759.80958.4768.3)( 23 +−+−= xxxxR billion dollars is the revenue when $x

billion is spent on advertising, 1.2 ≤ x ≤ 6.4.

b. 759.80916.95039.110)1(759.80)2(958.47)3(68.3)(' 22 −+−=+−+−= xxxxxR billion dollars per billion dollars (billion dollars of revenue per billion dollars of advertising) is the rate of change of revenue when $x billion is spent on advertising, 1.2 ≤ x ≤ 6.4.

c. Revenue: 3 2(5) 3.68(5) 47.958(5) 80.759(5) 166.98 $502.166R = − + − + ≈ billion

Rate of Change of Revenue: 2'(5) 11.039(5) 95.916(5) 80.759 $122.837R = − + − ≈ billion per billion dollar

d. '(5)

100% 24.46%(5)

R

R≈ per billion dollars spent on advertising.

36. a. 3 2( ) 0.015 0.848 20.062 8.741C u u u u= − + + dollars per hour is the production costs when u

units are produced, 0 ≤ u ≤ 90.

b. A uC u

u

u u u

uu u

u( )

( ) . . . .. . .

.= = − + + = − + +0 015 0848 20 062 8 7410 015 0848 20 062

8 7413 22

dollars per hour per unit is the average cost when u units are produced, 0 < u ≤ 90.

c. 2 1( ) 0.015 0.848 20.062 8.741A u u u u−= − + +

22

8.741( ) 0.015(2 ) 0.848 0 8.741( 1 ) 0.030 0.848A u u u u

u−′ = − + + − = − − dollars per hour per

unit per unit is the rate of the change of the average cost when u units are produced,

0 < u ≤ 90. Note that this derivative gives the rate of change of average hourly cost (dollars per hour per unit) as the number of units produces changes.

d. ′ = − − ≈ −A ( ) . ( ) ..

( )$0.15 0 030 15 0848

8 741

1544

2per hour per unit per unit produced

′ = − − ≈A ( ) . ( ) ..

( )$0.35 0 030 35 0848

8 741

3519


′ = − − ≈A ( ) . ( ) ..

( )$1.85 0 030 85 0848

8 741

8570




When 15 units are produced, the average hourly cost per unit is decreasing at a rate of approximately $0.44 per unit; that is, if production is increased from 15 to 16 units, the average hourly cost per unit will decrease by approximately $0.44. Increasing production will lower the average hourly cost per unit.

When 35 units are produced, the average hourly cost per unit is increasing at a rate of approximately $0.19 per unit; that is, if production is increased from 35 to 36 units, the average hourly cost per unit will increase by approximately $0.19. Increasing production will raise the average hourly cost per unit.

When 85 units are produced, the average hourly cost per unit is increasing at a rate of approximately $1.70 per unit; that is, if production is increased from 85 to 86 units, the average hourly cost per unit will increase by approximately $1.70. Increasing production will raise the average hourly cost per unit.

37. a. 2 49.6( ) 175 0.015 0.78 46P x x x

x = − − + +

dollars is the profit from the sale of one storm

window when x windows are produced each hour, x > 0.

b. 22

49( ) 0 0.015(2 ) 0.78 0 49( 1 ) 0.03 0.78 P x x x x

x− ′ = − − + + − = − + +

dollars per window gives the rate of change of the profit from the sale of one storm window when x windows are produced each hour, x > 0.

c. 2 49(80) 175 0.015(80) 0.78(80) 46 $94.79

80P = − − + + ≈

profit

d. 2

49(80) 0.03(80) 0.78 $1.61

(80)P′ == − + + ≈ − per window produced

When 80 units are produced each hour, the profit from the sale of one window is decreasing by $1.61 per unit produced. In other words, if production is increased from 80 to 81 windows per hour, then the average profit per window will decrease by approximately $1.61. This indicates that if the company wishes to maximize the profit per window, they should not increase production above 80 units per hour.

38. a. ( ) ( ) ( ) 68.95 0.125t x n x a x x x= + = + thousand copies gives the total number of copies of

a new book sold in the United States and abroad by the end of the xth week, x ≥ 0.

b. Rewrite t as 12( ) 68.95 0.125t x x x= + .

121 34.475

( ) 68.95 0.125 0.1252

t x xx

− ′ = + = +

thousand copies per week gives the rate of

change in the total number of copies of a new book sold in the United States and abroad by the end of the xth week, x > 0.

c. (52) 68.95 52 0.125(52) 503.706t = + ≈ thousand books

d. 34.475

(52) 0.125 4.90652

t′ = + ≈ thousand books per week

At the end of the first year, the total book sales are increasing at a rate of approximately 4,906 books per week



39. One possible answer: The graph of a cubic function with a positive 3ax term will increase, then decrease (or level off if the cubic function does not decrease at all), and finally increase again. The derivative graph for this cubic function will be a parabola with positive values at each end corresponding to the intervals of increase and either an interval of negative values (or the single slope value of 0 corresponding to the point that divides the two intervals of increase).

40. One possible answer: The derivative of the function 3 2y ax bx cx d= + + + can be broken down into four terms using the sum rule for derivatives:

( )3 2 3 2dy d d d d dyax bx cx d ax bx cx d

dx dx dx dx dx dx= + + + = + + + . The constant multiplier rule can

be applied to the first three terms and the constant rule can be applied to the fourth term

yielding 3 2 0dy d d d

a x b x c xdx dx dx dx

= + + +

. Finally, the power rule can be applied to

each of the derivatives to finish the process: ( ) ( ) ( )2 1 0 23 2 1 3 2dy

a x b x c x ax bx cdx

= + + = + + .

Section 3.3 Exponential and Logarithmic Rate-of-Change Formulas 1.

2.



3.

4.

5.

6.

7. ( ) 0 7 7x xh x e e′ = − = −

8. 5 3 5x xdye e

dx= + =

9. (ln2.1)(2.1 )xdg

dx=

Calculus Concepts Section 3.3: Exponential and Logarithmic Rate-of-Change Formulas 99


10. (ln3.5)(3.5 )xdy

dx=

11. '( ) 12(ln1.6)(1.6 )xh x =

12. 6(ln0.8)(0.8 )xdy

dx=

13. 40.05

( ) 10 14

x

f x = +

( ) ( )( )

4 40.05 0.054 4

40.054

( ) 10 ln 1 1

0.4969 1

0.4969(1.0509 )

x

x

x

f x ′ = + +

≈ +

≈

14. 120.06

( ) 24 112

x

f x = +

( ) ( )12 120.06 0.0612 12

( ) 24 ln 1 1x

f x ′ = + +

15. ( ) 4.2(ln0.8)(0.8 ) 0

4.2(ln0.8)(0.8 )

x

x

j x′ = +=

16. '( ) 7(ln1.3)(1.3 )x xj x e= −

17. 1 4

'( ) 4j xx x

= ⋅ =

18. 1 1

'( ) 1j xx x

−= − ⋅ =

19. 1 7

0 7dy

dx x x

−= − ⋅ =

20. 1 2

'( ) 3.7 2 3.7x xk x e ex x

= − ⋅ = −



21. a. ( )0.043( )t

A t e=

dA

dte e t= ln( ). .0 043 0 043

= 0 043 0 043. (ln ) .e e t

= 0043 0 043. .e t thousand dollars per year is the rate of change of the value of the investment after t years, t > 0.

b. 0.043(5)(5)

1.23986 thousand dollars

= $1239.86

A e=≈

c. 0.043(5)(5) 0.043

0.0533 thousand dollars per year

$53 per year

A e′ =≈≈

d. '(5)

100 4.3(5)

A

A⋅ ≈ % per year

22. a. 12

0.0970.062( ) 32 1 13

12

ttA t e = + +

thousand dollars is the value of the two investments after

t years, t ≥ 0.

b. ( )12

0.0970.062( ) 32 1 13

12

tt

A t e = + +

( )12 12

0.097 0.0970.062 0.062'( ) 32 ln 1 1 13 ln

12 12

ttA t e e

= + + +

0.097( ) 1.97889(1.06379 ) 1.261t tA t e′ ≈ + thousand dollars per year is the rate of change in

the value of the two investments after t years, t ≥ 0.

c. After 6 months, t = 0.5 year, and after 15 months, t = 1.25 years.

6 months:

12 12(0.5)0.097(0.5)0.062 0.062

'(0.5) 32 ln 1 1 13(0.097)12 12

= 3.365 thousand dollars per year

A e = + + +

15 months:

12 12(1.25)0.097(1.25)0.062 0.062

'(1.25) 32 ln 1 1 13(0.097)12 12

= 3.561 thousand dollars per year

A e = + + +

23. a. 10 10 10 10

10

1000ln( ) 1000(10) 10,000

10,000 dollars per 100 percentage points when the interest rate is 100 %

r r r

r

dAe e e e

dr

e r

= = =

=

b. 10(0.07)'(0.07) 10,000 $20,137.53A e= ≈ per 100 percentage points. Working with interest can get interesting when taking derivatives. The increase of $20,137.53 represents the rate of change when the interest rate (currently r = 0.07) increased to by 1 to 1.07. This means the interest rate is increased from 7% to 107%.



c. dA

dre e e er r r= = =1000 1000 01 10001 01 01 0 1ln( ) ( . ). . . . dollars per percentage point when the

interest rate is r%. The constant multiplier and exponent multiplier are both 1100 of what

they were in the function in part a.

d. 0.1(7)(7) 100 $201.38A e′ = ≈ per percentage point. This answer is 1100 of the answer to

part b.

24. a. Solving 25. = eh for h yields h = ln 2.5 ≈ 0.916 hours

b. dV

dheh= quarts per hour is the rate of growth of the dough after the dough has been

allowed to rise for h hours.

c. 24 min = 1 hr60 min24 min⋅ = 0.4 hr

V ( . ) .0 4 149≈ quarts per hour

Convert to quarts per minute by multiplying by 60 min1 hr :

V ( . )0 4 ≈ 0.025 quarts per minute

42 min = 1 hr60 min42 min⋅ = 0.7 hr

(0.7) 2.01 quarts per hour

0.034 quart per minute

V ≈≈

55 min = 1 hr60 min55 min⋅ = 55

60 hr

( )5560 2.5 quarts per hour

0.042 quart per minute

V ≈

≈

25. a. Solve ( )2.5 0.14 4.106x=

( )

2.54.106

0.142.5

ln ln 4.1060.14

2.5ln ln 4.106

0.14

2.5ln

0.14ln 4.106

2.041

x

x

x

x

x

=

=

=

=

≈

b. '( ) 0.14(ln4.106)(4.106 )xs x = million iPods per year is the rate of change of iPod sales,

where x is the number of years after the fiscal year that ended in September, 2002, 0 3.x≤ ≤



c. 2.0407'(2.041) 0.14(ln4.106)(4.106 ) 3.531s = ≈ million iPods per year; The iPod sales were

increasing by approximately 3.531 million iPods per year at the time the 2.5 millionth iPod was sold.

26. a. The cubic model is used to describe the population when 0.7 ≤ t ≤ 13. The exponential

model is used to describe the population when 13 < t ≤ 55.

b,c. 223.73 242 194 people per year when 0.7 13

( )45.5(ln0.847)(0.847 ) thousand people per year when 13 55t

t t tp t

t

− + + ≤ <′ = < ≤

gives the rate of the population of Aurora, where t is the number of years since the beginning of 1860.

d. 1870: (10) 241p′ = people per year

1873: (13)p′ does not exist (there is a jump in the graph of p(t) when t =13.)

1900: (40) .010p′ ≈ − thousand people per year or -10 people per year

27. a. 9 weeks: w( ) . . ln .7 113 7 37 7 2564= + ≈ grams; ′ = ≈w ( ).

.77 37

7105 grams per week

b. (9) (4) 5.977 grams

1.195 grams per week on average9 4 5 weeks

w w− ≈ ≈−

c. The older the mouse is, the more slowly it gains weight because the rate-of-change

formula, 7.37( ) tw t′ = , has the age of the mouse in the denominator.

28. a. 76.29

( )I xx

′ = million homes per year gives the rate of change of the projected number of

such homes x years after 1990, 8 ≤ x ≤ 15.

b. (14) 138.27 76.29ln14 63.064I = − + ≈ million homes

76.29

(14) 5.44914

I ′ = ≈ million homes per year

29. a. 9.9

'( )T dd

−= °F per day is the rate of change of the temperature at which mil must be

stored in order to remain fresh, where d is the number of days that the milk must be stored, d > 0.

b. As the number of days that the milk must be stored increases, the temperature at which the milk must be stored decreases more slowly. Consequently, the rate of change of the temperature approaches zero.

30. a. ( ) ( ) ( )

3250 75ln 50 1500

4750 75ln 50 dollars

t u c u s u

u u

u u

= += + + += + +

is the total cost to produce and ship u units each week, u > 0.



b. ′ = + + = +t uu u

( ) 075

5075

50 dollars per unit is the rate of change of the total cost to

produce and ship u units each week, u > 0.

c. (5000) 4750 75ln5000 50(5000) $255,389t = + + ≈

d. 75

'(5000) 50 50.0155000

t = + = dollars per unit

When 5000 units are produced and shipped each week, the total cost is increasing by approximately 50 dollars per unit. In other words, if production is increased from 5000 to 5001 units, the combined weekly production and shipping costs will increase by approximately $50.

31. a. 3.1219971999

1.2947.318 =−−

index points per year; this is the average rate of change between 1997

and 1999. b. ( ) 351.521 227.777lnC x x= − + points is the consumer price index for college tuition

between 1990 and 2000, where x is the number of years since 1980.

c. 227.777

'(18) 12.718

C = ≈ points per year

32. a. 3 2( ) 0.216 7.383 661.969 19,839.889M t t t t= − + + + dollars gives the median family income

t years after 1947, 0 ≤ t ≤ 50.

b. 3 2( ) 0.649 14.765 661.969M t t t′ = − + + dollars per year gives the rate of change in the

median family income t years after 1947, 0 ≤ t ≤ 50.

c. Year t ( )M t′′′′

( )100%

( )M t

M t

′′′′⋅⋅⋅⋅ Reelected

1972 25 626 1.7% per year Yes 1980 33 443 1.1% per year No 1984 37 320 0.7% per year Yes 1992 45 12.9 0.3% per year No 1996 49 −172 −0.4% per year Yes

d. There seems to be no relationship between re-election and the rate of change in median family income. This may be an indication that the model does not provide sufficient information to answer this question accurately because it models the 50-year trend in median family income rather than the median income close to each election year.

33. a. ( ) 3.960(3.584 )xR x = million dollars is the revenue realized by Apple from the sales of

iPods for fiscal years ending between September 30, 2002, and September 30, 2006, where x is the number of fiscal years since September 30, 2000.

b. '( ) 3.960(ln3.584)(3.584 )xR x = million dollars per year is the rate of change of the revenue

realized by Apple from the sales of iPods for fiscal years ending between September 30, 2002, and September 30, 2006, where x is the number of fiscal years since September 30, 2000.



c. Revenue: 5(5) 3.960(3.584 ) $2340.44R = ≈

Rate of Change: 5'(5) 3.960(ln3.584)(3.584 ) $2987.53R = = million per year

Percentage Rate of Change: '(5)

100 127.65(5)

R

R⋅ = % per year

The revenue realized by Apple for the fiscal year ending in September 30, 2005, from the sale of iPods was approximately $2340.44 million dollars. At that time, the revenue realized by Apple from the sale of iPods was increasing by approximately $2987.53 million per year. This rate of increase in the revenue realized by Apple from the sale of iPods was approximately 127.65 % per year.

34. a. ttV ln726.11034.56)( += percent is the percentage of households with TVs that also have

VCRs from 1990 through 2001, t years after 1987.

b. t

tV726.11

)(' = percentage points per year gives the rate of change in the percentage of

households with TVs that also have VCRs from 1990 through 2001, t years after 1987.

c. %1.86)13ln(726.11034.56)13( ≈+=V ; 90.013

726.11)13( ≈=′V percentage points per year

In 2000 approximately 86.1% of all households with TVs also had VCRs. In that same year, that percentage was growing by approximately 0.90 percentage points per year.

35. a. xy e= (ln )( ) (1)x x xdye e e e

dx→ = = =

b. ( ) ( )( ) ( )' ln= = → = = =x x xkx k k k k kxy e e y e e k e ke

36. One possible answer: The graph of a logarithmic model is characterized by a slower and

slower increase or decrease as x increases. If ln 0dy b b

y a b xdx x x

= + → = + = . If b > 0, then

dy

dx becomes less and less positive as x increases. If b < 0, then

dy

dx becomes less and less

negative as x increases. Section 3.4 The Chain Rule 1. a. f(x(2)) = f(6) = 140

b. When x = 6, df

dxf= ′ =( ) –276

c. When tdx

dtx= = ′ =2 2 13, ( ) .

d. When t = 2, x = x(2) = 6, so (6) (2) (–27)(1.3) 35.1df

f xdt

′ ′= = = −

Calculus Concepts Section 3.4: The Chain Rule 105


2. a. g(v(88)) = g(17) = 0.04

b. When x = 88, dv

dxv= ′ =( ) . .88 16

c. When x = 88, v = v(88) = 17, so dg

dvg= ′ =( ) . .17 0 005

d. When x = 88, v = v(88) = 17, so dg

dx

dg

dv

dv

dxg v= = ′ ′( ) ( )17 88 = =( . )( . ) . .0 005 16 0 008

3. Let t denote the number of days from today, w denote the weight in ounces, and v denote the

value of the gold in dollars. We know that 70.395$=dw

dvper troy ounce and 0.2dw

dt= troy

ounce per day. We seek the value of dvdt

:

dt

dw

dw

dv

dt

dv = = ($323.10 per troy ounce)(0.2 troy ounce per day)

= $64.62 per day The value of the investor’s gold is increasing at a rate of $64.62 per day. 4. Let t denote the number of days from today. Assuming that the gas in the tank is not yet being

used for customers, the amount of gas in the tank is g = –3.5t + 600 gallons, and the value of the gas in the tank is v = 1.51g dollars.

dv

dt

dv

dg

dg

dt=

= ($1.51 per gallon)(–3.5 gallons per day) = – $5.285 per day

The station is losing $5.29 of potential revenue per day. 5. a. R(476) = $10,000 Canadian

On November 25, 2002, 476 units of the commodity were sold, producing revenue of $10,000 Canadian.

b. C(10,000) = $633.47 On November 25, 2002, 10,000 Canadian dollars were worth 633.47 U.S. dollars.

c. $2.6dR

dx= Canadian per unit; On November 25, 2002, the revenue was increasing by 2.6

Canadian dollars per unit sold.

d. $0.6335dC

dr= U.S. per Canadian dollar; On November 25, 2002, the exchange rate was

$0.6335 U.S. per Canadian dollar.

e.

= ($0.6335 U.S. per Canadian dollar)($2.6 Canadian per unit)

= $1.65 U.S. per unit

dC dC dR

dx dr dx=

On November 25, 2002, revenue was increasing at rate of $1.65 U.S. per unit sold.



6. a. v(1) = 150 thousand pieces On the first day of this year, 150,000 pieces of mail were processed.

b. –0.2dv

dt= thousand pieces per day

The number of pieces of mail processed daily was decreasing by 200 pieces per day.

c. dE

dv= 12 hours per thousand pieces of mail. The rate of change of the number of

employee-hours is 12 hours per thousand pieces of mail.

d. dE dE dv

dt dv dt= = (12 hours per thousand pieces of mail)(–0.2 thousand pieces per day)

= -2.4 hours per day The number of daily employee-hours was decreasing by 2.4 hours per day.

7. a. In 2010, t = 10 and –0.02(10)

130(10) 12.009

1 12p

e= ≈

+ thousand people

In 2010 the city had a population of approximately 12,000 people.

b. g(p(10)) ≈ g(12.009) 32(12.009) – 0.001(12.009 )≈ ≈ 22 garbage trucks

In 2010 the city owned 22 garbage trucks.

c. Let –0.021 12 tu e= + and –0.02v t= . Then p u u ev= = +130 1 12–1 and .

′ = = =p tdp

dt

dp

du

du

dt

dp

du

du

dv

dv

dt( )

= +d

duu

d

dve

d

dttv( ) ( ) (–0. )–130 1 12 021 ( )= (–130 )( ) – .–2u ev12 0 02

= 3122

. e

u

v –0.02

–0.02 2

31.2 thousand people per year

(1 12 )

t

t

e

e=

+

′ =+

pe

e( )

.

( )

–0. ( )

– . ( )10

312

1 12

02 10

0 02 10 2 ≈ 0.22 thousand people per year

In 2010 the population was increasing at a rate of approximately 220 people per year.

d. ′ =g p p( ) – . ( )2 0 001 3 2

= 2 0 003 2– . p trucks per thousand people ′ ≈ ′g p g( ( )) ( . )10 12 009

22 – 0.003(12.009 ) 1.6= ≈ trucks per thousand people

In 2010 when the population was approximately 12,000, the number of garbage trucks needed by the city was increasing by 1.6 trucks per thousand people.



e. dg

dt

dg

dp

dp

dtg p p= = ′ ′( ( )) ( )10 10

≈ (1.6 trucks per thousand people)(0.22 thousand people per year) ≈ 0.34 trucks per year In 2010 the number of trucks needed by the city was increasing at a rate of approximately 0.34 truck per year, or 1 truck every 3 years.

f. See interpretations in parts a through e.

8. a. On November 25, 2002, sales were 476 mountain bikes, so

7779019.1)476( 476 ≈=p Canadian dollars. On November 25, 2002, sales of 476

mountain bikes produced profit of $7778.78 Canadian.

b. 65.4927$5786.1

019.1)7779()(

476

≈=≈ CpC

On November 25, 2002, a profit of $7779 Canadian was equivalent to $4928.

c. )019.1)(019.1(ln )(' xxp =

41.146$)019.1)(019.1(ln )476(' 476 ≈=p Canadian per mountain bike On November 25, 2002, when 476 mountain bikes were sold, the rate of change of profit was approximately $146.41 Canadian per mountain bike.

d. Because )(5786.1

1)( ppC

= , 5786.1

1=dp

dC U.S. dollars per Canadian dollar.

The exchange rate was 6335.05786.1

1 ≈ U.S. dollars per Canadian dollar.

e. )476('))476((' ppCdx

dp

dp

dC

dp

dC ==

≈ dollarper dollars 5786.1

1(146.41 dollars per mountain bike)

≈ $92.75 per mountain bike On November 25, 2002, the profit (in dollars) from the sale of mountain bikes was

increasing by approximately $92.75 per mountain bike.

f. See each part for an interpretation.

9. 2( ( )) 3(4 – 6 ) – 2c x t t=

dc

dt

dc

dx

dx

dt= = (6x)(–6) = 6(4 – 6t)(–6) = –144 + 216t



10. f t p e p( ( )) = 3 4 2

( )( )

( )( )2

2

4

4

3 8

3 8

24

t

p

p

df df dt

dp dt dp

e p

e p

pe

=

=

=

=

11. –0.5

4( ( ))

1 3 th p t

e=

+

Let 0.5 . Then 1 3uu t p e= = + .

dh

dt

dh

dp

dp

dt

dh

dp

dp

du

du

dt= =

= +d

dpp

d

duu

d

dtt( ) ( ) (–0. )–4 1 3 51 4

= (–4 )( )(–0. )–2p eu3 5

=6

2

e

p

u

= +6

1 3

5

0 5 2

e

e

t

t

–0.

– .( )

12. ( ( )) 7 5(4 ) 7 20w wg x w e e= + = +

Let u = 7 + 5x. Then12g u= .

dg

dw

dg

dx

dx

dw

dg

du

du

dx

dx

dw= =

= +d

duu

d

dxx

d

dwew( ) ( ) ( )

12 7 5 4

=

1

25 4

12u ew–

( )( )

=10e

u

w

=+

10

7 5

e

x

w=

+

10

7 5 4

e

e

w

w( )=

+

10

7 20

e

e

w

w



Alternate solution for the derivative:

Use u x ew= + = +7 5 7 20 . As before, g u=12 .

dg

dw

dg

du

du

dw=

= = +d

duu

d

dwew( ) ( )

12 7 20

=

1

220

12u ew–

( ) = 10e

u

w=

+

10

7 20

e

e

w

w

13. k t x x x x( ( )) . (ln ) (ln ) ln= − + −4 3 2 4 123 2

[ ] ( )dk

dxx

xx

x x= − + −4 3 3

12 2

14

102. (ln ) ln

( )212.9 ln 4ln 4x x

x x x= − +

14. f x t t( ( )) ln( )= +5 11

df

dt t t=

+=

+1

5 115

5

5 11( )( )

15. p t k k k( ( )) . ( . )= −7 9 104614 3 12 2

3 214 12 27.9(ln1.046)(1.046) (42 24 )k kdp

k kdk

−= −

16. r m ff f

f f f f f f( ( )).

( )( ) . ( )=

−+ − = − + −−91

3 91 3 34 2 2

4 2 4 2 2 4 2

[ ]dr

dff f f f

f ff f= − − + − = −

−+ −−91 2 12 6

18 212 64 2 3 3

4 2 33. ( )

.

( )

17. Inside function: u = 3.2x + 5.7 Outside function: f = u5

df

dx

df

du

du

dx

d

duu

d

dxx= = +( ) ( . . )5 32 57

= ( )( . )5 324u

= +5 32 57 324( . . ) ( . )x = +16 32 574( . . )x



18. Inside function: u x x= + +5 3 72 Outside function: f u= –1

df

dx

df

du

du

dx

d

duu

d

dxx x= = + +( ) ( )–1 25 3 7

= (– )( )–2u x10 3+

2 2

(10 3)

(5 3 7)

x

x x

− +=+ +

19. Inside function: u = x – 1 Outside function: fu

= 83 = 8u–3

df

dx

df

du

du

dx=

= d

duu

d

dxx( ) ( – )–8 13

= (–24 )( )–4u 1

= – 244u

= –

( – )

24

1 4x

20. Inside function: u = 4x + 7 Outside function: fu

u= =350350 1–

df

dx

df

du

du

dx=

= +d

duu

d

dxx( ) ( )–350 4 71

= (–350 )( )–2u 4

= +–350( ) ( )–24 7 4x

2

1400

(4 7)x

−=+

21. Inside function: u x x= 2 3– Outside function: f u u= =12

df

dx

df

du

du

dx

d

duu

d

dxx x= = ( ) ( – )

12 2 3

=

1

22 3

12u x

–( – )

=2 3

2

x

u

– = 2 3

2 32

x

x x

–

–



22. Inside function: u x x= +2 5 Outside function: f u u= =313

df

dx

df

du

du

dx=

= +d

duu

d

dxx x( ) ( )

13 2 5

=

+1

32 5

23u x

–( )

=+2 5

3 23

x

u

=+

+

2 5

3 52 23

x

x x( )= +

+

2 5

3 52 23

x

x x( )

23. Inside function: u = 35x Outside function: f = ln u

df

dx

df

du

du

dx=

=d

duu

d

dxx(ln ) ( )35

=

135

u( ) = 1

3535

x( ) = 1

x

24. Inside function: u = ln 6x = ln 6 + ln x Outside function: f u= 2 df

dx

df

du

du

dx

d

duu

d

dxx= = +( ) (ln ln )2 6

=( )2

1u

x

=( ln )2 6

1x

x

2ln 6x

x=

25. Inside function: u x x= +16 372 Outside function: f = ln u

2(ln ) (16 37 )

1(32 37)

df df du d du x x

dx du dx du dx

xu

= = +

= +

= ++1

16 3732 37

2x xx( )

= ++

32 37

16 372

x

x x



26. Inside function: u = 3.7x Outside function: f eu=

( ) (3.7 )udf df du d de x

dx du dx du dx= =

( )(3.7)ue=

3.73.7 xe=

27. Inside function: u = 0.6x Outside function: 72 uf e=

df

dx

df

du

du

dx=

(72 ) (0.6 )ud de x

du dx=

(72 )(0.6)ue=

0.6(72 )(0.6)xe=

0.643.2 xe=

28. Inside function: u x= 4 2 Outside function: f eu=

df

dx

df

du

du

dx

d

due

d

dxxu= = ( ) ( )4 2

= ( )( )e xu 8

= 8 4 2

xe x

29. Inside function: u = 0.08x Outside function: 1 58 uf e= +

df

dx

df

du

du

dx=

(1 58 ) (0.08 )ud de x

du dx= +

(58 )(0.08)ue=

0.08(58 )(0.08)xe= 0.084.64 xe=

30. Inside function: u = 1 + 3x Outside function: 1 58 uf e= +

df

dx

df

du

du

dx=

(1 58 ) (1 3 )ud de x

du dx= + +

(58 )(3)ue=

( )(1 3 )

(1 3 )

58 (3)

174

x

x

e

e

+

+

=

=



31. Inside function: 0.6 inside: 0.61 18

outside: 1 18x

w

w xu e

u e

== + = +

Outside function: 12

7.3fu

= + 112 7.3u−= +

df

dx

df

du

du

dx

df

du

du

dw

dw

dx= =

1(12 7.3) (1 18 )wd du e

du dw−= + + (0.6 )

dx

dx

2( 12 )(18 )(0.6)wu e−= −

0.6 2 0.612(1 18 ) (18 )(0.6)x xe e−= − +

0.6

0.6 2

129.6

(1 18 )

x

x

e

e

−=+

32. Inside function: 1.2 inside: 1.21 8.9

outside: 1 8.9x

w

w xu e

u e− = −

= + = +

Outside function: 137.589 37.5 89f u

u−= + = +

df

dx

df

du

du

dx

df

du

du

dw

dw

dx= =

1(37.5 89) (1 8.9 ) ( 1.2 )wd d du e x

du dw dx−= + + −

2( 37.5 )(8.9 )( 1.2)wu e−= − −

1.2 2 1.237.5(1 8.9 ) (8.9 )( 1.2)x xe e− − −= − + −

1.2

1.2 2

400.5

(1 8.9 )

x

x

e

e

−

−=+

33. Inside function: u = 123 3x x x x− = − Outside function: 3f u=

df

dx

df

du

du

dx=

= −d

duu

d

dxx x( ) ( )3

12 3

= −−

( )( )3 32 12

12u x

1 12 22 1

23( 3 ) ( 3)x x x−

= − − 2 13( 3 ) 3

2x x

x

= − −



34. Inside function: 2u x= Outside function: f u= 3

df

dx

df

du

du

dx=

( )( )( )

2

(3 ) 2

3 2 2 hint: 2 2 2

3 2

3 2

u

u

u

x

d dx

du dx

x

x

x

=

= ⋅ =

=

=

35. Inside function: u x= ln Outside function: f u= 2

df

dx

df

du

du

dx=

=d

du

d

dxxu( ) (ln )2

= (ln )2 2

1u

x

= (ln ) ln2 2

1x

x

= (ln ) ln2 2 x

x

36. Inside function: u = 2x Outside function: f = lnu

df

dx

df

du

du

dx=

=d

duu

d

dxx(ln ) ( )2

= 1

2 2u

x(ln ) = 1

22 2

xx(ln )

= ln2

37. Inside function: u = − Bx Outside function: f = Aeu

df

dx

df

du

du

dx=

( ) ( )ud dAe Bx

du dx= −

( )uAe B= −

BxABe−= −



38. Inside function: u Ae Bx= + −1 inside: =

outside: = 1 w

w Bx

u Ae

−

+

Outside function: fL

uLu= = −1

df

dx

df

du

du

dx

df

du

du

dw

dw

dx= =

= + −−d

duLu

d

dwAe

d

dxBxw( ) ( ) ( )1 1

= − −−( )( )( )Lu Ae Bw2

2(1 ) ( )( )Bx BxL Ae Ae B− − −= − + −

=+

−

−LABe

Ae

Bx

Bx( )1 2

39. a. 2( ( )) 0.75 –2.3 53.2 249.8 1.8S u x x x= + + + millions of dollars is the predicted sales for a

large firm, where x is the number of years in the future.

b. d

dxS u x

dS

du

du

dx( ( )) = = + + +d

duu

d

dxx x( . . ) (–2. . . )0 75 18 3 532 249 8

12 2

[ ]=

+ +0 751

22 3 2 532 0

12. – . ( ) .

–u x = +0 375 6 532. (–4. . )x

u

= +

+ +

– . .

– . . .

1725 19 95

2 3 532 249 82

x

x x millions of dollars per year is the rate of change in

predicted sales for a large firm, where x is the number of years in the future.

c. Answer will vary depending on the current year. If the year is 2007, then the answer for 2007 would be obtained by substituting x = 0 in the expression from part b, so the rate of change would be approximately 1.26 million dollars per year.

40. a. 0.09263( ) 0.216(0.09263) tp t e′ = percentage points per year is the rate of change in the rate of

change in the percentage of children living with their grandparents between 1970 and 2000, t years after 1970.

b. 0.09263(25)(25) 0.216(0.09263) 0.203p e′ = ≈ percentage points per year

c. (20) (0) 4.377 3.216 1.161

0.05820 0 20 20

p p− −≈ ≈ ≈−

percentage points per year



d.

The answer to part b is the slope of the line tangent to the graph of p at t = 25. The answer to part c is the slope of the secant line connecting the points on the graph of p corresponding to t = 0 and t = 20.

41. a. 0.62285( ) 0.0314(0.62285) qR q e′ = million dollars/quarter is the rate of change in the

revenue q quarters after the start of 1998.

b. Quarter Ending June 1998 June 1999 June 2000 R q( ) million dollars 3.0 4.2 18.8 ′R q( ) million dollars per quarter 0.07 0.82 9.92

( )( ) 100R q

R q′ ⋅ % per quarter 2.3 19.5 52.7

42. a. We use the formula developed in Activity 38: 2

( )(1 )

Bt

Bt

LABeP t

Ae

−

−′ =

+

with L = 62.7, A = 38.7, and B = 0.258.

2258.0

258.0

)7.381(

)258.0)(7.38)(7.62()('

t

t

e

etP

−

−

+=

2258.0

258.0

)7.381(

034.626t

t

e

e−

−

+≈ percentage points per year is

the rate of change in the percentage of households with TVs who subscribe to cable in the years from 1970 through 2002 where t is the number of years since 1970.

b. 2)30(258.0

)30(258.0

)7.381(

034.626)30(' −

−

+≈

e

eP ≈ 0.26 percentage point per year

c. The model predicts that, eventually, the percentage of households with TVs who subscribe to cable will be 62.7%. One possible answer concerning the reasonableness of this long-term behavior.


( )(1 )

Bx

Bx

LABem x

Ae

−

−′ =

+

with L = 49, A = 36.0660, and B = 0.206743.

0.206743 0.206743

2 0.206743 2 0.206743 2

(49)(36.0660)(0.206743) 365.363( )

(1 ) (1 36.0660 ) (1 36.0660 )

Bx x x

Bx x x

LABe e em x

Ae e e

− − −

− − −′ = = ≈+ + +

states per year is the rate of change of states with national P.T.A. membership, where x is the number of years since 1895, 0 ≤ x ≤ 36.

b. 0.206743(7)

49(7)

1 36.0660m

e−=+

≈ 5 states



c. 1915: ′m ( )20 ≈ 2.4 states per year 1927: ′m ( )32 ≈ 0.4 state per year


( )(1 )

Bt

Bt

LABec t

Ae

−

−′ =

+with L = 93,700,

A = 5095.9634, and B = 1.097175.

1.097175

2 1.097175 2

–1.097175

–1.097175 2

(93,700)(5095.9634)(1.097175)( )

(1 ) (1 5095.9634 )

523,892,033 deaths per week weeks after August 31, 1918

(1 5095.9634 )

Bt t

Bt t

t

t

LABe ec t

Ae e

et

e

− −

− −′ = =

+ +

≈+

–1.097175(4)

–1.097175(4) 2

523,892,033(4) 1575 deaths per week

(1 5095.9634 )

ec

e′ = ≈

+

b. (4) 1575 deaths per week

100% 100% 108% increase per week(4) 1458 deaths

c

c

′⋅ ≈ ⋅ ≈

c. (8) 25,331c′ ≈ deaths per week

(8) 25,331 deaths per week100% 100% 48% increase per week

(8) 52,473 deaths

c

c

′⋅ ≈ ⋅ ≈

d. Although the rate of change is larger, it represents a smaller proportion the total number of deaths that had occurred at that time.

45. a. ( )( ) 13865.113 1.035xf x = dollars is the projected tuition at a private 4-year college for the

years between 2000 and 2010, x years after 2000.

b. Because ae a ekx k x= ( ) , we solve for k in the equation 1.035ke = . Therefore,

k = ln 1.035 ≈ 0.0344 and 0.0344( ) 13865.113 xf x e≈

c. For the first formula:

( )'( ) 13865.113 1.035xd

f xdx

= ( )( )13865.113 ln1.035 1.035x=

≈ 476.98(1.035 )x dollars per year

For the second formula: Let u = 0.0344x. Then 13865.113uf e≈ .

df

dx

df

du

du

dx=

(13865.113 ) (0.0344 )ud de x

du dx= (13865.113 )(0.0344)ue= 476.98 ue≈

0.0344476.98 xe≈ dollars per year

d. Using the first model: 8'(8) 476.98(1.035 ) 632f ≈ ≈ dollars per year

Using the second model: ( )0.0344 8'(8) 476.98 632f e ⋅≈ ≈ dollars per year

The answers are identical.



46. a. A logistic model is probably a better model because of the leveling-off behavior, although

neither model should be used to extrapolate beyond one 24-hour day.

b. C te t

( ).

. .=

+ −1342 077

1 36797 0 258856 calls gives the total number of calls received since 5 A.M.,

t hours after 5 a.m., 3 ≤ t ≤ 24.

c. We use the formula developed in Activity 38: 2

( )(1 )

Bt

Bt

LABeC t

Ae

−

−′ =

+

with L = 1342.077, A = 36.797, and B = 0.258856.

0.258856 0.258856

2 0.258856 2 0.258856 2

(1342.077)(36.797)(0.258856) 12,783.365( )

(1 ) (1 36.797 ) (1 36.797 )

Bt t t

Bt t t

LABe e eC t

Ae e e

− − −

− − −′ = = ≈+ + +

calls per hour gives the rate of change in the total number of calls received since 5 A.M., t hours after 5 a.m., 3 ≤ t ≤ 24.

d. Noon: ′C ( )7 ≈ 42 calls per hour 10 p.m.: ′C ( )17 ≈ 74 calls per hour Midnight: ′C ( )19 ≈ 58 calls per hour 4 a.m.: ′C ( )23 ≈ 28 calls per hour

e. The rates of change give approximate hourly calls. This information could be used to determine how many dispatchers would be needed each hour.

47. a. The data are essentially concave up, suggesting a quadratic or exponential function. We

choose a quadratic model because it fits the data slightly better than an exponential model. 2( ) 7.763 47.447 1945.893t x x x= + + units is the average weekly production cost at a

manufacturing company during the xth quarter after January 1, 2000, for the period from January 2000 through December 2003.

b. 2( ( )) 196.25 44.45ln(7.763 47.447 1945.893)C t x x x= + + + dollars is the weekly production cost at a manufacturing company during the xth quarter after January 1, 2000, for the period from January 2000 through December 2003.

c. Jan–Mar 2004: C(t(17)) ≈ $574.81 per week Apr–June 2004: C(t(18)) ≈ $577.56 per week July–Sept 2004: C(t(19)) ≈ $580.27 per week Oct–Dec 2004: C(t(20)) ≈ $582.94 per week



d.

According to the graph, the cost appears never to decrease between January of 2000 and January of 2005.

e. d

dxC t x

dC

dt

dt

dx( ( )) =

= + + +d

dtt

d

dxx x( . . ln ) ( . . . )196 25 44 25 7 763 47 447 19458932

= + +44 25

115525 47 447 0. ( . . )

tx

=+

+ +44 25 15525 47 447

7763 47447 19458932

. ( . . )

. . .

x

x x dollars per quarter is the rate of change of

weekly manufacturing costs at a manufacturing company during the xth quarter after January 1, 2000, for the period from January 2000 through December 2003.

According to the graph, the rate of change is always positive. No, the cost will never decrease.

48. a. The data are essentially concave up, which indicates that a quadratic or exponential model

may be appropriate. After comparing the fit of both models, the quadratic model appears to fit the original data set best. An exponential model may fit better if the output data was shifted down.

b. 2( ) 177.356 – 342.240 5914.964u x x x= + units per week is the company’s production x

years after 1990, 0 ≤ x ≤ 13.

c. 2( ( )) 3250.23 74.95 ln(177.356 – 342.240 5914.964)C u x x x= + + dollars per week is the

company’s cost per week, x years after 1990, 0 ≤ x ≤ 13.

d. dC

dx

dC

du

du

dx=

= + +d

duu

d

dxx x( . . ln ) ( . – . . )3250 23 74 95 177 356 342 240 5914 9642



=

+74 95

354 712 342 240 0.

( . – . )u

x

2

74.95(354.712 – 342.240) dollars per week

177.356 – 342.240 5914.964 per yearx

x x

= +

is the rate of change of the company’s weekly cost x years after 1990, 0 ≤ x ≤ 13.

e. Use the expressions from parts c and d.

Year 2002 2003 2004 2005

x 12 13 14 15

C(u(x)) ($/week)

4015.95 4026.40 4036.31 4045.72

dCdx

($/week/year) 10.73 10.18 9.66 9.17

f, g.

Although a graph of C may not appear ever to decrease, a graph of dC

dx is negative

between x = 0 and x ≈ 0.965, indicating that cost was decreasing from the end of 1990 to (almost) the end of 1991. A close-up view of the graph of C between x = 0 and x = 2 confirms this.

49. One possible answer: Composite functions are formed by making the output of one function (the inside) the input of another function (the outside). It is imperative that the output of the inside and the input of the outside agree in the quantity that they measure as well as in the units of measurement.

50. One possible answer: The derivative rules tells us that

( ) ( ) ( ) ( ) 0bx bx bx bxd d d dae c ae c a e abe

dx dx dx dx+ = + = + = .

If a > 0 and b > 0 then the slope graph dy

dx will be an increasing, concave up, exponential

function with limx

dy

dx→∞= ∞ and lim 0.

x

dy

dx→−∞=



If a > 0 and b < 0 then the slope graph dy

dx will be an increasing, concave down, exponential

function with lim 0x

dy

dx→∞= and lim .

x

dy

dx→−∞= −∞

If a < 0 and b > 0 then the slope graph dy

dx will be a decreasing, concave down, exponential

function with limx

dy

dx→∞= −∞ and lim 0.

x

dy

dx→−∞=

If a < 0 and b < 0 then the slope graph dy

dx will be a decreasing, concave up, exponential

function with lim 0x

dy

dx→∞= and lim .

x

dy

dx→−∞= ∞

Section 3.5 The Product Rule 1. (2) (2) (2) (2) (2) 1.5(4) 6(3) 12h f g f g′ ′ ′= + = − + = 2. (100) (100) (100) (100) (100)r p q p q′ ′ ′= + = 0.5(160) + (4.65)(12) = 135.8 3. a. i. In 2007 there were 75,000 households in the city.

ii. In 2007 the number of households was declining at a rate of 1200 per year.

iii. In 2007, 52% of households owned a computer.

iv. In 2007 the percentage of households with a computer was increasing by 5 percentage points per year.

b. Input: the number of years since 1995 Output: the number of households with computers

c. N(2) = h(2)c(2) = (75,000)(0.52) = 39,000 households with computers. ′ = ′ + ′N h c h c( ) ( ) ( ) ( ) ( )2 2 2 2 2 = (–1200)(0.52) + (75,000)(0.05)

= 3126 households per year In 2007 there were 39,000 households with computers, and that number was increasing at a rate of 3126 households per year.

4. a. i. When the price is $6.25, there is demand for 1000 units.

ii. When the price is $6.25, the demand is changing at a rate of –50 units per dollar. (A price increase of $1.00 should reduce demand by approximately 50 units.)

b. R(x) = x D(x)

c. ( ) ( ) ( ) ( ) ( ) ( )d d

R x x D x x D x D x xD xdx dx ′ ′= + = +

′ = + ′R D D( . ) ( . ) . ( . )6 25 6 25 6 25 6 25 = 1000 + 6.25(–50) = 687.5 When the price is $6.25, the revenue is increasing at a rate of $687.50 per dollar of price. (Increasing the price by $1.00 should increase revenue by approximately $687.50.)



5. a. i. S( ).

$15.10 152 6

10 124= +

+≈

′ = + +S xd

dxx( ) [ . ( ) ]–15 2 6 1 1 = + + +0 2 6 1 12. (–1)( ) ( )–x

d

dxx

=+

– .

( )

2 6

1 2x dollars per week

2

–2.6(10) –$0.02

(10 1)S ′ = ≈

+per week

After 10 weeks, 1 share is worth $15.24, and the value is declining by $0.02 per week.

ii. 2(10) 100 0.25(10 ) 125N = + = shares

2( ) (100 0.25 ) 0.5d

N x x xdx

′ = + = shares per week after x weeks

(10) 0.5(10) 5N ′ = = shares per week After 10 weeks, the investor owns 125 shares and is buying 5 shares per week.

iii. V(10) = S(10)N(10) ≈ (15.24)(125) ≈ $1904.55

(10) (10) (10) (10) (10)V S N S N′ ′ ′≈ + ≈ (–0.02)(125) + (15.24)(5) ≈ $73.50 per week After 10 weeks, the investor’s stock is worth approximately $1905, and the value is increasing at a rate of $73.50 per week.

b.

22

2

2

( ) ( ) ( ) ( ) ( )

–2.6 2.6(100 0.25 ) 15 (0.5 )

1( 1)

0.65 260 1.3– 7.5 dollars per week after weeks

1( 1)

V x S x N x S x N x

x xxx

x xx x

xx

′ ′ ′= +

= + + + + +

+= + +++

6. a. Input: number of years since 2002 Output: yearly cost to educate all students at the school

b. i. S(3) = 100 ln(3 + 5) ≈ 208 students. In 2005 there were 208 students.

ii. ( ) 100ln( 5)d

S t tdt

′ = +1

100 ( 5)5

dt

t dt= +

+=

+100

5t

100(3) 12.5

3 5S ′ = =

+students per year.

In 2005 the number of students was increasing at a rate of 12.5 students per year.

iii. C( ) ( . ) $1736.3 1500 105 443= ≈ In 2005, the yearly cost to educate one student was approximately $1736.44.

Calculus Concepts Section 3.5: The Product Rule 123


iv. ( ) [1500(1.05) ] 1500(ln1.05)(1.05)t tdC t

dt′ = =

′ =C ( ) (ln . )( . )3 1500 105 1053 ≈ $84.72 per year In 2005 the yearly cost to educate one student was increasing at a rate of approximately $84.72 per year.

v. F(3) = S(3)C(t) ≈ (208)($1736.44) ≈ $361.000 In 2005 the yearly cost to educate all of the students at the school was approximately $361,000.

vi. (3) (3) (3) (3) (3)F S C S C′ ′ ′= + ≈ (12.5)(1736.44) + (208)(84.72) ≈ $39,300 per year In 2005, the yearly cost to educate all of the students at the school is increasing at a rate of approximately $39,300 per year.

c. ( ) ( ) ( ) ( ) ( )

100[1500(1.05 )] [100ln( 5)][1500(ln1.05)(1.05 )]

5

1150,000(1.05 ) (ln1.05) ln( 5) dollars per year years after 1996

5

t t

t

F t S t C t S t C t

tt

t tt

′ ′ ′= +

= + + +

= + + +

7. Let A(t) be the number of acres of corn and let B(t) be the number of bushels of corn per acre,

where t is the number of years from the current year. The total number of bushels of corn produced is given by C(t) = A(t)B(t), where A(0) = 500 acres, (0) 50A′ = acres per year,

B(0) = 130 bushels per acre, and (0) 5B′ = bushels per acre per year. The rate of change is:

(0) (0) (0) (0) (0)

= (50 acres/year)(130 bushels/acre) + (500 acres)(5 bushels/acre/year)

= 9000 bushels per year

C A B A B′ ′ ′= +

8. Let f(x) be the number of free throw opportunities in game x, and let p(x) be the proportion

(expressed as a decimal)of free throws that are made in game x. If the current game is game 0, we know that (0) 15, (0) 0.72, (0) –1f p f ′= = = , and (0) 0.005p′ = . Because each free throw is worth one point, the number of free throw points that this point guard makes in a game is found by calculating N(x) = f(x)p(x). The rate of change is:

(0) (0) (0) (0) (0)N f p f p′ ′ ′= + = (–1)(0.72) + (15)(0.005) = –0.645 free throw points per game

9. a. (17,000)(0.48) = 8160 voters

b. (8160)(0.57) ≈ 4651 votes for candidate A

c. Let v(t) be the proportion of registered voters who plan to vote, and let a(t) be the proportion who support candidate A, where t is the number of weeks from today and both quantities are expressed as decimals. Then the number of votes for candidate A is given by

N(t) = 17,000v(t)a(t), where v(0) = 0.48, a(0) = 0.57, (0) 0.07v′ = and (0) –0.03a′ = . The rate of change of N is:

( ) [17,000 ( ) ( )] 17,000 [ ( ) ( )] 17,000[ ( ) ( ) ( ) ()]d d

N t v t a t v t a t v t a t v t a tdt dt

′ ′ ′= = = +

When t = 0,

(0) 17,000[(0.07)(0.57) (0.48)(–0.03)] 434 votes for candidate A per weekN ′ = + ≈



10. Let A(t) be the proportion of passengers that pass by the store at the end of the tth year. Let P(t) be the number of passengers that pass by the store at the end of the tth year.

Given in the problem are:

.05%(1) 2% A'(1)

yearA = =

114 passengers(1) 52000 passengers P'(1)

dayP = =

The number of customers at the shop: N(t) = A(t) ⋅ P(t) The rate of change of the number of customers at the shop: N′(t) = A′ (t) ⋅ P(t) + A (t) ⋅ P′ (t)

'(1) '(1) (1) (1) '(1)

0.05% 114 passengers= 52000 passengers +2%

day day

= 28.28 passengers per day

N A P A P= ⋅ + ⋅

⋅ ⋅

Note: misprint in text “0.05 percent per year” should be “0.05 percent per day”

11. ( ) (ln ) (ln )x xd df x x e x e

dx dx ′ = +

1(ln )

(ln )

x x

xx

e x ex

ex e

x

= +

= +

12. ( ) ( 5) ( 5)x xd df x x e x e

dx dx ′ = + + +

1 ( 5)

( 6)

x x

x

e x e

x e

= + +

= +

13. 2 3 2 3

3 2 2

4 3 2

( ) (3 15 7) (32 49) (3 15 7) (32 49)

(6 15)(32 49) (3 15 7)(96 )

480 1920 672 294 735

d df x x x x x x x

dx dx

x x x x x

x x x x

′ = + + + + + + +

= + + + + +

= + + + +

14. ′ =f xd

dxxx( ) . [( . )(ln )]2 5 0 9

2.5 0.9 ln 0.9 lnx xd dx x

dx dx

= +

= +

2 5 0 9 0 9 0 9

1. (ln . )( . )(ln ) ( . )x xx

x

= +

2 5 0 9

10 9. ( . ) (ln . )(ln )x

xx



15. 2( ) (12.8 3.7 1.2) [29(1.7 )]xdf x x x

dx ′ = + +

2(12.8 3.7 1.2) [29(1.7 )]xdx x

dx+ + +

(25.6 3.7)[29(1.7 )]xx= + 2(12.8 3.7 1.2)[29(ln1.7)(1.7 )]xx x+ + +

16. 5 5( ) (5 29) (15 8) (5 29) (15 8)d d

f x x x x xdx dx

′ = + + + + +

4 55( 29) (5 29) (15 8) (5 29) (15)

dx x x x

dx = + + + + +

= + + + +25 29 15 8 15 5 294 5( ) ( ) ( )x x x

17. Note that f(x) = g(x)h(x), where g(x) 2 3(5.7 3.5 2.9) and x x= + + 2 –2( )=(3.8 5.2 7)h x x x+ + .

2 2 2( ) 3(5.7 3.5 2.9) (5.7 3.5 2.9)d

g x x x x xdx

′ = + + + +

= + + +3 57 35 2 9 114 352 2( . . . ) ( . . )x x x

2 –3 2( ) –2(3.8 5.2 7) (3.8 5.2 7)d

h x x x x xdx

′ = + + + +

= − + + +2 38 52 7 7 6 522( . . ) ( . . )–3x x x

′ = ′ + ′f x g x h x g x h x( ) ( ) ( ) ( ) ( )

2 2 2 –2[3(5.7 3.5 2.9) (11.4 3.5)](3.8 5.2 7)x x x x x= + + + + +

2 3 2 –3(5.7 3.5 2.9) [–2(3.8 5.2 7) (7.6 5.2)]x x x x x+ + + + + +

18. Note that f(x) = g(x)h(x), where 3 –1( ) 2 3 and ( ) (2.7 15)g x x h x x= + = +

2( ) 6g x x′ =

–2( ) –(2.7 15) (2.7 15)d

h x x xdx

′ = + +2

–2.7

(2.7 15)x=

+

( ) ( ) ( ) ( ) ( )f x g x h x g x h x′ ′ ′= +

2 –1 32

–2.7(6 )(2.7 15) (2 3)

(2.71 15.29)x x x

x

= + + + +

2 3

2

(6 ) –2.7(2 3)

(2.7 15) (2.71 15.29)

x x

x x

+= + + +

19. –2

–2

–2 –2

( ) [12.6(4.8 )( )]

12.6 [(4.8 )( )]

12.6 (4.8 ) (4.8 )

x

x

x x

df x x

dxd

xdx

d dx x

dx dx

′ =

=

= +



–2 –3

2 3 3

12.6 (ln4.8)(4.8 )( ) (4.8 )(–2 )

ln4.8 2 12.6(4.8 )12.6(4.8 ) – [ (ln4.8) – 2]

x x

xx

x x

xx x x

= +

= =

20. Note that f(x) = g(x)h(x), where 2 –0.09 –1( ) 8 13 and ( ) 39(1 15 )xg x x h x e= + = + . ′ =g x x( ) 16

Using the formula for the derivative of a logistic function, we know that

2

( )(1 )

Bx

Bx

LABeh x

Ae

−

−′ =

+ where L = 39, A = 15, and B = 0.09.

Thus –0.09

–0.09 2

52.65( )

(1 15 )

x

x

eh x

e′ =

+.

( ) ( ) ( ) ( ) ( )f x g x h x g x h x′ ′ ′= +

–0.09

–0.09 –1 2–0.09 2

52.65(16 )39(1 15 ) (8 13)

(1 15 )

xx

x

ex e x

e= + + +

+

–0.09 2

–0.09 –0.09 2

624 (52.65 )(8 13)

1 15 (1 15 )

x

x x

x e x

e e

+= ++ +

21. Note that f(x) = g(x)h(x), where g(x) = 79x,

h(x) –0.85 –1198(1 7.68 ) 15xe= + + , and ( ) 79g x′ =

Using the formula for the derivative of a logistic function, we know that

2

( )(1 )

Bx

Bx

LABeh x

Ae

−

−′ =

+ where L = 198, A = 7.68, and B = 0.85

Thus –0.85

–0.85 2

198(7.68)(0.85)( )

(1 7.68 )

x

x

eh x

e′ =

+.

–0.85

–0.85 –0.85 2

–0.85

–0.85 –0.85 2

( ) ( ) ( ) ( ) ( )

198 198(7.68)(0.85)79 15 (79 )

1 7.68 (1 7.68 )

15,642 102,110.9761,185

1 7.68 (1 7.68 )

x

x x

x

x x

f x g x h x g x h x

ex

e e

xe

e e

′ ′ ′= +

= + + + +

= + ++ +

22. Note that f(x) = g(x)h(x), where 33 15.7( ) ln(15.7 ) and ( ) xg x x h x e= = .

′ =g xd

dxx( ) ln( . )157 3

= +d

dxx(ln . ln )157 3 = 3

x



′ =h xd

dxe x( ) .15 7 3 ( )315.7 315.7x d

e xdx

=

=

e xx15 7 23

471. ( . ) = 471 2 15 7 3. .x e x

′ = ′ + ′f x g x h x g x h x( ) ( ) ( ) ( ) ( )

= +3

157 47115 7 3 2 15 73 3

xe x x ex x. .ln( . )( . )

= +

e

xx xx15 7 2 33 3

471 157. . ln( . )

23. Note that f(x) = g(x)h(x), where g(x)1

430(0.62 ) and ( ) 6.42 3.3(1.46 )x xh x−

= = +

.

′ =g x x( ) (ln . )( . )430 0 62 0 62

2

2

3.3(ln1.46)(1.46 )( ) 6.42 3.3(1.46 ) 3.3(ln1.46)(1.46 )

6.42 3.3(1.46 )

xx x

xh x

− − ′ = − + =

+

′ = ′ + ′f x g x h x g x h x( ) ( ) ( ) ( ) ( )

= + −

+ +430 0 62 0 62

1430 0 62

33 146 146

6 42 3 3 146 6 42 3 3 1 462

(ln . )( . ) ( . ). (ln . )( . )

. . ( . ) . . ( . )

x xx

xx

24. Note that ( ) ( ) ( )f x g x h x= where g x x( ) ln= +19 12 2 and h x x( ) ln= −17 3 4 .

′ = =g xx x

( ) ( )12

22

12 and ′ = − = −

h xx x

( ) ( )3

44

3

′ = ′ + ′f x g x h x g x h x( ) ( ) ( ) ( ) ( ) 2 3

( ) (17 3ln 4 ) (19 12ln 2 )

204 36ln 4 36ln 2 57 261 36ln 4 36ln 2

f x x xx x

x x x x

x x

−′ = − + +

− − − − − − −= =

25. 12

12

( ) 4 3 2 4 (3 2) 93

14 3 2 4 (3 2) (3) 0

2

64 3 2

3 2

d d df x x x x x

dx dx dx

x x x

xx

x

−

′ = + + + +

= + + + +

= + ++

26. Note that ( ) ( ) ( )f x g x h x= where g x x( ) ( )= 4 3 and h x x( ) =−1

2 .

′ =g x x( ) (ln )( )4 3 3 and ′ = − −h x x( )

1

2

32



′ = ′ + ′f x g x h x g x h x( ) ( ) ( ) ( ) ( ) = + −

−4 3 3

14 3

1

2

32(ln )( ) ( )x x

xx = + −4 3 3 2 3

3

(ln )( ) ( )x x

x x

27. Note that ( ) ( ) ( )f x g x h x= where ( )g x x= and h is a logistic function with L = 14,

A = 12.6, and B = 0.73 and derivative of the form 2

( )(1 )

Bx

Bx

LABeh x

Ae

−

−′ =

+:

( )0.73

20.73

128.772( )

1 12.6

x

x

eh x

e

−

−′ =

+

( )

( )

0.73

0.73 20.73

0.73

0.73 20.73

14 128.772( ) ( ) ( ) ( ) ( ) (1) ( )

1 12.6 1 12.6

14 128.772

1 12.6 1 12.6

x

x x

x

x x

ef x g x h x g x h x x

e e

xe

e e

−

− −

−

− −

′ ′ ′= + = ++ +

= ++ +

28. Note that ( ) ( ) ( )f x g x h x= where g x x( ) ( )= − −2 2 and h x x x( ) = − +3 17 42 .

′ = − − = −−

−g x xx

( ) ( ) ( )( )

2 2 12

2

33

and ′ = −h x x( ) 6 17

′ = ′ + ′f x g x h x g x h x( ) ( ) ( ) ( ) ( )

= −−

− + +−

−2

23 17 4

1

26 17

32

2( )( )

( )( )

xx x

xx =

− + −−

+ −−

6 34 8

2

6 17

2

2

3 2

x x

x

x

x( ) ( )

29. a. 20.73(1.2912 ) 8( ) ( 0.026 3.842 538.868)

100

x

E x x x+= − − + women received

epidural pain relief during childbirth at an Arizona hospital between 1981 and 1997, x years after 1980.

0.73(1.2912 ) 8

( ) ( 0.052 3.842)100

x

E x x+′ = − − +

20.73(ln1.2912)(1.2912 )( 0.026 3.842 538.868)

100

x

x x− − + women per year is the rate of

change in the number of women who received epidural pain relief during childbirth at an Arizona hospital between 1981 and 1997, x years after 1980.

b. Increasing by (17) 14.4p′ ≈ percentage points per year

c. Decreasing by approximately 5 births per year (′ ≈ −b ( ) . )17 4 7

d. Increasing by (17) 64E′ ≈ women per year.

e. Profit = $57 (17) $17,043E⋅ ≈ (using a value of 299 for the number of births)

or $17,071 (using an unrounded number of births)



30. a. P(m) = 0.04m + 1.10 dollars per package is the price of a certain brand of tissue paper during the mth month of the year.

2( ) –0.95 0.24 279.91S m m m= + + packages of tissue paper are sold during the mth month

of the year.

b. 2( ) ( ) ( ) (0.04 1.10)(–0.95 0.24 279.91)R m P m S m m m m= = + + + dollars is the revenue from

the sale of packages of tissue paper during the mth month of the year.

c.

2'( ) '( ) ( ) ( ) '( ) (0.04)(–0.95 0.24 279.91) (0.04 1.10)(–1.9 0.24)R m P m S m P m S m m m m m= + = + + + + +dollars per month is the rate of change in the revenue from the sale of packages of tissue paper during the mth month of the year.

d. February: (2) $6.86R′ ≈ per month

August: (8) $12.40R′ ≈ − per month

September: (9) $16.41R′ ≈ − per month

31. a. Multiply the number of CDs, 6250(0.9286 )x , by the price, x.

( ) 6250 (0.9286 )xR x x≈ dollars is the monthly revenue from CD sales when the price of a

CD is x dollars.

b. The profit for each CD is x – 7.5 dollars. Multiply the number of CDs, 6250(0.9286 )x , by

the per-CD profit, x – 7.5.

( ) 6250(0.9286) ( – 7.5)xP x x= dollars is the monthly profit from CD sales when the price

of a CD is x dollars.

c. ′ =R xd

dxx x( ) [ ( . ) ]6250 0 9286

6250 (0.9286) (0.9286)x xd dx x

dx dx

= +

[ ]= +6250 1 0 9286 0 9286 0 9286( )( . ) (ln . )( . )x xx

= +6250 0 9286 1 0 9286( . )( ln . )x x dollars of revenue per dollar of price is the rate of change in the monthly revenue from CD sales when the price of a CD is x dollars.

′ =P xd

dxxx( ) [( . ) ( – . )]6250 0 9286 7 5

6250 0.9286 ( – 7.5) (0.9286 ) ( – 7.5)x xd dx x

dx dx

= +

= +6250 0 9286 0 9286 7 5 0 9286 1[(ln . )( . )( – . ) ( . )( )]x xx

= +6250 0 9286 1 7 5 0 9286( . ) [ ( – . )(ln . )]x x dollars of profit per dollar of price is the rate of change in the monthly revenue from CD sales when the price of a CD is x dollars.

d. Evaluate the expressions for( ) and ( )R x P x′ ′ to complete the table.



Price Rate of change of revenue (dollars of revenue per dollar of price)

Rate of change of profit (dollars of profit per

dollar of price) $13 88.27 1413.83

$14 –82.15 1148.76

$20 –684.05 105.17

$21 –732.93 –0.06

$22 –771.34 –90.79

e. Because (13) 0 and (14) 0R R′ ′> < , the highest revenue will be achieved with a price

between $13 and $14.

f. Because ( ) 0 for 13 20, (21) 0, and (21) 0P x x P P′ ′ ′> ≤ ≤ ≈ < , the price corresponding to the

maximum profit is approximately $21.

32. a. F(d) = ( )

( )100

m dP d

6.53(0.994) 22

( ) (99.9(1.108 ))100

ddF t

+=

million people live in the Midwest d decades

after 1970.

c. '( ) '( ) ( ) ( ) '( )F d P d m d P d m d= ⋅ + ⋅

( )( )( ) ( )( )6.53(0.994) 22 6.53(ln0.994)(0.994)99.9 ln(1.108 1.108 99.9 1.108

100 100

d dd d += +

million people per decade is the rate of change in the number of people living in the Midwest d decades after 1970.

d. 1990: '(2) 3.531F ≈ million people per decade

1995: '(2.5) 3.714F ≈ million people per decade

2000: '(3) 3.907F ≈ million people per decade

33. a. ( ) 71.459(1.050 )xC x = dollars is the cost to produce x units in an hour, 10 ≤ x ≤ 90.

b. ( ) 71.459(ln1.050)(1.050 )xC x′ = dollars per unit produced is the rate of change of the cost

to produce x units in an hour, 10 ≤ x ≤ 90.

c. A xC x

xxx( )

( ). ( . )( )–= = 71459 1050 1 dollars per unit is the average cost to produce one unit

when x units are produced each hour, 10 ≤ x ≤ 90.



d. ′ =A xd

dxxx( ) [ . ( . )( )]–71459 1050 1

= 71459 1050 1. [( . )( )]–d

dxxx

–1 –171.459 1.050 ( ) (1.050 ) ( )x xd d

x xdx dx

= +

–1 –271.459 (ln1.050)(1.050 )( ) (1.050 )(– )x xx x = +

= −

71459 1050

1050 12

. ( . )ln .x

x xdollars per unit per hourly unit produced is the rate of

change in the average cost to produce one unit when x units are produced, 10 ≤ x ≤ 90. e. 15 units: (15) –$0.18A′ ≈ per unit per hourly unit produced

35 units: (35) $0.23A′ ≈ per unit per hourly unit produced

85 units: (85) $1.97A′ ≈ per unit per hourly unit produced

f.

Yes, the average cost is decreasing when ( ) 0A x′ < , for an hourly

production of between 0 and 20 units.

g. The graph of A′ crosses the x-axis near x = 20.5. Practically speaking, the average cost is decreasing at 20 units and increasing at 21 units, so a production level of 21 units is the one at which average cost first increases.

34. a. 208.8209.0)( += xxm million men is the number of men 65 years or older in the United

States x years after 1970, 0 ≤ x ≤ 30.

100

07.20085.10223.0)(

2 +−= xxxp percent (expressed as a decimal) is the percentage of

men 65 or older below poverty level in the United States x years after 1970, 0 ≤ x ≤ 30.

b.

+−+==100

07.20085.10223.0)208.8209.0()()()(

2 xxxxpxmxn million men 65 or older

below poverty level in the United States x years after 1970, 0 ≤ x ≤ 30.

c. ′ = ′ + ′n x m x p x m x p x( ) ( ) ( ) ( ) ( )

−++

+−=100

085.10446.0)208.8209.0(

100

07.20085.10223.0209.0

2 xx

xx

1990: 009.0)20(' −≈n million men per year

2000: 05.0)30(' ≈n million men per year



35. a. 2( ) 0.025 2.099 58.821t x x x= − + + million gives the number of households with TVs in the

U.S. between 1970 and 2002, x years after 1970.

b. –0.618

0.815( )

1 31,239.748 xv x

e=

+ percent (expressed as a decimal) gives the percentage of

U.S. households with TVs that also have VCRs between 1970 and 2002, x years after 1970.

c. n x t x v x( ) ( ) ( )= 2–0.618

0.815( 0.025 2.099 58.821)

1 31,239.748 xx x

e= − + +

+ million households

owned TVs with VCRs x years after 1970, 0 ≤ x ≤ 32.

d. 2( ) ( 0.025 2.099 58.821)

0.050 2.099 million households per year years after 1970

dt x x x

dxx x

′ = − + +

= − +

Note that v is a logistic function with L = 0.815, A = 31,239.748, and B = 0.618 and

derivative of the form 2

( )(1 )

Bx

Bx

LABev x

Ae

−

−′ =

+:

–0.618

–0.618 2

–0.618

–0.618 2

(0.815)(31,239.748)(0.618)( )

(1 31,239.748 )

15,658.143

(1 31,239.748 )

x

x

x

x

ev x

e

e

e

′ =+

≈+

′ = ′ + ′n t t x v x t x v x( ) ( ) ( ) ( ) ( )

( ) –0.618

0.8150.050 2.099

1 31,239.748 xx

e

= − +

+

( )–0.618

2–0.618 2

15,658.1430.025 2.099 58.821

(1 31,239.748 )

x

x

ex x

e

+ − + + +

million households per year is the rate of change in the number of households with TVs and VCRs x years after 1970, 0 ≤ x ≤ 32.

e. 1980: (10) 0.6n′ ≈ million households per year

1985: (15) 8.4n′ ≈ million households per year

1990: (20) 5.5n′ ≈ million households per year

36. a. We choose to model the amount spent on agricultural research and services with a cubic

function. Remember that, although we show only 3 decimal places of accuracy, we use all the places available in our technology to calculate answers.

182.22999.0060.0004.0)( 23 ++−= ttttA billion dollars is the amount of federal funds

spent for agricultural research and services in the U.S. t years after 1990, 0 ≤ t ≤ 22.

We choose to model the purchasing power of the dollar with a cubic model.

850.0028.0004.0)10(606.2)( 234 +−+−= − ttttp constant 1982 dollars is the purchasing

power of the dollar t years after 1990, -2 ≤ t ≤ 10.



b. f(t) = A(t)p(t)

)850.0028.0004.0)10(606.2)(182.22999.0060.0004.0( 23423 +−+−++−= − tttttt

billion constant 1982 dollars is the amount spent on agricultural research and services t years after 1990.

c. f(t) = A(t)p(t) ( ) ( ) ( ) ( ) ( )f t A t p t A t p t′ ′ ′= +

1992 ROC: (2) (2) (2) (2) (2) .050f A p A p′ ′ ′= + ≈ billion dollars per year

1992 PROC:

(2)100 2.395 % per year

(2)

f

f

′⋅ ≈

2000 ROC: (10) (10) (10) (10) (10) .168f A p A p′ ′ ′= + ≈ billion dollars per year

2000 PROC:

(10)100 7.047 % per year

(10)

f

f

′⋅ ≈

d. One possible answer. One possible answer: Considering expenditures in constant dollars allows us to analyze the change in expenditures apart from the effects of inflation.

37. a. 3 2( ) 151.516 2060.988 8819.062 195,291.201E x x x x= − + − + students is the enrollment in

ninth through twelfth grades in South Carolina between 1980-81 and 1989-90, where x is the number of years since the 1980-91 school year.

3 2( ) 14.271 213.882 1393.655 11,697.292D x x x x= − + − + students dropping out from ninth

through twelfth grades in South Carolina between 1980-81 and 1989-90, where x is the number of years since the 1980-91 school year.

b. ( )

( ) 100( )

D xP x

E x= ⋅ % is the percent of students dropping out from ninth through twelfth

grades in South Carolina between 1980-81 and 1989-90, where x is the number of years since the 1980–81 school year.

c. 2( ) 42.813 427.763 1393.655D x x x′ = − + − ; 2( ) 454.548 4121.976 8819.062E x x x′ = − + −

Write 1( ) 100 ( )[ ( )]P x D x E x −= .

( )1( ) 100 ( )[ ( )]d

P x D x E xdx

−′ =

1 1100 ( ) [ ( )] 100 ( ) [ ( )]

d dD x E x D x E x

dx dx− − = +

1 2100 ( )[ ( )] 100 ( )( 1)[ ( )] ( )D x E x D x E x E x− −′ ′= + −

2

100 ( ) 100 ( ) ( )

( ) [ ( )]

D x D x E x

E x E x

′ ′⋅= −

2

3 2

100( 42.813 427.763 1393.655)

151.516 2060.988 8819.062 195,291.201

x x

x x x

− + −=− + − +

−

3 2 2

3 2 2

100( 14.271 213.882 1393.655 11,697.292)( 454.548 4121.976 8819.062)

( 151.516 2060.988 8819.062 195,291.201)

x x x x x

x x x

− + − + − + −− + − +



percentage points per year is the rate of change in the percent of students dropping out from ninth through twelfth grades in South Carolina between 1980-81 and 1989-90, where x is the number of years since the 1980–81 school year.

d. X

( )P x′′′′ (percentage point

per year)

X ( )P x′′′′

(percentage point per year)

0 –0.44 5 –0.19 1 –0.38 6 –0.19 2 –0.32 7 –0.22 3 –0.26 8 –0.29 4 –0.21 9 –0.41

In the 1980–81 school year, the rate of change was most negative with a value of –0.44 percentage point per year. This is the most rapid decline during this time period. The rate of change was least negative in the 1985–86 school year with a value of –0.187 percentage point per year.

e. Negative rates of change indicate that the number of dropouts in South Carolina was falling during the 1980s. This means that more students were staying in school.

38. a. ( ) 430.073(1.299966 )xi x = dollars per job is a house painter’s income x years after 1997,

0 ≤ x ≤ 6.

b. f(x) = i(x) j(x) 1104.25430.073(1.299966 ) 44,835.085(1.299966 )( )x x x

x−= ≈ dollars gives

the painter’s total income each year x years after 1997, 0 ≤ x ≤ 6.

c. 1( ) 44,835.085 [(1.299966 )( )]xdf x x

dx−′ =

1 144,835.085 1.299966 ( ) (1.299966 ) ( )x xd d

x xdx dx

− − = +

1 244,835.085[(ln1.299966)(1.299966 )( ) (1.299966 )( )]x xx x− −= + −

2

ln1.299966 144,835.085(1.299966 )x

x x

= −

dollars per year is the rate of change in

the painter’s total income each year x years after 1990, 0 < x ≤ 6.

d. f(6) ≈ $36,063

e. (6)f ′ ≈ $3450 per year

39. One possible answer: For a function constructed by multiplying two other functions, the input

is the same as the common input of the two other functions.



40. One possible answer: The graphical method is the least accurate method for finding an

instantaneous rate of change. This method requires the user to sketch an accurate graph and tangent line as well as estimate two points on the line to determine its slope. The numerical method provides a better estimate of the instantaneous rate of change and will sometimes provide the exact instantaneous rate of change. The algebraic method computes the exact instantaneous rate of change at a point or for an entire function. The disadvantage to the algebraic method is that when f(x) is any function other than a polynomial, higher level algebra

is required to simply the difference quotient ( ) ( )f x h f x

h

+ −.

Section 3.6 Limiting Behavior Revisited: ˆL'Hopital's Rule 1. 3 3 2

2lim2 3 2(2 ) 3 7x

xx

→− = − = using direct substitution

2. 2 2 2 2

2lim 3 3( 2) 12 12.135x

xx e e e− −

→−+ = − + = + ≈ using direct substitution

3. 0

0lim ln( 1) ln(1) 1 0 1x

xe x e

→− + = − = − = using direct substitution

4. 3

9 9lim 3

3x x→= = using direct substitution

5. 0

lim lnx

x+→

= −∞ so 0

1 1lim 0

lnx x+→= =

−∞

6. 2

1lim

2x x−→= −∞

− and

2

1lim

2x x+→= ∞

−, so

2

1lim

2x x→ − does not exist.

7. 1

limln 0n

n→

= and 1

lim 1 0n

n→

− = . Therefore, 1

lnlim

1n

n

n→ − is of the indeterminate form

0.

0

Applying L’Hopital’s Rule: 1 1 1

1ln 1lim lim lim 111 1n n n

n nn→ → →

= = =−

8. 2 2(1)

2

1

1lim 3.695

2 2(1) 2

t

t

e ee

t→= = ≈

9. 4

1lim 1 0x

x→

− = and 3

1lim 1 0x

x→


31

1lim

1x

x

x→

−−

is of the indeterminate form 0

.0

Applying L’Hopital’s Rule: 4 3

3 21 1 1

1 4 4 4lim lim lim

1 3 3 3x x x

x xx

x x→ → →

− = = =−

.

10. lim x

xe

→∞= ∞ and 2lim

xx

→∞= ∞ . Therefore,

2lim

x

x

e

x→∞ is of the indeterminate form .

∞∞

Applying L’Hopital’s Rule: 2

lim lim lim2 2

x x x

x x x

e e e

x x→∞ →∞ →∞= = = ∞ .



11. 0

lim lnt

t t+→

is of the indeterminate form 0 .⋅ −∞ Applying L’Hopital’s Rule:

0

lim lnt

t t+→

=3

2 12

1 30 0 0 02 2

1ln 2lim lim lim lim 2 2(0) 0

12

t t t t

t tt ttt t

+ + + +− −→ → → →

−= = = − = − =−

.

12. ( )2lim 2 n

nn e

→−∞− is of the indeterminate form 0.−∞ ⋅ Applying L’Hopital’s Rule:

( )2

2 2 4 4lim 2 lim lim lim 0n

n n nn n n n

n nn e

e e e− − −→−∞ →−∞ →−∞ →−∞

− − − = = = = − − .

13. 2lim x

xx e−

→∞ is of the indeterminate form 0.∞ ⋅ Applying L’Hopital’s Rule:

2

2 2 2lim lim lim lim 0x

x x xx x x x

x xx e

e e e−

→∞ →∞ →∞ →∞= = = = .

14.

0 0 0lim ln lim lim ln 1t t

t t te t e t

+ + +

− −

→ → →= ⋅ = ⋅∞ = ∞ which is not indeterminate

15. 2

3 6 3(2) 6lim 0

2 2 2x

x

x→

− −= =+ +

using direct substitution

16. 2

7lim 2 35 0x

x x→

− − = and 2

7lim7 0

xx x

→− = . Therefore,

2

27

2 35lim

7x

x x

x x→

− −−

is of the indeterminate

form 0

.0


27 7

2 35 2 2 2(7) 2 12lim lim

7 7 2 7 2(7) 7x x

x x x

x x x→ →

− − − −= = = −− − −

.

17. 5

lim 1 2 0x

x→

− − = and 2

5lim 25 0x

x→


1 2lim

25x

x

x→

− −−

is of the indeterminate form

0.

0 Applying L’Hopital’s Rule:

12

5 5

1( 1) 1 1 12lim lim

2 404 1 4(5) 5 1x x

x

x x x

−

→ →

−= = =

− −.

18. 3

4lim 4 2 0x

x→

+ − = and 4

lim 4 0x

x→


4lim

4 2x

x

x→

−+ −

is of the indeterminate form

0.

0 Applying L’Hopital’s Rule:

2 23 3

24 43

1lim lim3( 4) 3(8) 3(4) 12.

1( 4)

3

x xx

x−→ →

= + = = =+

19. 2

2lim2 5 2 0x

x x→

− + = and 2

2lim5 7 6 0x

x x→

− − = . Therefore, 2

22

2 5 2lim

5 7 6x

x x

x x→

− +− −

is of the

indeterminate form 0

.0


22 2

2 5 2 4 5 3lim lim

5 7 6 10 7 13x x

x x x

x x x→ →

− + −= =− − −

20. 2 2

2 23

3 2 3(3) 2 29lim

2 3 2(3) 3 21x

x

x→

+ += =+ +

by direct substitution.

21. 00 0 0

2 2 2lim(3 ) lim(3 ) lim 3(0) 0

x xx x xx x

e e e→ → →

= ⋅ = ⋅ =

Calculus Concepts Section 3.6: Limiting Behavior Revisited: L’Hôpital’s Rule 137


22. 220 0

4lnlim 4 ln limx x

xx x

x+ + −→ →= .

0lim 4lnx

x+→

= −∞ and 2

0limx

x+

−

→= ∞ so

20

4lnlimx

x

x+ −→ is of the

indeterminate form .∞−∞

Applying L’Hopital’s Rule:

32

2 30 0 0 0

44ln 4lim lim lim lim 2 0

2 2x x x x

x xx xx x x+ + + +− −→ → → →

= = ⋅ = − =− −

23. ( )lim .6 0x

x→∞= and ( )lim ln

xx

→∞= ∞ so ( )( )lim .6 lnx

xx

→∞ is of the indeterminate form 0⋅∞ .

Applying L’Hopital’s Rule: ( )( ) ( )( )( )( )

( )1 .6ln

lim .6 ln lim lim lim.6 ln.6ln.6 .6 1

x

xx xx x x x

x xxx− −→∞ →∞ →∞ →∞

−= = =

−= 0.

24. ( ) 2 6lim 3 lim

x xx x

xx

e e→∞ →∞

=

. lim6x

x→∞

= ∞ and lim x

xe

→∞= ∞ so ( ) 2

lim 3xx

xe→∞

is of the indeterminate

form .∞∞


lim lim 0.x xx x

x

e e→∞ →∞= =

25. 2lim3 2 4x

x x→∞

+ + = ∞ and 2lim5 1x

x x→∞

+ + = ∞ . Therefore, 2

2

3 2 4lim

5 1x

x x

x x→∞

+ ++ +

is of the

indeterminate form .∞∞

Applying L’Hopital’s Rule:

2

2

3 2 4 6 2 6 3lim lim lim

5 1 10 1 10 5x x x

x x x

x x x→∞ →∞ →∞

+ + += = =+ + +

.

26. 2lim 4 7x

x→∞

+ = ∞ and 3lim 2 3x

x→∞

+ = ∞ . Therefore, 2

3

4 7lim

2 3x

x

x→∞

++

is of the indeterminate form .∞∞


3 2

4 7 8 4lim lim lim 0

2 3 6 3x x x

x x

x x x→∞ →∞ →∞

+ = = =+

.

27. 4lim3x

x→∞

= ∞ and 3lim5 6x

x→∞


3

3lim

5 6x

x

x→∞ + is of the indeterminate form .

∞∞


3 2

3 12 4lim lim lim

5 6 15 5x x x

x xx

x x→∞ →∞ →∞= = = ∞

+.

28. 3lim 4x

x→∞

= ∞ and 3lim5 6x

x→∞


3

4lim

5 6x

x

x→∞ + is of the indeterminate form .

∞∞


3 2

4 12 4 4lim lim lim

5 6 15 5 5x x x

x x

x x→∞ →∞ →∞= = =

+.

29. One possible answer: Consider the case 0

0. We know that 1)

0lim 0h

h

c→= for any non-zero real

number and 2) 0

limh

c

h→ is increasing (or decreasing) without bound for any non-zero real

number. If we apply these two statements repeatedly with c approaching 0, we end with an

apparent contradiction. Similar arguments can be applied for ∞∞

and 0 ⋅∞ .

30. One possible answer: Consider two functions n and m with input variable x that satisfy the

condition that as x approaches some constant c, n and m both approach 0. So the limit as x

approaches c of ( )( )

n xm x

is of the indeterminate form 00 . As x approaches c, ( )n x approaches 0,



so 1( )n x

increases (or decreases) without bound. Likewise, as x approaches c, ( )m x approaches

0, so 1( )m x

increases (or decreases) without bound. Using a little algebra, we can rewrite

( )( )

n xm x

as 1( )1( )

m x

n x

. As x approaches c, 1( )1( )

lim m x

x cn x

→is of the indeterminate form ∞

∞ . Similarly, we can

rewrite ( )( )

n xm x

as ( )1( )

( )m x

n x ⋅ . And, ( )1( )

lim ( )m xx c

n x→

⋅ is of the indeterminate form 0 ⋅∞ .

Chapter 3 Concept Review 1. a. x ≈ 0.8

b. positive slope: 0.8 < x < 2 negative slope: -3 < x < 0, 0 < x < 0.8

c. x = 0

d. f’(-2) ≈ -4, f’(1) ≈ 1.1, See Answer Key page A-31 in Text for figure.

e. See Answer Key page A-31 in Text for figure. 2. a. Note that D is a logistic function with L = 8.101, A = 214.8, and B = 0.797 and derivative

of the form 2797.0

797.0

2 )8.2141(

)797.0)(8.214(101.8

)1()('

t

t

Bt

Bt

e

e

Ae

LABetD

−

−

−

−

+=

+= pounds per person per year

t years after 1980

b. (10) 0.4D′ ≈ pound per person per year

In 1990 the average annual per capita consumption of turkey in the United States was increasing by 0.4 pound per person per year.

c. 00007.0)21(' ≈D pound per person per year. There was essentially no growth in the per

capita consumption of turkey in 2001.

3. a. 3.50619962000

48656890 =−−

billion per year

b.

c. 4.496$)18(' ≈A billion per year

Calculus Concepts Chapter 3 Concept Review 139


4. a. Rewrite P( t) as 1( ) 100 ( )[ ( )]P t N t A t −= .

Use the Product Rule to find the derivative.

( )1( ) 100 ( )[ ( )]d

P t N t A tdt

−′ =

1 1100 ( ) [ ( )] ( ) [ ( )]

d dN t A t N t A t

dt dt− − = +

( )1 2100 ( )[ ( )] ( )( 1)[ ( )] ( )N t A t N t A t A t− −′ ′= + −

2

100 ( ) 100 ( ) ( )

( ) [ ( )]

N t N t A t

A t A t

′ ′= − percentage points per year t years after 1980

b. Input units: years Output units: percentage points per year

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Chapter 3 Determining Change: Derivatives · Chapter 3 Determining Change: Derivatives Section 3.1 Drawing Rate-of-Change Graphs 1. The slopes are negative to the left of A and positive