Heaussler 12 Edicion

ISM: Introductory Mathematical Analysis Section 11.3

389

89. 1 12 21( )f x x x x

x−

= + = +

312 21 1 1 1 1( )

2 2 2 2 2xf x x x

x x x x x− − −′ = − = − =

Thus 1 1 1( ) 0

2 2 2x x xf xx x x x x x− − −′− = − = .

90. (1 ) p cz b w bw= + −

(1 ) p

c c

dwdz b bdw dw

= + −

Rewriting the right side and factoring out 1 + b

gives (1 )(1 )1

p

c c

dwdz b bbdw dw b

+= + −

+,

(1 )1

p

c c

dwdz bbdw dw b

⎡ ⎤= + −⎢ ⎥+⎣ ⎦

.

91. 3 3y x x= − 2( ) 3 3y x x′ = −

( )22 3 2 3 9xy =

′ = − =

The tangent line at (2, 2) is given by y – 2 = 9(x – 2), or y = 9x – 16.

8

–2

–5 5

92. 1/ 33y x x= =

2 / 33 2

1 1( )3 3

y x xx

−′ = =

81

12xy =−′ =

The tangent line at (−8, −2) is given by 12 ( 8),

12y x+ = + or

1 4 .12 3

y x= −

5–15

1

–3

Principles in Practice 11.3

1. Here 5dPdp

= and ∆p = 25.5 – 25 = 0.5.

5(0.5) 2.5dPP pdp

∆ ≈ ∆ = =

The profit increases by 2.5 units when the price is changed from 25 to 25.5 per unit.

2. ( )216 16 16 16(2 ) 16 32dy d t t t tdt dt

= − = − = −

0.516 32(0.5) 16 16 0

t

dydt =

= − = − =

The graph of y(t) is shown. 5

00 1

When t = 0.5, the object is at the peak of its flight.

3. ( )2 24( ) 3 4 (2 ) 4 83

V r r r r r′ = π + π = π + π

When r = 2, 2( ) 4 (2) 8 (2) 32V r′ = π + π = π and

3 24 32 80( ) (2) 4 (2) 163 3 3

V r π= π + π = + π = π .

The relative rate of change of the volume when

r = 2 is 803

(2) 32 6 1.2(2) 5

VV′ π

= = =π

. Multiplying 1.2

by 100 gives the percentage rate of change: (1.2)(100) = 120%.

Problems 11.3

1. 2( ) 2 3s f t t t= = +

If 1,t∆ = then over [1, 2] we have

(2) (1) 14 5 9.2 1 1

s f ft

∆ − −= = =

∆ −

If 0.5,t∆ = then over [1, 1.5] we have

(1.5) (1) 9 5 8.1.5 1 0.5

s f ft

∆ − −= = =

∆ −

Continuing this way, we obtain the following table:

Chapter 11: Differentiation ISM: Introductory Mathematical Analysis

390

t∆ 1 0.5 0.2 0.1 0.01 0.001

st

∆∆ 9 8 7.4 7.2 7.02 7.002

We estimate the velocity when t = 1 to be 7 m/s. With differentiation we get 4 3,dsv tdt

= = +

14(1) 3 7 m/s.

t

dsdt =

= + =

2. ( ) 2 5y f x x= = + .

If ∆x = 1, then over [3, 4] we have

(4) (3) 13 11 0.28891

y f fx x

∆ − −= = ≈

∆ ∆

If ∆x = 0.5, then over [3, 3.5] we have

(3.5) (3) 12 11 0.29500.5

y f fx x

∆ − −= = ≈

∆ ∆

Continuing in this way we obtain the following table:

x∆ 1 0.5 0.2 0.1 0.01 0.001

yx

∆∆ 0.2889 0.2950 0.2988 0.3002 0.3014 0.3015

We estimate the rate of change to be 0.3015. 1Note: The actual rate of change is 0.3015.11

⎛ ⎞≈⎜ ⎟

⎝ ⎠

3. 2( ) 2 4s f t t t= = −

a. When t = 7, then 22(7 ) 4(7) 70s = − = m.

b. 2(7.5) (7) [2(7.5) 4(7.5)] 70 25

0.5 0.5s f ft

∆ − − −= = =

∆ m/s

c. 4 4.dsv tdt

= = − If t = 7, then v = 4(7) − 4 = 24 m/s

4. 1( ) 12

s f t t= = +

a. When t = 2, 1 (2) 1 22

s = + = m.

b. 12 (2.1) 1 2(2.1) (1) 0.5

0.1 0.1s f ft

⎡ ⎤+ −∆ − ⎣ ⎦= = =∆

m/s

c. 12

dsvdt

= = . If t = 2, then 12

v = m/s


391

5. 3( ) 2 6s f t t= = +

a. When t = 1, 32(1) 6 8s = + = m.

b.

3

(1.02) (1)0.02

2(1.02) 6 8

0.026.1208 m/s

s f ft

∆ −=

∆⎡ ⎤+ −⎣ ⎦=

=

c. 26dsv tdt

= = . If t = 1, then

26(1) 6v = = m/s

6. 2( ) 3 2 1s f t t t= = − + +

a. When t = 1, ( )23 1 2(1) 1 0s = − + + = m.

b.

2

(1.25) (1)0.25

3(1.25) 2(1.25) 1 04.75 m/s

0.25

s f ft

∆ −=

∆⎡ ⎤− + + −⎣ ⎦= = −

c. 6 2.dsv tdt

= = − + If t = 1, v = –4 m/s

7. 4 3( ) 2s f t t t t= = − +

a. When t = 2, ( )4 32 2 2 2 2s = − + = m.

b.

4 3

(2.1) (2)0.1

(2.1) 2(2.1) 2.1 210.261 m/s

0.1

s f ft

∆ −=

∆⎡ ⎤− + −⎣ ⎦= =

c. 3 24 6 1.dsv t tdt

= = − + If t = 2, then

( ) ( )3 24 2 6 2 1 9v = − + = m/s

8. 4 7 / 2( ) 3s f t t t= = −

a. When t = 0, 4 7 / 23 0 0 0.s = ⋅ = =

b. ( ) ( ) ( )4 7 / 21 11 4 44

1 14 4

3 0(0)

1 m/s64

f fst

⎡ ⎤⋅ − −− ⎢ ⎥∆ ⎣ ⎦= =∆

=

c. 3 5 / 2712 .2

dsv t tdt

= = − If t = 0, then

3 5 / 2712(0) (0) 0 m/s.2

v = − =

9. 3225

2dy xdx

= . If x = 9, 25 (27) 337.502

dydx

= = .

10. 2dA rdr

= π . If r = 3, 2 (3) 6dAdr

= π = π .

11. 0 0.27(1 0) 0.27e

dTdT

= + − =

12. 24dV rdr

= π

When 46.3 10 ,r −= ×

4 2

6

4 [6.3 10 ] 158.76

4.988 10 .

dVdr

− −8

−

= π × = π×10

≈ ×

13. c = 500 + 10q, 10dcdq

= . When q = 100,

10dcdq

= .

14. c = 5000 + 6q, 6dcdq

= . When q = 36, 6dcdq

= .

15. 0.1(2 ) 3 0.2 3dc q qdq

= + = + . When q = 5,

0.2(5) 3 4.dcdq

= + =

16. 0.2 3dc qdq

= + . When q = 3, 3.6dcdq

= .

17. 2 50dc qdq

= + . Evaluating when q = 15, 16 and

17 gives 80, 82 and 84, respectively.


392

18. 20.12 4.4dc q qdq

= − +

Evaluating when q = 5, 25, and 1000 gives 2.4, 54.4 and 119,004.4, respectively.

19. 5000.01 5c qq

= + +

20.01 5 500c cq q q= = + +

0.02 5dc qdq

= +

506

q

dcdq =

=

1007

q

dcdq =

=

20. 10002c

q= +

2 1000c cq q= = +

2dcdq

= for all q

21. 3 20.00002 0.01 6 20,000c cq q q q= = − + +

20.00006 0.02 6dc q qdq

= − +

If q = 100, then 4.6dcdq

= . If q = 500, then

11dcdq

= .

22. 3 20.002 0.5 60 7000c cq q q q= = − + +

20.006 60dc q qdq

= − +

If q = 15, then 46.35.dcdq

= If q = 25, then

38.75.dcdq

=

23. 0.8r q=

0.8drdq

= for all q.

24. 21 115 1530 30

r q q q q⎛ ⎞= − = −⎜ ⎟⎝ ⎠

11515

dr qdq

= −

For q = 5, 443

drdq

= ; for q =15, 14drdq

= ; for

q = 150, 5drdq

= .

25. 2 3250 45r q q q= + −

2250 90 3dr q qdq

= + − . Evaluating when

q = 5, 10 and 25 gives 625, 850 and 625, respectively.

26. 260 0.2r q q= −

60 0.4dr qdq

= −

Evaluating when q = 10 and 20 gives 56 and 52, respectively.

27. 6.750 0.000328(2 ) 6.750 0.000656dc q qdq

= − = −

20006.750 0.000656(2000) 5.438

q

dcdq =

= − =

10, 484.69 6.750 0.000328cc qq q

−= = + −

10, 484.69(2000) 6.750 0.000328(2000)2000

0.851655

c −= + −

=

28. 20.79 0.04284 0.0003dc q qdq

= − + −

700.7388

q

dcdq =

=

29. 0.93 5,000,000PR = 0.935,000,000P R−=

1.934,650,000dP RdR

−= −

30. 10,500dvdt

= − for all t.


393

31. a. 1.5dy xdx

= − −

61.5 6 7.5

x

dydx =

= − − = −

b. Setting –1.5 – x = –6 gives x = 4.5.

32. 2( ) 0.4 4 5c f q q q= = + +

0.8 4dc qdq

= +

If q = 2, then 5.6dcdq

= . Over the interval [2, 3],

(3) (2) 20.6 14.6 63 2 1

c f fq

∆ − −= = =

∆ −.

33. a. 1y′ =

b. 1

4yy x′=

+

c. (5) 1y′ =

d. 1 1 0.111

5 4 9= ≈

+

e. 11.1%

34. a. 3y′ = −

b. 3 3

7 3 3 7yy x x′ −= =

− −

c. (6) 3y′ = −

d. 3 3 0.2727

3(6) 7 11= ≈

−

e. 27.27%

35. a. 6y x′ =

b. 26

3 7y xy x′=

+

c. (2) 6(2) 12y′ = =

d. 12 12 0.632

12 7 19= ≈

+

e. 63.2%

36. a. 29y x′ = −

b. 2

39

5 3y xy x

′ −=

−

c. (1) 9y′ = −

d. 9 9 4.5

5 3 2−

= − = −−

e. −450%

37. a. 23y x′ = −

b. 2

33

8y xy x′ −=

−

c. (1) 3y′ = −

d. 3 3 0.429

8 1 7−

= − ≈ −−

e. –42.9%

38. a. 2 3y x′ = +

b . 22 3

3 4y xy x x′ +=

+ −

c. ( 1) 2( 1) 3 1y′ − = − + =

d. 1 1 0.167

1 3 4 6= − ≈ −

− −

e. –16.7%

39. 20.3 3.5 9c q q= + +

0.6 3.5dc qdq

= +

If q = 10, then 0.6(10) 3.5 9.5.dcdq

= + = If

q = 10, then c = 74 and

9.5(100) (100) 12.8%74

dcdq

c= ≈ .


394

40. 1100 100y xx

−= =

22

100100dy xdx x

−= − = −

If x = 10, 100 1100

dydx

= − = − and 1(100) (100) 10%

10yy′ −

= = − .

41. a. 30 0.6dr qdq

= −

b. If q = 10, 30 6 24 4 0.09

300 30 270 45rr′ −= = = ≈

−.

c. 9%

42. a. 10 0.4dq qdr

= −

b. If q = 25, 210 0.4(25) 0.

10(25) 0.2(25)rr′ −= =

−

c. 0%

43. 0.568

0.4320.864 0.432

2W tW tt

−′= =

44. a.

0.3

1.3

1.31855.241

1 1855.24

1.3I

I

RR I′= =

0.3

1.3

1.31101.292

2 1101.29

1.3I

I

RR I′= =

b. They are equal.

c. 1

1

1( )

n

nnC xf x n

f x xC x

−′= =

12

2

( )( )

n

nnC xg x n

g x xC x

−′= =

The rates are equal.

45. The cost of q = 20 bikes is 20(150) $3000qc = = . The marginal cost, $125, is the approximate cost of one

additional bike. Thus the approximate cost of producing 21 bikes is $3000 + $125 = $3125.


395

46. The relative rate of change of c is dcdq

c, which is given to be

1q

: 1

dcdq

c q= . Thus

dc c cdq q

= = , and the marginal cost

function dcdq

⎛ ⎞⎜ ⎟⎝ ⎠

and the average cost function ( )c are equal.

47. $5.07 per unit

48. 11,275 people per year


1. (2 0.15 ) (225 20 )dR dx xdx dx

= − + (225 20 ) (2 0.15 )dx xdx

+ + −

= (2 – 0.15x)(20) + (225 + 20x)(–0.15) = 40 – 3x – 33.75 – 3x = 6.25 – 6x

6.25 6dR xdx

= −

2. 2 31( )3

T x x x= −

( )T x′ 22x x= −

When the dosage is 1 milligram the sensitivity is 2(1) 2(1) 1 1T ′ = − = .

Problems 11.4

1. ( ) (4 1)(6) (6 3)(4)f x x x′ = + + + = 24x + 6 + 24x + 12 = 48x + 18 = 6(8x + 3)

2. ( ) (3 1)(7) (7 2)(3) 42 1f x x x x′ = − + + = −

3. 2 3 2 2 3 2 3 2 3 2( ) (5 3 )(3 4 ) ( 2 )( 3) 15 20 9 12 3 6 12 33 20s t t t t t t t t t t t t t t t′ = − − + − − = − − + − + = − + −

4. 2 2( ) (3 )(10 ) (5 2)(1) 15 30 2Q x x x x x x′ = + + − = + −

5. ( ) ( )2 2( ) 3 4 (2 5) 5 1 (6 )f r r r r r r′ = − − + − + 3 2 3 26 15 8 20 6 30 6r r r r r r= − − + + − + 3 212 45 2 20r r r= − − +

6. ( ) ( )2 2( ) 2 3 (6 4) 3 4 1 (4 )C I I I I I I′ = − − + − + 3 2 3 212 8 18 12 12 16 4I I I I I I= − − + + − + ( )3 22 12 12 7 6I I I= − − +

7. Without the product rule we have

( )2 2 4 2( ) 2 5 2 5f x x x x x= − = −

3( ) 8 10f x x x′ = −

8. Without the product rule we have

( )3 2 5 4 3( ) 3 2 2 3 6 6f x x x x x x x= − + = − +

4 3 2( ) 15 24 18f x x x x′ = − +


396

9. ( ) ( )2 23 2 (4 1) 2 3 (2 3)y x x x x x x′ = + − − + − − +

( )3 2 24 12 8 3 2x x x x x= + − − − + ( )3 2 24 2 6 6 3 9x x x x x+ − − + − −

3 28 15 20 7x x x= + − −

10. 2 2

2 2 3 2 2 3

3 2

( ) (3 5 2 )(1 8 ) (2 4 )( 5 4 )3 5 2 24 40 16 10 5 20 8 4 16

32 66 26 7

x x x x x x xx x x x x x x x x xx x x

φ′ = − + − + + − − +

= − + − + − − − + + + −

= − + − −

11. 2 2 3

4 3 2 4 3

4 3 2

( ) ( 3 7)(6 ) (2 4)(2 3)6 18 42 4 6 8 1210 24 42 8 12

f w w w w w ww w w w w ww w w w

′ = + − + − +

= + − + + − −

= + − − −

12. ( ) ( )2 2( ) 3 ( 1 2 ) 3 (3 2 )f x x x x x x x′ = − − − + − − −

2 3 2 2 33 5 2 9 3 3 6 2 2x x x x x x x x= − − + + − − − + + 3 24 6 12 9x x x= − − +

13. ( ) ( )2 21 9 6y x x′ = − − ( )33 6 5 (2 ) 4(8 2)x x x x+ − + − +

4 2 4 29 15 6 6 12 10 32 8x x x x x x= − + + − + − − 4 215 27 22 2x x x= − − −

14. ( ) ( )4 2( ) 4 5 3 8 5 (2) (2 2)(16 )h x x x x x⎡ ⎤′ = + − + +⎢ ⎥⎣ ⎦

4 2 220 3(16 10 32 32 )x x x x= + − + + 4 220 144 96 30x x x= + + −

15. 1/ 2 1/ 2

1/ 2 1/ 2 1/ 2

1/ 2 1/ 2

3 1( ) (5 2)(3) (3 1) 52 23 15 515 62 2 23[45 12 5 ]4

F p p p p

p p p

p p

−

−

−

⎡ ⎤⎛ ⎞′ = − + − ⋅⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤= − + −⎢ ⎥⎣ ⎦

= − −

16. 1/ 2 2 / 3 1/ 2 1/ 3 1/ 2 1/ 2

1/ 6 1/ 3 2 / 3 1/ 2 1/ 2 1/ 6 1/ 3 1/ 2

1/ 2 1/ 3 1/ 6 1/ 2 2 / 3

1 3 1( ) ( 5 2) ( 3 ) 53 2 2

1 5 2 3 15 1 33 5 153 3 3 2 2 2 21 ( 135 40 5 18 4 18)6

g x x x x x x x x

x x x x x x x x

x x x x x

− − −

− − − −

− − −

⎛ ⎞ ⎛ ⎞′ = + − − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + − − − + + + − −

= − + + + − −

17. 273

y = ⋅ is a constant function, so 0y′ = .

18. 3 26 11 6y x x x= − + − 23 12 11y x x′ = − +


397

19. 3 26 47 31 28y x x x= + + − 218 94 31y x x′ = + +

20. 2 2

2

(4 1)(2) (2 3)(4) 8 2 8 12(4 1) (4 1)

14(4 1)

dy x x x xdx x x

x

+ − − + − += =

+ +

=+

21. 2 2

2

( 1)(5) (5 )(1) 5 5 5( )( 1) ( 1)

5( 1)

x x x xf xx x

x

− − − −′ = =

− −

= −−

22. 2

2 2

(5 )( 5) ( 5 )( 1)( )(5 )

25 5 5 25(5 ) (5 )

x xH xx

x xx x

− − − − −′ =−

− + −= = −

− −

23. 55

13 13( )33

f x xx

−−= = −

66

13 65( ) ( 5 )3 3

f x xx

−′ = − − =

24. ( )25( ) 27

f x x= −

5 10( ) (2 )7 7

f x x x′ = =

25. 2( 1)(1) ( 2)(1)

( 1)x xy

x− − +

′ =−

21 2

( 1)x x

x− − −

=−

23

( 1)x= −

−

26. ( )2

2

( 3)(6 5) 3 5 1 (1)( )

( 3)

w w w wh w

w

− + − + −′ =

−

2 2

26 13 15 3 5 1

( 3)w w w w

w− − − − +

=−

2

23 18 14

( 3)w w

w− −

=−


398

27. ( )

( )

2

22

4 ( 2) (6 2 )(2 )( )

4

z z zh z

z

− − − −′ =

−

( ) ( )2 2 2

2 22 2

2 8 12 4 2 12 8

4 4

z z z z z

z z

− + − + − += =

− −

( )( )

2

22

2 6 4

4

z z

z

− +=

−

28. 2 2

2 2

3 2 3 2

2 2

2

2 2

(3 5 3)(4 5) (2 5 2)(6 5)(3 5 3)

12 35 37 15 (12 40 13 10)(3 5 3)

5 24 25(3 5 3)

x x x x x xzx x

x x x x x xx x

x xx x

+ + + − + − +′ =+ +

+ + + − + + −=

+ +− + +

=+ +

29. ( ) ( )

( )

2 2

22

5 (16 2) 8 2 1 (2 5)

5

x x x x x xy

x x

− − − − + −′ =

−

( )( )

3 2 3 2

22

16 82 10 16 44 12 5

5

x x x x x x

x x

− + − − + −=

− ( )2

22

38 2 5

5

x x

x x

− − +=

−

30. ( )( ) ( )

( )

2 2 3 2

22

1 3 2 1 (2 )( )

1

x x x x x xf x

x

+ − − − +′ =

+

4 3 2 4 3

2 23 2 3 2 2 2 2

( 1)x x x x x x x

x− + − − + −

=+

( )( )

3

22

3 4

1

x x x

x

+ −=

+

31. ( ) ( )

( )

2 2

22

2 3 2 (2 4) 4 3 (4 3)

2 3 2

x x x x x xy

x x

− + − − − + −′ =

− +

( )( )

3 2 3 2

22

4 14 16 8 4 19 24 9

2 3 2

x x x x x x

x x

− + − − − + −=

− +

( )2

22

5 8 1

2 3 2

x x

x x

− +=

− +


399

32. The quotient rule can be used, or we can write

( )4

3 14 1( ) 43 3

zF z z zz

−+= = + ,

so ( )4

2 22

1 3 4( ) 3 43 3

zF z z zz

− −′ = − = .

33. ( ) ( )

( ) ( )

100 99 99

2 2100 100

7 (0) (1) 100 100( )7 7

x x xg xx x

+ −′ = = −

+ +

34. 559 9

22y x

x−−

= = −

6452

y x−′ =

35. 3 3

2 18 8( ) 8v vu v v vv v v

−−= = − = −

32

2 24 2( 4)( ) 2 8 2 vu v v v vv v

− +⎛ ⎞′ = + = + =⎜ ⎟

⎝ ⎠

36. 1 12 25 1 5

88xy x x

x−− ⎛ ⎞= = −⎜ ⎟

⎝ ⎠

312 2

1 3 32 2 2

1 1 5 1 1 5 58 2 2 16 16

xy x xx x x

− − ⎛ ⎞ +⎛ ⎞ ⎜ ⎟′ = + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

37. 5 2 13 3 3

13

2 2

33 1 3 1 3x x x xy x x x

x x

−− − − −= = = − −

2 1 4 23 3 3 3

1 43 3

2 1 2 15 53 3 3 3

y x x x xx x

− −′ = − + = − +

43

215 2 1

3

x x

x

− +=

38. ( )( ) ( )( )

( )

2.1 0.7 0.3 1.1

22.1

2 1 0.3 2 4.2

2 1

x x x xy

x

−+ − −′ =

+

1.4 0.7 1.4 1.1

2.1 20.6 0.3 4.2 8.4

(2 1)x x x x

x

−+ − +=

+

( )( )

1.8 2.1

20.7 2.1

0.3 1 28 12

2 1

x x

x x

+ −=

+

39. 2 2( 8)(0) (4)(1) (3 1)(2) (2 )(3)

( 8) (3 1)x x xy

x x− − + −

′ = − +− +

2 24 2

( 8) (3 1)x x= +

− +

40. 2 42

2 42

(3 5)(5) (5 1)(3)( ) 6 6(3 5)

286 6(3 5)

x xq x x xx

x xx

−

−

− − +′ = + +−

= − +−

41. 2[( 2)( 4)](1) ( 5)(2 2)

[( 2)( 4)]x x x xy

x x+ − − − −

′ =+ −

( )2 2

2

2 8 2 12 10

[( 2)( 4)]

x x x x

x x

− − − − +=

+ −

( )2

2

10 18

[( 2)( 4)]

x x

x x

− − +=

+ −

42. 2(9 1)(3 2) 27 15 2

4 5 4 5x x x xy

x x− + + −

= =− −

( )2

2

(4 5 )(54 15) 27 15 2 ( 5)

(4 5 )

x x x xy

x

− + − + − −′ =

−

2 2

2270 141 60 135 75 10

(4 5 )x x x x

x− + + + + −

=−

2135 216 50

(4 5 )x x

x− −

= −−


400

43. ( )s t′( )( ) ( )( )

( ) ( )

2 3 2 4 2

22 3

1 7 (2 3) 3 5 3 14

1 7

t t t t t t t t

t t

⎡ ⎤− + + − + − +⎢ ⎥⎣ ⎦=⎡ ⎤− +⎢ ⎥⎣ ⎦

( )( )6 5 4 3 2

22 3

3 12 6 21 14 21

1 7

t t t t t t

t t

− − + + − − −=

⎡ ⎤− +⎢ ⎥⎣ ⎦

44. 3 217( )

5 10 4f s

s s s=

− +

( )( )

( )

2 2

2 23 2 3 2

0 17 15 20 4 17 15 20 4( )

5 10 4 5 10 4

s s s sf s

s s s s s s

⎡ ⎤− − + − +⎣ ⎦′ = = −− + − +

45.

2( 1) 332( 1)13 3

2 2

x xx xx xy x x

x x

− −−−−

= − = −− −

3 22 23 3

( 1)( 2) 3 2x xx x

x x x x x x+ +

= + = +− − − +

3 2 2

2( 3 2 )(1) ( 2)(3 6 2)3

[ ( 1)( 2)]x x x x x xy

x x x− + − + − +

′ = +− −

3 2

22 3 12 43[ ( 1)( 2)]x x xx x x+ − +

= −− −

46.

2

2 25 2 5 2

3 3 32 22 2 4 2

1 33 12 3 12 3 125 5 7 10

xx x xy x x x

x x x x

+ −+ +

− −= − + = − + = − +

+ + + +

4 2 2 3 5 32 2

4 2 2 2 2 2( 7 10)(2 ) ( 3)(4 14 ) 2 12 6236 36

( 7 10) [( 2)( 5)]x x x x x x x x xy x x

x x x x+ + − − + − + +′ = − + = − +

+ + + +

47. 2 2( )(1) ( )( 1) 2( )

( ) ( )a x a x af x

a x a x− − + −

′ = =− −

48. Simplifying, 1 1

1 1( ) x a ax a xf xax a xx a

− −

− −+ +

= ⋅ =−−

2 2( )(1) ( )( 1) 2( )

( ) ( )a x a x af x

a x a x− − + −

′ = =− −

49. ( )( )2 34 2 5 7 4y x x x x= + − + +

( )( ) ( )2 2 34 2 5 3 7 7 4 (8 2)y x x x x x x′ = + − + + + + +

( 1) ( 3)(10) ( 4)( 6) 6y′ − = − + − − = −


401

50. 3

4 1xy

x=

+

4 2 3 3

4 2( 1)(3 ) ( )(4 )

( 1)x x x xy

x+ −

′ =+

2(2)(3) ( 1)( 4) 1( 1)

2(2)y − − −′ − = =

51. 6

1y

x=

−

2 2( 1)(0) (6)(1) 6

( 1) ( 1)xy

x x− −

′ = = −− −

26 3(3)

22y′ = − = −

The tangent line is 33 ( 3)2

y x− = − − , or 3 152 2

y x= − + .

52. 1 225 5xy x x

x− −+

= = +

2 32 3

1 1010y x xx x

− −′ = − − = − −

(1) 1 10 11y′ = − − = −

The tangent line is y − 6 = −11(x − 1) or y = −11x + 17.

53. ( )4 2(2 3) 2 5 4y x x x⎡ ⎤= + − +⎢ ⎥⎣ ⎦

( )3(2 3) 2 4 10y x x x⎡ ⎤′ = + −⎢ ⎥⎣ ⎦ ( )4 22 5 4 (2)x x⎡ ⎤+ − +⎢ ⎥⎣ ⎦

(0) (3)(0) [2(4)](2) 16y′ = + =

The tangent line is y – 24 = 16(x – 0), or y = 16x + 24.

54. 2 3 21 1

( 4) 4x xy

x x x x+ +

= =− −

( ) ( )( )

3 2 2

23 2

4 (1) ( 1) 3 8

4

x x x x xy

x x

− − + −′ =

−

2( 8)(1) (3)(–4) 4 1(2)

64 16( 8)y − −′ = = =

−

The tangent line is 3 1 ( 2)8 16

y x+ = − , or 1 1

16 2y x= − .


402

55. 2 6

xyx

=−

2 2(2 6)(1) (2) 6

(2 6) (2 6)x xy

x x− − −

′ = =− −

If x = 1, then 1 1

2 6 4y = = −

− and

26 6 3

16 8( 4)y − −′ = = = −

−.

Thus 3814

3 1.52

yy

−′= = =−

.

56. 11

xyx

−=

+

2 2(1 )( 1) (1 )(1) 2

(1 ) (1 )x xy

x x+ − − −

′ = = −+ +

When x = 5, then 1

1823

112

yy

−′= =−

.

57. 32

1s

t=

+. When t = 1, then s = 1 m.

3 2 2

3 2 3 2( 1)(0) 2(3 ) 6

( 1) ( 1)ds t t tvdt t t

+ −= = = −

+ +

If t = 1, then 6 1.54

v = − = − m/s.

58. 237

tst+

=+

2

2 2

2

2 2 2 2

( 7)(1) ( 3)(2 )( 7)

7 6 (7 )(1 )( 7) ( 7)

ds t t tvdt t

t t t tt t

+ − += =

+− − + −

= =+ +

v = 0 when t = −7 or t = 1. Since t is positive, we choose t = 1.

59. p = 50 − 0.01q 250 0.01r pq q q= = −

50 0.02dr qdq

= −

60. 500pq

=

500r pq= =

0drdq

=

61. 108 3

2p

q= −

+

108 32qr pq q

q= = −

+

2( 2)(108) (108 )(1) 3

( 2)dr q qdq q

+ −= −

+

2216 3

( 2)q= −

+

62. 75050

qpq+

=+

2 75050

q qr pqq+

= =+

( )2

2

( 50)(2 750) 750 (1)

( 50)

q q q qdrdq q

+ + − +=

+

2

2100 37,500( 50)

q qq

+ +=

+

63. 0.672dCdI

=

64. 0.712dCdI

=

65. 1/ 2 1/ 33 2C I I= + +

1/ 2 2 / 33 2

1 2 1 202 3 2 3

dC I IdI I I

− −= + + = +

When I = 1, then 1 2 7 .2 3 6

dCdI

= + =

3 2

1 21 12 3

dS dCdI dI I I

= − = − −

When I = 1, then 7 11 1 .6 6

dCdI

− = − = −


403

66. 3 14 6

dCdI I

= −

25

4360I

dCdI =

= , so 25

43 17160 60I

dSdI =

= − =

67. ( )( ) ( )

( )

38 12

2

4 1.2 0.2 16 0.8 0.2

4

I II I I I IdC

dI I

⎛ ⎞+ + − − + −⎜ ⎟⎝ ⎠=+

360.615

I

dCdI =

≈ , so 1 0.615 0.385dSdI

≈ − = when I = 36.

68. ( )( ) ( )

( )

310 12

2

5 0.75 0.4 20 0.5 0.4

5

I II I I I IdC

dI I

⎛ ⎞+ + − − + −⎜ ⎟⎝ ⎠=+

1000.393

I

dCdI =

≈ , so 1 0.393 0.607dSdI

≈ − = when I = 100.

69. Simplifying gives 1210 0.7 0.2C I I= + −

a. 12 0.10.7 0.1 0.7dC I

dI I−= − = −

0.11 0.3dS dCdI dI I

= − = +

25

0.10.3 0.325I

dSdI =

= + =

b. dCdIC

when I = 25 is 0.150.7

0.02610 0.7(25) 0.2(5)

−≈

+ −

70. Simplifying S gives

( )( )2 42 8 42 2

I II IS II I

+ −− −= = = −

+ +

Thus 1/ 21 1 .2 2

dS IdI I

−= =

150

1 0.040822 150I

dSdI =

= ≈⋅

and 150

1 0.04082 0.9592.I

dCdI =

≈ − ≈

71. 2 2

2 2 2( 2)(2 ) (1) 4 6 ( 4)6 6

( 2) ( 2) ( 2)dc q q q q q q qdq q q q

+ − + += ⋅ = ⋅ =

+ + +


404

72. We assume that ( ) 0d cdq

= . Thus 2

(1)0

dcdqq cdc d c

dq dq q q

⋅ −⎛ ⎞= = =⎜ ⎟

⎝ ⎠.

This implies that 0dcq cdq⋅ − = ,

dcq cdq⋅ = ,

dc c cdq q

= = , so the marginal cost function dcdq

⎛ ⎞⎜ ⎟⎝ ⎠

and the average

cost function ( )c are equal.

73. 900

10 45xy

x=

+

2(10 45 )(900) (900 )(45)

(10 45 )dy x xdx x

+ −=

+

22

(100)(900) (1800)(45) 910(100)x

dydx =

−= =

74. 0.05RT VA xV

=+

2( )(0.05) (0.05 )( )(RT)

( )d A xV V x

dV A xV+ −

=+

20.05

( )A

A xV=

+

Both numerator and denominator are always positive, so (RT) 0ddV

> . This rate of change means that if V

increases by one unit, RT increases.

75. 0.7355

1 0.02744xy

x=

+

2(1 0.02744 )(0.7355) (0.7355 )(0.02744)

(1 0.02744 )dy x xdx x

+ −=

+

20.7355

(1 0.02744 )x=

+

76. (1 ) (2 )( )

(2 )(1 ) (2 )a x b n xf x

a n x b n x+ − +

=+ + − +

For convenience let c = 2 + n.

Then (1 ) 1 (1 )( )(1 ) (1 )

a x bcx a x bcxf xac x bcx c a x bx

+ − + −= = ⋅

+ − + −.

21 [ (1 ) ]( ) [ (1 ) ]( )( )

[ (1 ) ]a x bx a bc a x bcx a bf x

c a x bx+ − − − + − −

′ = ⋅+ −

2 21 1 ( 1)

[ (1 ) ] [ (1 ) ]abc ab c ab

c ca x bx a x bx− + − +

= ⋅ = ⋅+ − + −

2 21 [ 1(2 ) 1] (1 )

2 [ (1 ) ] [ (1 ) ] (2 )n ab n ab

n a x bx a x bx n− + + − +

= ⋅ =+ + − + − +

( ) A Bxg xC Dx+

=+


405

2

2

2

( )( ) ( )( )( )( )

( )

( )

C Dx B A Bx Dg xC Dx

CB BDx AD BDxC Dx

BC ADC Dx

+ − +′ =+

+ + −=

+−

=+

Thus, ( )g x′ has the form given. When ( )g x′ is defined for ,CxD

⎛ ⎞≠⎜ ⎟⎝ ⎠

its sign is constant.

77. 2

(1)dcdqq cdc d c

dq dq q q

⋅ −⎛ ⎞= =⎜ ⎟

⎝ ⎠. When q = 20 we have

2 220(125) 20(150)

(20) 1150 120

dcdqq c

dcdq q

c c

⋅ − −

= = = −

78.

2

(3)(2 1)( 4) (3 1)(2)( 4) (3 1)(2 1)(1)

18 50 3

dy x x x x x xdx

x x

= − − + + − + + −

= − +


1. By the chain rule,

( )24 (6 ) (8 )(6) 48dy dy dx d dx t x xdt dx dt dx dt

= ⋅ = ⋅ = = .

Since x = 6t, 48(6 ) 288dy t tdt

= = .

Problems 11.5

1. (2 2)(2 1)dy dy du u xdx du dx

= ⋅ = − − ( )22 2 (2 1)x x x⎡ ⎤= − − −⎢ ⎥⎣ ⎦ ( )22 2 2 (2 1)x x x= − − − 3 24 6 2 2x x x= − − +

2. ( )( )2 26 8 7 3dy dy du u xdx du dx

= ⋅ = − − ( )( )6 4 2 22 3 42 147 4 7 3x x x x= − + − −

3. 3 3 32 2 2( 1)

(2 )dy dy dwdx dw dx w w x

⎛ ⎞= ⋅ = − − = =⎜ ⎟

−⎝ ⎠

4. 4 3

3/ 4 4 35 4 34

1 5 4(5 4 )4 4 ( 3)

dy dy dz x xz x xdx dz dx x x

− −= ⋅ = − =

⎛ ⎞− +⎜ ⎟⎝ ⎠

5. 2 22 2

( 1) ( 1) 2(3 ) 3 .( 1) ( 1)

dw dw du t tu udt du dt t t

⎡ ⎤ ⎡ ⎤+ − −= ⋅ = =⎢ ⎥ ⎢ ⎥

+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ If t = 1, then

1 1 0,1 1

u −= =

+ so 2

1

23(0) 04t

dwdt =

⎡ ⎤= =⎢ ⎥⎣ ⎦.


406

6. 12 (4 )

2dz dz du u sds du ds u

⎛ ⎞= ⋅ = +⎜ ⎟

⎝ ⎠. If s = –1, then

u = 1, so 1

5 ( 4) 102s

dzds =−

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

7. (6 8)(4 )dy dy dw w xdx dw dx

= ⋅ = − . If x = 0, then

0dydx

= .

8. ( )29 2 7 (5)dy dy du u udx du dx

= ⋅ = − + . If x = 1, then

u = 3, so 1

(82)(5) 410x

dydx =

= =

9. 56(3 2) (3 2)dy x xdx

′ = + ⋅ +

5 56(3 2) (3) 18(3 2)x x= + = +

10. ( ) ( )32 24 4 4dy x xdx

′ = − ⋅ −

( ) ( )3 32 24 4 (2 ) 8 4x x x x= − = −

11. 3 4 3

3 4 2

2 3

5(3 2 ) (3 2 )

5(3 2 ) (6 )30 (3 2 )

dy x xdx

x xx x

′ = + ⋅ +

= +

= +

12. 2 3 2

2 3

2 3

4( ) ( )

4( ) (2 1)4(2 1)( )

dy x x x xdx

x x xx x x

′ = + +

= + +

= + +

13. ( ) ( )993 2 3 22 100 8 8dy x x x x x xdx

′ = ⋅ − + ⋅ − +

( ) ( )993 2 2200 8 3 16 1x x x x x= − + − +

( )( )992 3 2200 3 16 1 8x x x x x= − + − +

14. ( ) ( )

4242

2 1 1 2 12 2

xy x

+= = +

( )( )

32 2

32 3 2

1 4 2 1 (2 1)2

2(2 1) (4 ) 8 2 1

dy x xdx

x x x x

′ = ⋅ + +

= + = +

15. ( ) ( )42 23 2 2dy x xdx

−′ = − − ⋅ −

( ) ( )4 42 23 2 (2 ) 6 2x x x x− −

= − − = − −

16. 3 13 3

2 3 13

12(2 8 ) (2 8 )

12(6 8)(2 8 )

dy x x x xdx

x x x

−

−

′ = − − ⋅ −

= − − −

17. 2 12 / 7 2

2 12 / 7

52 ( 5 2) ( 5 2)7

10 (2 5)( 5 2)7

dy x x x xdx

x x x

−

−

⎛ ⎞′ = − + − ⋅ + −⎜ ⎟⎝ ⎠

= − + + −

18. ( ) ( )524 334 7 7 4

2y x x x

−⎛ ⎞′ = − − −⎜ ⎟⎝ ⎠

( )( )523 46 7 4 7x x x

−= − − −

19. ( )122 25 5y x x x x= − = −

( )1221 5 (10 1)

2y x x x

−′ = − −

( )1221 (10 1) 5

2x x x

−= − −

20. ( )122 23 7 3 7y x x= − = −

( ) ( )1 12 22 21 3 7 (6 ) 3 3 7

2y x x x x

− −′ = − = −

21. 144 2 1 (2 1)y x x= − = −

3 34 41 1(2 1) (2) (2 1)

4 2y x x− −′ = − = −

22. ( )133 2 28 1 8 1y x x= − = −

( ) ( )2 23 32 21 168 1 (16 ) 8 1

3 3y x x x x

− −′ = − = −


407

23. ( ) ( )2523 352 1 2 1y x x= + = +

( ) ( ) ( )3 35 53 2 2 32 122 1 3 1

5 5y x x x x

− −⎛ ⎞′ = + = +⎜ ⎟⎝ ⎠

24. 5 5 5 5 / 337 ( 3) 7( 3)y x x= − = −

5 2 / 3 4

4 5 2 / 3

57 ( 3) (5 )3

175 ( 3)3

y x x

x x

′ = ⋅ −

= −

25. ( ) 122

6 6 2 12 1

y x xx x

−= = − +

− +

( ) 226( 1) 2 1 (4 1)y x x x−

′ = − − + −

( ) 226(4 1) 2 1x x x−

= − − − +

26. ( ) 1443 3 2

2y x

x

−= = +

+

( ) ( ) ( )2 24 3 3 43( 1) 2 4 12 2y x x x x− −

′ = − + = − +

27.

( )( ) 22

22

1 33

y x xx x

−= = −

−

( ) 322 3 (2 3)y x x x−

′ = − − −

( ) 322(2 3) 3x x x−

= − − −

28. 44

1 (2 )(2 )

y xx

−= = ++

5 54(2 ) (1) 4(2 )y x x− −′ = − + = − +

29. 2 1/ 22

4 4(9 1)9 1

y xx

−= = ++

2 3/ 2

2 3/ 2

14 (9 1) (18 )2

36 (9 1)

y x x

x x

−

−

⎛ ⎞′ = − +⎜ ⎟⎝ ⎠

= − +

30.

( )( )

23

23

2

2

3 3 33

y x xx x

−= = −

−

( )53223 3 (6 1)

3y x x x

−⎛ ⎞′ = − − −⎜ ⎟⎝ ⎠

( )53–22(6 1) 3x x x= − − −

31. 133 3 37 7 (7 ) 7y x x x x= + = +

2 23 33 31 7(7 ) (7) 7(1) (7 ) 7

3 3y x x− −′ = + = +

32. 1 12 212 (2 ) (2 )

2y x x x

x−

= + = +

312 21 1(2 ) (2) (2 ) (2)

2 2y x x− −⎛ ⎞ ⎛ ⎞′ = + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

312 2(2 ) (2 )x x− −

= −

33. 2 4 55( 4) (1) ( 4) (2 )y x x x x⎡ ⎤′ = − + −⎣ ⎦

4( 4) [5 2( 4)]x x x x= − + − 4( 4) (7 8)x x x= − −

34. 3 4

3 3

4( 4) (1) ( 4) (1)

( 4) (4 4) ( 4) (5 4)

y x x x

x x x x x

⎡ ⎤′ = + + +⎣ ⎦= + + + = + +

35. 122 24 5 1 4 (5 1)y x x x x= + = +

12

12

2

2

14 (5 1) (5) 5 1(8 )2

10 (5 1) 8 5 1

y x x x x

x x x x

−

−

⎛ ⎞′ = + + +⎜ ⎟⎝ ⎠

= + + +


408

36. 123 2 3 24 1 4 (1 )y x x x x= − = −

3 2 1/ 2 2 2

42 2

2

14 (1 ) ( 2 ) 1 (12 )2

4 12 11

y x x x x x

x x xx

−⎡ ⎤⎛ ⎞′ = − − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= − + −−

37. ( ) ( )3 22 22 1 (5) (5 ) 3 2 1 (2 2)y x x x x x x⎡ ⎤′ = + − + + − +⎢ ⎥

⎣ ⎦

( ) ( )22 25 2 1 2 1 3 (2 2)x x x x x x⎡ ⎤= + − + − + +⎢ ⎥⎣ ⎦

( ) ( )22 25 2 1 7 8 1x x x x= + − + −

38. ( ) ( ) ( )3 42 3 2 34 1 3 1 (2 )y x x x x x⎡ ⎤′ = − + −⎢ ⎥

⎣ ⎦

( ) ( ) ( )( )3 33 3 3 3 32 1 6 1 2 7 1 1x x x x x x x⎡ ⎤= − + − = − −⎢ ⎥⎣ ⎦

39. 3 3 4 2(8 1) 4(2 1) (2) (2 1) 3(8 1) (8)y x x x x⎡ ⎤ ⎡ ⎤′ = − + + + −⎣ ⎦ ⎣ ⎦

2 38(8 1) (2 1) [(8 1) 3(2 1)]x x x x= − + − + + 2 38(8 1) (2 1) (14 2)x x x= − + +

2 316(8 1) (2 1) (7 1)x x x= − + +

40. 5 2 4

4

4

(3 2) [2(4 5)(4)] (4 5) [5(3 2) (3)](3 2) (4 5)[8(3 2) 15(4 5)](3 2) (4 5)(84 59)

y x x x xx x x xx x x

′ = + − + − +

= + − + + −

= + − −

41. 11

2

11

2

11

13

3 ( 2)(1) ( 3)(1)122 ( 2)

3 5122 ( 2)

60( 3)( 2)

x x xyx x

xx xx

x

⎡ ⎤− + − −⎛ ⎞′ = ⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦⎡ ⎤−⎛ ⎞= ⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦

−=

+

42. 3 3

2 52 ( 2)(2) 2 (1) 1284

2 ( 2) ( 2)x x x xy

x x x

⎡ ⎤+ −⎛ ⎞′ = =⎢ ⎥⎜ ⎟+⎝ ⎠ + +⎢ ⎥⎣ ⎦

43.

12

21 2 ( 3)(1) ( 2)(1)2 3 ( 3)

x x xyx x

− ⎡ ⎤− + − −⎛ ⎞′ = ⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦

12

2 25 2 5 3

3 22( 3) 2( 3)x xx xx x

−− +⎛ ⎞= =⎜ ⎟+ −⎝ ⎠+ +


409

44. ( ) ( )

( )

23 2 22

2 22

2 (16 ) 8 3 (2 )1 8 33 2 2

x x x xxyx x

− ⎡ ⎤+ − −⎛ ⎞ ⎢ ⎥−

′ = ⎜ ⎟ ⎢ ⎥⎜ ⎟+ ⎢ ⎥⎝ ⎠ +⎣ ⎦

( )

232

2 22

1 8 3 383 2 2

x xx x

−⎛ ⎞−

= ⎜ ⎟⎜ ⎟+⎝ ⎠ +

( ) ( )2 43 32 2

38

3 8 3 2

x

x x=

− +

45. ( ) ( )

( )

3 22 2

62

4 (2) (2 5) 3 4 (2 )

4

x x x xy

x

⎡ ⎤+ − − +⎢ ⎥

⎣ ⎦′ =+

( ) ( ){ }( )

22 2

62

4 4 (2) (2 5)[3(2 )]

4

x x x x

x

+ + − −=

+

( ) ( )2 2 2

4 42 2

2 8 12 30 10 30 8

4 4

x x x x x

x x

+ − + − + += =

+ +

( )( )

2

42

2 5 15 4

4

x x

x

− − −=

+

46. 2 3 4

2 2

3 2

2 2

3 2

2 2

(3 7)[4(4 2) (4)] (4 2) (6 )(3 7)

(4 2) [16(3 7) 6 (4 2)](3 7)

(4 2) (24 12 112)(3 7)

x x x xyx

x x x xx

x x xx

+ − − −′ =+

− + − −=

+

− + +=

+

47.

3 4 5 2

6

(3 1) 5(8 1) (8) (8 1) 3(3 1) (3)

(3 1)

x x x xy

x

⎡ ⎤ ⎡ ⎤− − − − −⎣ ⎦ ⎣ ⎦′ =−

2 4

6(3 1) (8 1) [(3 1)(40) (8 1)(9)]

(3 1)x x x x

x− − − − −

=−

4

4(8 1) (48 31)

(3 1)x x

x− −

=−


410

48. 2 2 1/ 33 ( 2) ( 2) [( 2) ( 2)]y x x x x= − + = − +

2 2 / 3 2

2 2 / 3

2 2 / 3

1/ 3 2 / 3

1[( 2) ( 2)] [(1)( 2) 2( 2)( 2)]31[( 2) ( 2)] ( 2)[ 2 2( 2)]31[( 2) ( 2)] ( 2)(3 2)31 ( 2) ( 2) (3 2)3

y x x x x x

x x x x x

x x x x

x x x

−

−

−

− −

′ = − + − + − +

= − + − − + +

= − + − +

= − + +

49. ( ) ( )( )122 4 2 46 5 2 5 6 5 2 5y x x x x

⎡ ⎤= + + = + +⎢ ⎥

⎣ ⎦

( ) ( ) ( ) ( )1 12 22 4 3 416 5 2 5 4 5 (10 )

2y x x x x x

−⎡ ⎤′ = + ⋅ + + +⎢ ⎥

⎣ ⎦

( )( ) ( ) ( )1 12 22 4 3 46 5 2 5 2 5 (10 )x x x x x

−⎡ ⎤= + + + +⎢ ⎥

⎣ ⎦

( )( ) ( ) ( )1 12 22 4 2 412 5 2 5 5 (5)x x x x x

−⎡ ⎤= + + + +⎢ ⎥

⎣ ⎦

Factoring out ( )124 5x

−+ gives

( ) ( )( ) ( )124 2 2 412 5 5 2 5 (5)y x x x x x

− ⎡ ⎤′ = + + + +⎢ ⎥⎣ ⎦

( ) ( )124 4 212 5 10 2 25x x x x

−= + + +

50. 23 4 (2)(7 1)(7) (7 1) (1)y x x x⎡ ⎤′ = − + + +⎣ ⎦

2 23 4 147 28 1 588 112 1x x x x⎡ ⎤= − + + = − − −⎣ ⎦

51. 2( 4)(1) ( 1)(1) 8 7 18 2 8

4 4( 4)t t ty

t+ − − −⎛ ⎞⎛ ⎞′ = + − ⋅⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠+

2 25 58 (8 7) 15 8

( 4) ( 4)t t

t t= + − − = − +

+ +


411

52. 3 4 3

2 2(2 6)(7 5) 14 10 42 30

(2 4) (2 4)x x x x xy

x x+ − − + −

= =+ +

2 3 2 4 3

4

3 2 4 3

4

4 3 3 2 4 3

3

4 3 2

(2 4) (56 30 42) (14 10 42 30)[2(2 4)(2)](2 4)

(2 4)[(2 4)(56 30 42) 4(14 10 42 30)](2 4)

112 60 84 224 120 168 56 40 168 120(2 4)

4(14 51 30 21 72)

x x x x x x xyx

x x x x x x xx

x x x x x x x xx

x x x x

+ − + − − + − +′ =+

+ + − + − − + −=

+− + + − + − − − +

=+

+ − − += 3(2 4)x +

53. 3 5 3 2 2 3 2 3 4 2

3 10( 5) [(2 1) (2)( 3)(1) ( 3) (3)(2 1) (2)] (2 1) ( 3) [5( 5) (3 )]

( 5)x x x x x x x x xy

x− + + + + + − + + −

′ =−

54. ( ) ( ) ( ) ( )1

22 22 2 21

2

2

(9 3) 2(2) 4 1 (8 ) 4 1 ( 2) 2 4 1 (9)

(9 3)

x x x x x x x xy

x

−⎡ ⎤− + − + − + − + −⎢ ⎥⎣ ⎦′ =

−

55. ( )32 23(5 6) (5) 4 1 (2 )dy dy du u x xdx du dx

⎡ ⎤⎡ ⎤= ⋅ = + +⎢ ⎥⎣ ⎦ ⎣ ⎦

When x = 0, then 0dydx

= .

56. (4 4)(6)(2)dz dz dy dx ydt dy dx dt

= ⋅ ⋅ = −

When t = 1, then x = 2 and y = 7. Thus 1

(24)(6)(2) 288t

dzdt =

= = .

57. ( )223 7 8 (2 7)y x x x′ = − − −

If x = 8, then slope 23(64 56 8) (16 7) 0y= ′ = − − − = .

58. 12( 1)y x= +

121 ( 1)

2y x −′ = +

If x = 8, 16

y′ = .


412

59. ( )232 8y x= −

( )( )

13

13

2

2

2 48 (2 )3

3 8

xy x xx

−′ = − =

−

If x = 3, then 12 43(1)

y′ = = . Thus the tangent line

is y – 1 = 4(x – 3), or y = 4x – 11.

60. 2 23( 3) (1) 3( 3)y x x′ = + = +

If x = −1, 23(2) 12.y′ = =

The tangent line is y − 8 = 12(x + 1) or y = 12x + 20.

61. ( )

121

22

( 1) (7 2) (7) 7 2(1)

( 1)

x x xy

x

−+ + − +

′ =+

( )7 12 7 2

2

( 1) 7 2

( 1)x

x x

x+

+ − +=

+

If x = 1, then ( )( )7 1

2 32 3 14 6

y−

′ = = − . The

tangent line is 3 1 ( 1)2 6

y x− = − − , or

1 56 3

y x= − + .

62. ( ) 323 3 1y x−

= − +

( ) 423( 3) 3 1 (6 )y x x−

′ = − − +

If x = 0, then 0y′ = . The tangent line is

y + 3 = 0(x – 0), or y = –3.

63. ( )32 9y x= + and ( )226 9y x x′ = + . When

x = 4, then 3(25)y = and 26(4)(25)y′ = , so 2

36(4)(25) 24(100) (100) (100) 96%

25(25)yy′

= = =

64. 2 31

( 1)y

x=

− and 2 4

6( 1)

xyx

′ = −−

When x = 2, 127

y = and 412 4 ,

273y′ = − = − so

4(100) 27(100) 400%27

yy′⎛ ⎞

= − ⋅ = −⎜ ⎟⎝ ⎠

65. q = 5m, p = –0.4q + 50; m = 6 dr dr dqdm dq dm

= ⋅

20.4 50 ,r pq q q= = − + 0.8 50,dr qdq

= − + . For

m = 6, then q = 30, so 6

24 50 26.m

drdq =

= − + =

Also, 5.dqdm

= Thus 6

(26)(5) 130.m

drdm =

= =

66. ( )21 20020

q m m= −

p = –0.1q + 70; m = 40 dr dr dqdm dq dm

= ⋅

20.1 70r pq q q= = − + , so 0.2 70dr qdq

= − + . If

m = 40, then q = 320, so

4064 70 6

m

drdq =

= − + = .

1 (200 2 )20

dq mdm

= − . When m = 40, 6dqdm

= .

Thus 40

(6)(6) 36m

drdm =

= = .

67. 2

2

10

9

mqm

=+

5253

pq

=+

; m = 4

dr dr dqdm dq dm

= ⋅

5253qr pq

q= =

+, so

2 2( 3)(1) (1) 1575525

( 3) ( 3)dr q qdq q q

+ −= ⋅ =

+ +.


413

If m = 4, then q = 32, so 4

1575 91225 7m

drdq =

= = .

( ) ( )1 12 22 2 21

22

9 (20 ) 10 9 (2 )

9

m m m m mdqdm m

−+ − ⋅ +

=+

( ) ( )122 2 3

2

9 20 9 10

9

m m m m

m

− ⎡ ⎤+ + −⎢ ⎥⎣ ⎦=+

( )32

3

2

10 180

9

m m

m

+=

+

When m = 4, then

32

10(64) 180(4) 1360 272125 25(25)

dqdm

+= = = . Thus

4

9 272 13.997 25m

drdm =

= ⋅ ≈ .

68. 2

100

19

mqm

=+

450010

pq

=+

; m = 9

dr dr dqdm dq dm

= ⋅

450010

qr pqq

= =+

, so 245,000

( 10)drdq q

=+

.

If m = 9, then q = 90, so 9

92m

drdq =

= .

( )322

1900

19

dqdm

m=

+. When m = 9, then

1910

dqdm

= .

Thus 9

9 19 8.552 10m

drdm =

= ⋅ = .

69. a. ( )122

2

10 20 (2 )2 20

dp qq qdq q

− −= − + =

+

b. 2 20

2100 20

qdpqdq

p q

−

+=

− +

2 220 100 20

q

q q= −

⎛ ⎞+ − +⎜ ⎟⎝ ⎠

2 2100 20 20

q

q q= −

+ − −

c. 2100 20r pq q q q= = − +

drdq

( )122 21100 20 (2 ) 20(1)

2q q q q

−⎡ ⎤= − ⋅ + + +⎢ ⎥

⎣ ⎦

22

2100 20

20

q qq

= − − ++

70. kpq

= ; q = f(m)

dr dr dqdm dq dm

= ⋅

r = pq = k, so 0drdq

= . Thus 0 0dr dqdm dm

= ⋅ = .

71. (12 0.4 )( 1.5)dc dc dq qdp dq dp

= ⋅ = + −

When p = 85, then q = 772.5, so

85481.5.

p

dcdp =

= −

72. 3250( ) 1

250f t

t⎛ ⎞= − ⎜ ⎟+⎝ ⎠

2

2250 250( ) 3

250 (250 )f t

t t

⎡ ⎤⎛ ⎞′ = − −⎢ ⎥⎜ ⎟+⎝ ⎠ +⎢ ⎥⎣ ⎦

2

2250 250(100) 3350 35025 1349 490

15 .4802

f ⎡ ⎤⎛ ⎞′ = − −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

Thus when t increases from 100 to 101, the proportion discharged increases by

approximately 15 .

4802


414

73. ( ) ( ) ( )

1 12 22 2 21

2

2

3 (10 ) 5 3 (2 )

3

q q q q qdcdq q

−⎡ ⎤+ − +⎢ ⎥

⎣ ⎦=+

Multiplying numerator and denominator by ( )122 3q + gives

( )( )

32

2 2

2

3 (10 ) 5 ( )

3

q q q qdcdq

q

+ −=

+ ( )( )

( )3 32 2

23

2 2

5 65 30

3 3

q qq q

q q

++= =

+ +.

74. a. 680 4360dS EdE

= − . If E = 16, 6520dSdE

= .

b. Solving 680 4360 5000E − = gives 680 9360, 13.8.E E= ≈

75. ( )2 8 74 10 (2 ) 10dV dV dr r tdt dr dt

− −⎡ ⎤= ⋅ = π +⎣ ⎦ . When t = 10, then ( )8 2 710 10 10 (10)r − −= + 6 6 610 10 2(10)− − −= + = .

Thus 26 8 7

104 2(10) 10 (2)(10) 10

t

dVdt

− − −

=

⎡ ⎤ ⎡ ⎤= π +⎣ ⎦ ⎣ ⎦ ( )12 7 194 4(10) 3 10 48 (10)− − −⎡ ⎤⎡ ⎤= π = π⎢ ⎥⎣ ⎦ ⎣ ⎦

76. a. 1 12 21 (2 ) (2 ) (2 )

2dp VI V V VIdI

ρ ρ ρ ρ− −= =

b.

12

12

(2 ) 12(2 )

dpdI V VIp IVI

ρ ρ

ρ

−

= =

77. a. 3 3( ) 0.001416 0.01356 1.696 34.9xd I x x xdx

= − + + −

If x = 65, ( ) 256.238.xd Idx

= −

b. If x = 65, ( ) 256.238 0.01578

16, 236.484

dxdx

x

I

I−

≈ ≈ −

If x = 65, the percentage rate of change is ( ) 25,623.8 1.578%.

16,236.484

dxdx

x

I

I−

⋅ = = −

78. (P + a)(v + b) = k kv b

P a+ =

+

kv bP a

= −+

1( )v k P a b−= + −

22( 1)( )

( )dv kk P adP P a

−= − + = −+


415

79. By the chain rule, dc dc dqdp dq dp

= ⋅ . We are given that 1100 100q pp

−= = , so 22

100100dq pdp p

− −= − = . Thus

2100dc dc

dp dq p

⎡ ⎤−= ⎢ ⎥

⎢ ⎥⎣ ⎦. When q = 200, then

100 1200 2

p = = and we are given that 0.01dcdq

= . Therefore

( )212

1000.01 4dcdp

⎡ ⎤−⎢ ⎥= = −⎢ ⎥⎢ ⎥⎣ ⎦

.

80. a. When m = 12, then q = 3000, so r = 1500.

Thus 1500 1 $0.503000 2

rpq

= = = = .

b. ( )

121

21000 3 (50) 50 (1000 3 ) (3)

1000 3

q q qdrdq q

−+ − +

=+

3000

2750 1110,000 40q

drdq =

= =

c. dr dr dqdm dq dm

= ⋅ . From part (b) we know drdq

. Now,

312 23(2 ) (2 1) (2) (2 1) (2)

2dq m m mdm

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

, so 12

610m

dqdm =

= .

Thus 12

11 67161040 4m

drdm =

= ⋅ = .

81. ( ) ( )dy dy dx f x g tdt dx dt

= ⋅ = ′ ′ . We are given that g(2) = 3, so x = 3 when t = 2. Thus

2 (2) 2(3) (2) 10(4) 40

t x g t

dy dy dx f gdt dx dt= = =

= ⋅ = ′ ′ = = .

82. a. 2

324 5 19 19 19lim lim 0 018 18 1835q q

cqq→∞ →∞

⎛ ⎞⎜ ⎟= + + = + + =⎜ ⎟+⎝ ⎠

b. 2

324 1951835

qc cq qq

= = + ++

( )( )122 21

22

35(324) 324 35 (2 ) 191835

q q q qdcdq q

−+ − +

= ++

173

q

dcdq =

=

c. From part (b) the increase in cost of the additional unit is approximately $300. Since the corresponding revenue increases by $275, the move should not be made.


416

83. 86,111.37

84. 5.25

Chapter 11 Review Problems

1. 2( ) 2f x x= −

( )2 2

0 0

2 ( ) 2( ) ( )( ) lim limh h

x h xf x h f xf xh h→ →

⎡ ⎤− + − −+ − ⎣ ⎦′ = =

( )2 2 2 2

0 0

2 2 2 2lim limh h

x hx h x hx hh h→ →

⎡ ⎤− − − − − − −⎣ ⎦= =

0 0

(2 )lim lim (2 ) 2h h

h x h x h xh→ →

− += = − + = −

2. 2( ) 2 3 1f x x x= − +

0

( ) ( )( ) limh

f x h f xf xh→

+ −′ =

( )2 2

0

2( ) 3( ) 1 2 3 1limh

x h x h x x

h→

⎡ ⎤+ − + + − − +⎣ ⎦=

( )2 2 2

0

2 4 2 3 3 1 2 3 1limh

x hx h x h x x

h→

⎡ ⎤+ + − − + − − +⎣ ⎦=

2

0 0

4 2 3 (4 2 3)lim limh h

hx h h h x hh h→ →

+ − + −= =

0lim (4 2 3) 4 3h

x h x→

= + − = −

3. ( ) 3f x x=

0 0

3( ) 3( ) ( )( ) lim limh h

x h xf x h f xf xh h→ →

+ −+ −′ = =

0

3( ) 3 3( ) 3lim

3( ) 3h

x h x x h xh x h x→

+ − + += ⋅

+ +

( ) ( )0 0

3( ) 3 3lim lim3( ) 3 3( ) 3h h

x h x hh x h x h x h x→ →

+ −= =

+ + + +

0

3lim3( ) 3h x h x→

=+ +

3 3 33 3 2 3 2x x x x

= = =+

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Heaussler 12 Edicion