Page 1

Calculus An Applied Approach Brief International Metric Edition Solutions Manual 10th Edition Larson

SOLUTIONS MATHEMATICS TEST BANKS


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C H A P T E R 2 Differentiation Section 2.1

The Derivative and the Slope of a Graph ............................................ 72

Section 2.2

Some Rules for Differentiation ............................................................89

Section 2.3

Rates of Change: Velocity and Marginals...........................................96

Section 2.4

The Product and Quotient Rules ........................................................102

Quiz Yourself .............................................................................................................113 Section 2.5

The Chain Rule ...................................................................................115

Section 2.6

Higher-Order Derivatives...................................................................123

Section 2.7

Implicit Differentiation.......................................................................129

Section 2.8

Related Rates ......................................................................................137

Review Exercises ........................................................................................................141 Test Yourself .............................................................................................................154

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C H A P T E R Differentiation

2

Section 2.1 The Derivative and the Slope of a Graph Skills Warm Up 1. P(3, 1), Q(3, 6) 6 −1 ; m is undefined. 3−3 x = 3

m =

2 Δx 3 x + 3 xΔx + ( Δx)    = lim Δx → 0 Δx

= 3x2 Δx → 0

3. P(1, 5), Q( 4, −1)

8.

−1 − 5 −6 = = −2 m = 4 −1 3

+ Δx ) − x 2 2

Δx

Δx → 0

= lim

x + 2 xΔx + ( Δx) − x 2

= lim

Δx → 0

y = − 2x + 7

= lim

4. P(3, 5), Q( −1, − 7)

Δx 2 xΔx + ( Δx)

2

Δx Δx( 2 x + Δx ) Δx

Δx → 0

−7 − 5 −12 = = 3 −1 − 3 −4

2

Δx → 0

y − 5 = − 2x + 2

= 2x

9. f ( x) = 3x

y − 5 = 3( x − 3)

Domain: ( −∞, ∞)

y = 3x − 4 2 xΔx + ( Δx) Δx

(x

lim

2

y − 5 = −2( x − 1)

Δx → 0

1 1 = 2 x( x + Δx) x

7. lim

y = 2

5. lim

2

Δx → 0

2− 2 = 0 −5 − 2

2

3

Δx

Δx → 0

= lim 3 x 2 + 3xΔx + ( Δx)

y − 2 = 0( x − 2)

m =

lim

2

2. P( 2, 2), Q = ( −5, 2)

m =

3x 2 Δx + 3 x( Δx) + ( Δx) 2

6.

Δx( 2 x + Δx) Δx = lim 2 x + Δx

1 x −1 Domain: ( −∞, 1) ∪ (1, ∞)

10. f ( x) =

= lim

Δx → 0

Δx → 0

= 2x

11. f ( x) =

1 x3 5

− 2 x 2 + 13 x − 1

Domain: ( −∞, ∞)

6x x + x Domain: ( −∞, 0) ∪ (0, ∞)

12. f ( x) =

1.

2.

y

3

y

x x

72

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DOWNLOAD SOLUTIONS MANUAL Section 2.1 y

3.

The Derivative and the Slope of a Graph

73

14. 2010: m ≈ 500 2012: m ≈ 500 The slope is the rate of change in millions of dollars per year of sales for the years 2010 and 2012 for Fossil. 15. t = 3: m ≈ 8 x

t = 7: m ≈ 1 t = 10: m ≈ −10

4.

y

The slope is the rate of change of the average temperature in degrees Fahrenheit per month in Bland, Virginia, for March, July, and October. 16. (a) At t1 , f ′(t1 ) > g ′(t1 ), so the runner given by f is running faster. (b) At t2 , g ′(t2 ) > f ′(t2 ), so the runner given by g is

5.

running faster. The runner given by f has traveled farther.

y

(c) At t3 , the runners are at the same location, but the runner given by g is running faster. (d) The runner given by g will finish first because that runner finishes the distance at a lesser value of t. x

6.

17. f ( x) = −1 at (0, −1)

f (0 + Δx) − f (0) Δx −1 − ( −1) = Δx 0 = Δx = 0

msec =

y

m = lim msec = lim 0 = 0 Δx → 0

7. The slope is m = 1. 8. The slope is m =

4. 3

9. The slope is m = 0. 10. The slope is m =

1. 4

11. The slope is m = − 13. 12. The slope is m = −3.

Δx → 0

18. f ( x) = 6 at ( − 2, 6)

f ( −2 + Δx) − f ( −2) Δx 6−6 = Δx 0 = Δx = 0

msec =

m = lim msec = lim 0 = 0 Δx → 0

Δx → 0

13. 2009: m ≈ 118 2011: m ≈ 375 The slope is the rate of change in millions of dollars per year of revenue for the years 2009 and 2011 for Under Armour.

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DOWNLOAD SOLUTIONS MANUAL 74

Chapter 2

Differentiation

19. f ( x) = 13 − 4 x at (3, 1)

msec =

22. f ( x) = 11 − x 2 at (3, 2)

f (3 + Δ x) − f (3) Δx

msec =

13 − 4(3 + Δ x) − 1 =  Δx

=

13 − 12 − 4Δ x − 1 Δx

=

=

− 4Δ x = Δx

Δx 2 2 11 − (3 + Δx) − 11 − (3)    Δx

11 − (9 + 6Δx + Δx 2 ) − 2 Δx 11 − 9 − 6Δx + ( Δx) − 2 2

=

= −4

m = lim msec = lim ( − 4) = − 4 Δx → 0

=

Δx → 0

=

20. f ( x) = 6 x + 3 at (1, 9) msec =

f (3 + Δx) − f (3)

Δx → 0

6(1 + Δ x ) + 3 − 9 =  Δx

=

6Δ x Δx

Δx Δx( − 6 − Δx)

m = lim msec = lim ( − 6 − Δx) = − 6

Δx

6 + 6Δ x + 3 − 9 Δx

2

Δx = − 6 − Δx

f (1 + Δ x ) − f (1)

=

Δx − 6Δx − ( Δx)

Δx → 0

23. f ( x ) = x 3 − 4 x at ( −1, 3)

msec = =

= 6

f ( −1 + Δx) − f ( −1) Δx

(−1 +

3 3 Δx) − 4( −1 + Δx) − ( −1) − 4( −1)   Δx

=

−1 + 3Δx − 3( Δx) + ( Δx) + 4 − 4Δx − 3 Δx

=

−Δx − 3( Δx) + ( Δx) Δx

2

m = lim msec = lim 6 = 6 Δx → 0

Δx → 0

3

2

21. f ( x) = 2 x 2 − 3 at ( 2, 5)

msec =

f ( 2 + Δ x) − f ( 2)

=

Δx

2( 2 + Δ x)2 − 3 − 5  =  Δx

(

)

2 4 + 4Δ x + ( Δ x)2 − 3 − 5   =  Δx 8 + 8Δ x + 2( Δ x)2 − 3 − 5  =  Δx =

2Δ x( 4 + Δ x )

m = lim msec = lim ( 2( 4 + Δ x)) = 8 Δx → 0

Δx → 0

2

Δx

= −1 − 3Δx + ( Δx)

)

2

(

m = lim msec = lim −1 − 3Δx + ( Δx) Δx → 0

Δx → 0

2

) = −1

24. f ( x) = 7 x − x 3 at ( − 3, 6) msec = =

Δx

= 2( 4 + Δ x)

(

Δx −1 − 3Δx + ( Δx)

3

=

f ( − 3 + Δx) − f ( − 3) Δx 3 3 7( − 3 + Δx) − ( − 3 + Δx) − 7( − 3) − ( − 3)    Δx

(

− 21 + 7 Δx − − 27 + 27 Δx − 9( Δx) + ( Δx) 2

Δx

− 20Δx + 9( Δx ) − ( Δx) 2

= =

(

Δx

Δx − 20 + 9Δx − ( Δx) Δx

= − 20 + 9Δx − ( Δx )

2

Δx → 0

)−6

3

)

2

(

m = lim msec = lim − 20 + 9Δx − ( Δx) Δx → 0

3

2

) = − 20

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.1 25. f ( x ) = 2

f ( 4 + Δx) − f ( 4) Δx

=

2 4 + Δx − 2 4 Δx

=

2 4 + Δx − 4 2 4 + Δx + 4 ⋅ Δx 2 4 + Δx + 4

= = =

4( 4 + Δx) − 16

(

16 + 4Δx − 16

(

) )

Δx 2 4 + Δx + 4 4Δx

)

4 2 4 + Δx + 4

4 2 4 + Δx + 4 4 1 = = 2 2 4 +4

m = lim msec = lim

Δx → 0

26. f ( x ) = msec =

=

9 + Δx − 3 Δx

= =

9 + Δx − 9

Δx Δx

( (

=

m = lim msec Δx → 0

= lim

Δx → 0

1 9 +3

= =

1 9 + Δx + 3

Δx → 0

= lim

Δx → 0

Δx → 0

9 + Δx + 3 9 + Δx + 3

) )

9 + Δx + 3

1 9 + Δx + 3

29. f ( x) = − 5 x

= lim

9 + Δx + 3 Δx

Δx → 0

= 0

= lim 8+1

9 + Δx − 3 ⋅ Δx

=

f ( x + Δx) − f ( x) Δx −2 − ( −2) = lim Δx → 0 Δx 0 = lim Δx → 0 Δx = lim 0

f ′( x) = lim

Δx → 0

f (8 + Δx) − f (8) Δx 8 + Δx + 1 − Δx

28. f ( x) = − 2

f ′( x) = lim

x + 1 at (8, 3)

=

Δx → 0

= 0

Δx → 0

Δx 2 4 + Δx + 4

Δx → 0

f ( x + Δx) − f ( x) Δx → 0 Δx 3−3 = lim Δx → 0 Δx 0 = lim Δx → 0 Δx = lim 0

f ′( x) = lim

Δx 2 4 + Δx + 4

(

75

27. f ( x) = 3

x at ( 4, 4)

m sec =

=

The Derivative and the Slope of a Graph

= lim

Δx → 0

f ( x + Δx) − f ( x) Δx −5( x + Δx) − ( −5 x) Δx −5 x − 5Δx + 5 x Δx −5Δx Δx −5

= −5 30. f ( x) = 4 x + 1

f ( x + Δx) − f ( x) Δx 4( x + Δx) + 1 − ( 4 x + 1) lim Δx → 0 Δx 4 x + 4Δx + 1 − 4 x − 1 lim Δx → 0 Δx 4 Δx lim Δx → 0 Δx lim 4

f ′( x) = lim

Δx → 0

= = = =

Δx → 0

= 4

1 6

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DOWNLOAD SOLUTIONS MANUAL 76

Chapter 2

31. g ( s ) =

1 3

Differentiation 32. h(t ) = 6 − 12 t

s + 2

g ′( s ) = lim

Δs → 0

= lim

g ( s + Δs ) − g ( s ) Δs 1 3

(s

+ Δs ) + 2 − Δs

Δs → 0

= lim

1s 3

+ 13 Δs + 2 − 13 s − 2 Δs

Δs → 0

= lim

( 13 s + 2)

h′(t ) = lim

Δt → 0

= lim

h(t + Δt ) − h(t ) Δt 6−

1 2

Δs 1 = lim Δs → 0 3 1 = 3

= lim

Δs → 0

)

6 − 12 t − 12 Δt − 6 + 12 t Δt

Δt → 0

1 Δs 3

(

+ Δt ) − 6 − 12 t Δt

Δt → 0

= lim

(t

− 12 Δt

Δt 1 = lim − Δt → 0 2 1 = − 2 Δt → 0

33. f ( x ) = 4 x 2 − 5 x

f ′( x) = lim

Δ x→0

f ( x + Δ x) − f ( x) Δx

4( x + Δ x) − 5( x + Δ x) − ( 4 x 2 − 5 x)  = lim  Δ x→0 Δx 2

(

)

4 x 2 + 2 xΔ x + ( Δ x)2 − 5 x − 5Δ x − 4 x 2 + 5 x   = lim  Δ x→0 Δx 4 x 2 + 8 xΔ x + 4( Δ x) − 5 x − 5Δ x − 4 x 2 + 5 x Δx 2

= lim

Δ x→0

5  4Δ x 2 x + Δ x −  4  = lim Δ x→0 Δx 5  = lim 4 2 x + Δ x −  = 8 x − 5 Δ x→0  4 34. f ( x ) = 2 x 2 + 7 x

f ′( x) = lim

Δx →0

f ( x + Δ x) − f ( x) Δx

2( x + Δ x) 2 + 7( x + Δ x) − ( 2 x 2 + 7 x)  = lim  Δx →0 Δx

(

)

2 x 2 + 2 xΔ x + ( Δ x)2 + 7 x + 7 Δ x − 2 x 2 − 7 x   = lim  Δx →0 Δx 2 x 2 + 4 xΔ x + 2( Δ x) + 7 x + 7 Δ x − 2 x 2 − 7 x Δx 2

= lim

Δx →0

7  2Δ x 2 x + Δ x +  2  = lim Δx →0 Δx 7  = lim 2 2 x + Δ x +  = 4 x + 7 Δx →0  2

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DOWNLOAD SOLUTIONS MANUAL Section 2.1 35. h(t ) =

The Derivative and the Slope of a Graph

77

t −3

h(t + Δt ) − h(t ) t

h′(t ) = lim

Δt → 0

= lim

t + Δt − 3 − Δt

t −3

= lim

t + Δt − 3 − Δt

t −3

Δt → 0

Δt → 0

= lim

Δt

Δt → 0

= lim

Δt

Δt → 0

(

t + Δt − 3 − (t − 3) t + Δt − 3 + t + Δt − 3 +

1 t + Δt − 3 +

= lim

Δt → 0

=

1 2 t − 13

=

t −3 2(t − 3)

36. f ( x) =

x + 2

f ′( x) = lim

Δx → 0

t −3

)

t −3

)

Δt

(

f ( x + Δx) − f ( x) Δx x + Δx + 2 − Δx

x+ 2

= lim

x + Δx + 2 − Δx

x+ 2

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

=

2

t −3 t −3

t −3

= lim

Δx → 0

t + Δt − 3 + t + Δt − 3 +

Δx

(

Δx

(

x + Δx + 2 − ( x + 2) x + Δx + 2 +

x+ 2

)

x+ 2

)

Δx x + Δx + 2 +

1 x + Δx + 2 +

x + Δx + 2 + x + Δx + 2 +

x+ 2 x+ 2

x+ 2

1 x + 2

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DOWNLOAD SOLUTIONS MANUAL 78

Chapter 2

Differentiation

37. f (t ) = t 3 − 12t

f ′(t ) = lim

Δt → 0

= lim

f ( t + Δt ) − f ( t ) Δt

(t

+ Δt ) − 12(t + Δt ) − (t 3 − 12t ) 3

Δt

Δt → 0

= lim

t + 3t Δt + 3t ( Δt ) + ( Δt ) − 12t − 12Δt − t 3 + 12t Δt

= lim

3t 2 Δt + 3t ( Δt ) + ( Δt ) − 12Δt Δt

3

Δt → 0

2

2

3

2

Δt → 0

= lim

Δt → 0

(

3

)

Δt 3t 2 + 3tΔt + ( Δt ) − 12 2

Δt

(

)

= lim 3t 2 + 3tΔt + ( Δt ) − 12 Δt → 0 2

2

= 3t − 12 38. f (t ) = t 3 + t 2

f ′(t ) = lim

Δt → 0

= lim

f (t + Δt ) − f (t ) Δt

(t

+ Δt ) + ( t + Δ t ) − ( t 3 + t 2 ) 3

2

Δt

Δt → 0

t 3 + 3t 2 Δt + 3t ( Δt ) + ( Δt ) + t 2 + 2tΔt + ( Δt ) − t 3 − t 2 = lim Δt → 0 Δt 2

3

3t 2 Δt + 3t ( Δt ) + ( Δt ) + 2tΔt + ( Δt ) Δt 2

= lim

Δt → 0

= lim

Δt → 0

(

3

Δt 3t 2 + 3tΔt + ( Δt ) + 2t + Δt 2

Δt

(

= lim 3t 2 + 3tΔt + ( Δt ) + 2t + Δt Δt → 0

2

2

2

)

)

2

= 3t + 2t 39. f ( x) =

1 x+2

f ′( x) = lim

f ( x + Δx) − f ( x)

Δx 1 1 − x + Δ x + x + 2 2 = lim Δx → 0 Δx x + 2 x + Δx + 2 1 1 ⋅ − ⋅ x + Δ x + x + x + x + Δx + 2 2 2 2 = lim Δx → 0 Δx x + 2 − ( x + Δx + 2) ( x + Δx + 2)( x + 2) = lim Δx → 0 Δx −Δx = lim Δx → 0 Δx( x + Δx + 2)( x + 2) Δx → 0

= lim

Δx → 0

= −

−1 x x + Δ + 2)( x + 2) ( 1

(x

+ 2)

2

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DOWNLOAD SOLUTIONS MANUAL Section 2.1

41. f ( x) =

1 s −4 g ( s + Δs ) − g ( s ) g ′( s ) = lim Δs → 0 Δs 1 1 − s s s + Δ − − 4 4 = lim Δs → 0 Δs s − 4 − ( s + Δs − 4) ( s + Δs − 4)( s − 4) = lim Δs → 0 Δs s − 4 − ( s + Δs − 4) ( s + Δs − 4)( s − 4) ⋅ 1 = lim Δs → 0 ( s + Δs − 4)( s − 4) Δs

40. g ( s ) =

= lim

−Δs Δs( s + Δs − 4)( s − 4)

= lim

−1 ( s + Δs − 4)( s − 4)

Δs → 0

Δs → 0

= −

− 4)

79

at ( 2, 2)

1 x2 2

f ( x + Δx) − f ( x)

f ′( x) = lim

Δx

Δx → 0

= lim

1 2

( x + Δx) −

1 2

(x

Δx → 0

= lim

2

+ 2 xΔx + ( Δx) Δx

xΔx + ( Δx)

= lim

Δx → 0

= lim

Δx → 0

2

)−

1 x2 2

2

Δx Δx( x + Δx)

Δx → 0

= lim

1 x2 2

Δx 2

Δx → 0

(x

Δx + Δx)

= x

m = f ′( 2) = 2 y − 2 = 2( x − 2)

1

(s

The Derivative and the Slope of a Graph

y = 2x − 2

2

4

(2, 2) −6

6

−4

42. f ( x) = − 18 x 2 at ( − 4, − 2)

f ′( x) = lim

f ( x + Δx) − f ( x) Δx

Δx → 0

(

− 18 ( x + Δx) − − 18 x 2 2

= lim

Δx → 0

= lim

Δx

(

− 18 x 2 + 2 xΔx + ( Δx) Δx

Δx → 0

= lim

− 18 x 2 −

1 xΔx 4

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

1 8

2

)+

(Δx)2

1 x2 8

+ 18 x 2

Δx

Δx → 0

= lim

)

− 14 xΔx −

1 8

(Δx)

Δx Δx − 14 x − 18 Δx

(

Δx

2

)

(− 14 x − 81 Δx)

= − 14 x m = f ′( − 4) = − 14 ( − 4) = 1

y − ( − 2) = 1 x − ( − 4) y +2 = x+4 y = x+2

3 −9

9

(−4, −2)

−9

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DOWNLOAD SOLUTIONS MANUAL 80

Chapter 2

Differentiation

43. f ( x) = ( x − 1) at ( − 2, 9) 2

f ′( x) = lim

Δx → 0

= lim

f ( x + Δx) − f ( x ) Δx

(x

Δx → 0

+ Δx − 1) − ( x − 1) Δx 2

2

x 2 + 2 xΔx − 2 x + ( Δx) − 2Δx + 1 − x 2 + 2 x − 1 Δx 2

= lim

Δx → 0

2 xΔx + ( Δx) − 2Δx Δx → 0 Δx Δx( 2 x + Δx − 2) = lim Δx → 0 Δx = lim ( 2 x + Δx − 2) 2

= lim

Δx → 0

11

= 2x − 2

(−2, 9)

m = f ′( −2) = 2( −2) − 2 = −6 y − 9 = −6  x − ( −2) y = −6 x − 3

−12

12

−5

44. f ( x) = 2 x 2 − 5 at ( −1, − 3)

f ′( x) = lim

Δx → 0

f ( x + Δx ) − f ( x ) Δx

2( x + Δx) − 5 − ( 2 x 2 − 5) 2

= lim

Δx → 0

Δx

2 x 2 + 4 xΔx + 2( Δx) − 5 − 2 x 2 + 5 = lim Δx → 0 Δx 2

4 xΔx + 2( Δx) Δx → 0 Δx Δx( 4 x + 2Δx) = lim Δx → 0 Δx = lim ( 4 x + 2Δx)

2

= lim

Δx → 0

= 4x

m = f ′( −1) = 4( −1) = − 4 y − ( −3) = − 4( x − ( −1)) y + 3 = − 4x − 4 y = − 4x − 7

2 −6

6

(−1, −3)

−6

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DOWNLOAD SOLUTIONS MANUAL Section 2.1 45. f ( x) =

81

x + 1 at ( 4, 3)

f ′( x) = lim

Δx → 0

f ( x + Δx) − f ( x) Δx x + Δx + 1 −

= lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

=

The Derivative and the Slope of a Graph

(

)

x +1

Δx x + Δx − x ⋅ Δx x + Δx − x Δx Δx

( (

x + Δx +

x + Δx + x + Δx + x

)

x

)

Δx x + Δx +

1 x + Δx +

x x

x

1 x

2

m = f ′( 4) =

1 2 4

=

1 4

5

(4, 3)

1 y − 3 = ( x − 4) 4 1 y = x+ 2 4 46. f ( x ) =

−2

7 −1

x + 3 at (6, 3)

f ( x + Δx) − f ( x) Δx x + Δx + 3 − x + 3 = lim Δx → 0 Δx x + Δx + 3 − x + 3 x + Δx + 3 + = lim ⋅ Δx → 0 Δx x + Δx + 3 + x + Δx + 3 − ( x + 3) = lim Δx → 0 Δx x + Δx + 3 + x + 3

f ′( x) = lim

Δx → 0

(

= lim

Δx → 0

= lim

Δx → 0

= =

2

Δx

(

x+3 x+3

)

Δx x + Δx + 3 +

1 x + Δx + 3 +

x+3

)

x+3

1 x+3

x+3 2( x + 3)

m = f ′(6) =

1 1 = 6 2 6+3

1 ( x − 6) 6 1 y − 3 = x −1 6 1 y = x+ 2 6

7

y −3 =

(6, 3)

−4

8 −1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 82

Chapter 2

47. f ( x) =

f ′( x) =

=

= =

Differentiation

1  1  at  − , −1 5x  5  f ( x + Δx) − f ( x) lim Δx → 0 Δx 1 1 − 5( x + Δx) 5 x lim Δx → 0 Δx x − ( x + Δx) 5 x( x + Δx) lim Δx → 0 Δx x − ( x + Δx) 1 ⋅ lim Δx → 0 5 x( x + Δx ) Δx

= lim

−Δx 5 x ⋅ Δx ⋅ ( x + Δx)

= lim

−1 5 x( x + Δx)

Δx → 0

Δx → 0

1 5x2 1 1  1 m = f ′ −  = − = − = −5 2 1 5     1 5  5 −   25   5   1  y − ( −1) = − 5 x −  −    5   = −

−3

1  y + 1 = − 5 x +  5  y + 1 = − 5x − 1 y = − 5x − 2 48. f ( x) =

1 3

(

(

− 1 , −1 5

−3

1 at ( 2, −1) x−3

f ′( x) = lim

Δx →0

= lim

Δx →0

= lim

Δx →0

= lim

Δx →0

= lim

Δx →0

= lim

Δx →0

f ( x + Δ x) − f ( x) Δx 1 1 − x + Δx − 3 x − 3 Δx 1 1 x −3 x + Δx − 3 ⋅ − ⋅ x + Δx − 3 x − 3 x − 3 x + Δx − 3 Δx x − 3 − ( x + Δ x − 3) ( x + Δ x − 3)( x − 3) Δx −Δx ( x + Δ x − 3)( x − 3)Δ x

(x

1 −1 = − 2 + Δ x − 3)( x − 3) x − ( 3)

m = f ′( 2) = − y − ( −1) = −1( x − 2) y + 1 = −x + 2 y = −x + 1

1

(2 − 3)

2

3

= −1 −2

(2, −1)

7

−3

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.1

Δx → 0

= lim

f ( x + Δx) − f ( x) Δx 2 − 14 ( x + Δx) − − 14 x 2 − 14 x 2 −

1 xΔx 2

Δx → 0

= lim

1 4

f ′( x) = lim

Δ x →0

(Δx)2

− 12 xΔx −

1 4

(Δx)

Δx Δx − 12 x − 14 Δx

(

f ( x + Δ x) − f ( x) Δx

( x + Δ x) 2 − 7 − ( x 2 − 7)  = lim  Δ x →0 Δx

)

x 2 + 2 xΔ x + ( Δ x) − 7 − x 2 + 7 Δ x →0 Δx Δ x( 2 x + Δ x ) = lim Δ x →0 Δx = lim ( 2 x + Δ x) = 2 x 2

1 x2 4

+

= lim

Δx

Δx → 0

= lim

(

Δx

Δx → 0

= lim

83

50. f ( x ) = x 2 − 7

1 49. f ( x ) = − x 2 4 f ′( x) = lim

The Derivative and the Slope of a Graph

2

)

Δ x →0

Δx 1   1 = lim  − x − Δx  Δx → 0 2 4  1 = − x 2 Since the slope of the given line is −1, Δx → 0

Since the slope of the given line is − 2, 2x = − 2 x = −1 and f ( −1) = − 6. At the point ( −1, − 6), the tangent line parallel to

2 x + y = 0 is

− 12 x = −1

y − ( − 6) = − 2( x − ( −1))

x = 2 and f ( 2) = −1.

At the point ( 2, −1), the tangent line parallel to

y = − 2 x − 8.

x + y = 0 is y − ( −1) = −1( x − 2) y = − x + 1.

1 51. f ( x) = − x3 3 f ′( x) = lim

Δx →0

f ( x + Δ x) − f ( x) Δx Δx

Δx →0

= lim

(

− 13 ( x + Δ x) − − 13 x3 3

= lim

(

− 13 x3 + 3 x 2 Δ x + 3 x( Δ x) + ( Δ x) 2

Δx

Δx →0

= lim

− 13 x3

− x Δ x − x( Δ x ) − 2

2

(

2

Δ x − x − xΔ x −

Δx →0

(

2

Δx

= lim − x − xΔ x − Δx →0

1 3

( Δ x )3

3

)+

1 x3 3

+ 13 x3

Δx

Δx →0

= lim

)

1 3

1 3

( Δ x)2 )

( Δ x)2 )

= − x2

Since the slope of the given line is − 9, − x2 = − 9 x2 = 9 x = ± 3 and f (3) = − 9 and f ( − 3) = 9.

At the point (3, − 9), the tangent line parallel to 9 x + y − 6 = 0 is

y − ( − 9) = − 9( x − 3) y = − 9 x + 18. At the point ( − 3, 9), the tangent line parallel to 9 x + y − 6 = 0 is y − 9 = − 9( x − ( − 3)) y = − 9 x − 18.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 84

Chapter 2

Differentiation

52. f ( x) = x3 + 2

f ( x + Δx) − f ( x) Δx

f ′( x) = lim

Δx → 0

= lim

+ Δx) + 2 − ( x3 + 2)

(x

3

Δx

Δx → 0

= lim

x + 3 x Δx + 3 x( Δx) + ( Δx) + 2 − x3 − 2 Δx

= lim

3 x 2 Δx + 3( Δx) + ( Δx) Δx

3

Δx → 0

2

2

2

Δx → 0

= lim

(

Δx 3 x 2 + 3 xΔx + ( Δx)

Δx → 0

(

Δx

= lim 3 x + 3 xΔx + ( Δx) Δx → 0

2

2

3

3

2

)

)

= 3x 2 The slope of the given line is 3x − y − 4 = 0

y = 3x − 4 m = 3.

3x 2 = 3 x2 = 1 x = ±1 x = 1 and f (1) = 3 x = −1 and f ( −1) = 1 At the point (1, 3), the tangent line parallel to 3x − y − 4 = 0 is y − 3 = 3( x − 1) y − 3 = 3x − 3 y = 3x.

At the point ( −1, 1), the tangent line parallel to 3x − y − 4 = 0 is y − 1 = 3( x − ( −1)) y − 1 = 3( x + 1) y − 1 = 3x + 3 y = 3x + 4.

53. y is differentiable for all x ≠ −3.

At ( −3, 0), the graph has a node. 54. y is differentiable for all x ≠ ± 3.

At ( ± 3, 0), the graph has a cusp. 55. y is differentiable for all x ≠ − 12 .

At

(

− 12 ,

56. y is differentiable for all x > 1.

The derivative does not exist at endpoints. 57. y is differentiable for all x ≠ ± 2.

The function is not defined at x = ± 2. 58. y is differentiable for all x ≠ 0.

The function is discontinuous at x = 0.

)

0 , the graph has a vertical tangent line.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.1

The Derivative and the Slope of a Graph

85

y

59. Since f ′( x) = −3 for all x, f is a line of the form

5

f ( x) = −3x + b. f (0) = 2, so 2 = ( −3)(0) + b, or b = 2.

2 1

Thus, f ( x) = −3x + 2.

−4 −3 −2 −1

2

3

4

x

−2 −3

60. Sample answer: Since f ( −2) = f ( 4) = 0,

( x + 2)( x − 4)

= 0.

A function with these zeros is f ( x) = x 2 − 2 x − 8. f ( x + Δx) − f ( x) Δx ( x + Δx)2 − 2( x + Δx) − 8 − ( x 2 − 2 x − 8)  = lim  Δx → 0 Δx 2 x 2 + 2 xΔx + ( Δx) − 2 x − 2Δx − 8 − x 2 + 2 x + 8 = lim Δx → 0 Δx 2 2 xΔx + ( Δx) − 2Δx = lim Δx → 0 Δx = lim 2 x + Δx − 2

Then f ′( x) = lim

Δx → 0

y 1 −3

−1

1 2 3

5 6

x

−2

Δx → 0

= 2 x − 2. So f ′(1) = 2(1) − 2 = 0. Sketching f ( x) shows that

−8 −9

f ′( x) < 0 for x < 1 and f ′(0) > 0 for x > 1. 2

61. −2

2

−2

−2

x

− 32

−1

− 12

0

1 2

1

3 2

2

f ( x)

1

0.5625

0.25

0.0625

0

0.0625

0.25

0.5625

1

f ′( x)

−1

−0.75

−0.5

−0.25

0

0.25

0.5

0.75

1

Analytically, the slope of f ( x ) = m = lim

Δx → 0

= lim

f ( x + Δx ) − f ( x) Δx 1 x + Δx 2 − 1 x 2 ( ) 4 4

Δx → 0

= lim

1 x 2 4

2 2 + 2 x( Δx) + ( Δx )  − 14 x 2  Δx 1 x 2 + 1 xΔx + 1 Δx 2 − 1 x 2 4 2 4( ) 4

Δx

Δx → 0

= lim

1 xΔx 2

Δx → 0

= lim

Δx → 0

=

+

1 4

( Δx)

2

Δx

Δx → 0

= lim

is

Δx

Δx → 0

= lim

1x 4

2

Δx

( 12 xΔx + 14 Δx) Δx

( 12 x + 14 Δx)

1 x. 2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 86

Chapter 2

Differentiation 4

62. −6

6

−4

− 32

− 12

3 2

−2

f ( x)

−6

− 2.53

− 0.75

− 0.1

0

0.1

0.75

2.53

6

9

5.0625

2.25

0.5625

0

0.5625

2.25

5.0625

9

3 2

2

f ′( x)

−1

1 2

x

Analytically, the slope of f ( x) = m = lim

Δ x →0

= lim

(x

3 4

(x

Δ x →0

= lim

+ Δ x) − 3

+ 3 x 2 Δ x + 3 x( Δ x ) + ( Δ x ) 2

Δx

Δ x →0

= lim

3 x3 4

9 x 2Δ x 4

+

+

9x 4

9 x 2Δ x 4

Δ x →0

= lim

Δ x →0

= lim

Δ x →0

=

( Δ x)

2

+

3 4

3

)−

( Δ x)

3

3 x3 4

3 x3 4

Δx

Δ x →0

= lim

2

3 x3 4

Δx 3

1

3 x 3 is 4

f ( x + Δ x ) − f ( x) Δx 3 4

0

+

(

9 x2 4

( Δ x)

( Δ x)

2

+

3x 4

+

3 4

( Δ x)

Δx

Δ x 94 x 2 +

(

9x 4

9 xΔ x 4

Δx +

9 xΔ x 4

+

3 4

(Δ x)

2

2

3

)

)

9 x2. 4

63.

2

−2

2

−2

− 32

−1

− 12

0

4

1.6875

0.5

0.0625

0

−0.0625

−0.5

−1.6875

−4

−6

−3.375

−1.5

−0.375

0

−0.375

−1.5

−3.375

−6

−2

x

f ( x)

f ′( x)

1 2

1

Analytically, the slope of f ( x ) = − 12 x 3 is m = lim

Δx → 0

= lim

f ( x + Δx) − f ( x) Δx 3 − 12 ( x + Δx) + 12 x3

Δx 2 3 − 12  x3 + 3 x 2 Δx + 3 x( Δx) + ( Δx)  +   = lim Δx → 0 Δx 2 3 − 12 3 x 2 Δx + 3 x( Δx) + ( Δx)    = lim Δx → 0 Δx 1 2 = lim − 3 x 2 + 3 x( Δx) + ( Δx)   Δx → 0 2  3 2 = − x . 2 Δx → 0

1 x3 2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.1 64.

The Derivative and the Slope of a Graph

87

2 −2

2

−8

−2

− 32

−1

− 12

0

f ( x)

−6

−3.375

−1.5

−0.375

0

−0.375

−1.5

−3.375

−6

f ′( x)

6

4.5

3

1.5

0

−1.5

−3

−4.5

−6

x

1 2

3 2

1

2

Analytically, the slope of f ( x ) = − 32 x 2 is m = lim

Δx → 0

f ( x + Δx) − f ( x) Δx

(

− 32 ( x + Δx) − − 32 x 2 2

= lim

Δx

Δx → 0

= lim

− 32  x 2 

Δx → 0

)

2 + 2 xΔx + ( Δx)  +  Δx

3 x2 2

2 − 32 2 xΔx + ( Δx)    = lim Δx → 0 Δx 3 = lim − ( 2 x + Δx) Δx → 0 2 = − 3 x.

65. f ′( x) = lim

Δx → 0

= lim

f ( x + Δx) − f ( x) Δx

(x

8

+ Δx) − 4( x + Δx) − ( x 2 − 4 x) 2

Δx

Δx → 0

12

−6

2 xΔx + ( Δx) − 4Δx Δx → 0 Δx = lim ( 2 x + Δx − 4) 2

−4

= lim

Δx → 0

= 2x − 4 The x-intercept of the derivative indicates a point of horizontal tangency for f. 66. f ′( x) = lim

Δx → 0

= lim

Δx → 0

f ( x + Δx ) − f ( x ) Δx 6Δx − 2 xΔx − ( Δx) Δx

2 + 6( x + Δ x ) − ( x + Δ x ) − ( 2 + 6 x − x 2 ) 2

= lim

Δx → 0

2

Δx

= lim (6 − 2 x − Δx) = 6 − 2 x Δx → 0

13

14

−7 −1

The x-intercept of the derivative indicates a point of horizontal tangency for f.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Chapter 2

Differentiation

67. f ′( x ) = lim

f ( x + Δx ) − f ( x) Δx

88

Δx → 0

= lim

(x

+ Δx) − 3( x + Δx) − ( x 3 − 3x) 3

Δx

Δx → 0

x + 3 x Δx + 3x( Δx) + ( Δx) − 3 x − 3Δx − x3 + 3 x Δx 3

= lim

Δx → 0

2

2

3

3 x 2 Δx + 3x( Δx) + ( Δx) − 3Δx Δx → 0 Δx 2

3

= lim

(

)

= lim 3x 2 + 3 xΔx + ( Δx ) − 3 Δx → 0

2

2

= 3x − 3 4

−6

6

−4

The x-intercepts of the derivative indicate points of horizontal tangency for f. 68. f ′( x) = lim

Δx → 0

= lim

f ( x + Δx) − f ( x) Δx

(x

+ Δx) − 6( x + Δx) − ( x3 − 6 x 2 ) 3

2

Δx

Δx → 0

= lim

2

2

(

3

Δx

Δx → 0

= lim 3x + 3 xΔx + ( Δx) − 12 x − 6Δx Δx → 0

(

x − 3x Δx + 3 x( Δx) + ( Δx) − 6 x 2 + 2 xΔx + ( Δx) 3

2

2

2

)−x

3

+ 6x2

)

2

= 3 x − 12 x 4 −4

8

−36

The x-intercepts of the derivative indicate points of horizontal tangency for f. 69. True. The slope of the graph is given by f ′( x) = 2 x,

which is different for each different x value. 70. False. f ( x) = x is continuous, but not differentiable at

73. The graph of f ( x ) = x 2 + 1 is smooth at (0, 1), but the

graph of g ( x) = x + 1 has a node at (0, 1). The function g is not differentiable at (0, 1). 6

x = 0. 71. True. See page 122. 72. True. See page 115.

−4

4 −1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.2

Some Rules for Differentiation

89

Section 2.2 Some Rules for Differentiation Skills Warm Up 1 1 5.   x −3 4 = 4 x3 4  4

1. (a) 2 x 2 , x = 2

2( 22 ) = 2( 4) = 8 (b)

(5 x ) 2 ,

6.

x = 2 2

5( 2) = 10 2 = 100

(c) 6 x −2 , x = 2 6( 2)

2. (a)

()

−2

1

(3 x )

7. 3 x 2 + 2 x = 0

3 2

=

x ( 3 x + 2) = 0 x = 0

, x = 2

2

1

3( 2) (b)

1 4

= 6

1 1 1 1 (3) x 2 − 2  x −1 2 + x −2 3 = x 2 − x −1 2 + x −2 3 3 3 3  2 1 1 = x2 − + 3 x 3 x2

2

=

3 x + 2 = 0 → x = − 23

1 1 = 62 36

x3 − x = 0

8.

x( x 2 − 1) = 0

1 , x = 2 4 x3

x( x + 1)( x − 1) = 0 x = 0

1 1 1 = = 3 4 8 32 4( 2 ) () (c)

(2 x) 4x

x + 1 = 0 → x = −1 x −1 = 0 → x =1

−3

−3

2( 2) −2 4( 2)

(x

4−3

=

4( 2)

−2

22

=

4( 4

3

)

=

x + 10 = 0 → x = −10 x − 2 = 0 → x = 2

(

1 2

(3) x 2

3 12 x 2

=

3 2 x 2

1. y = 3

10.

)

3 2

x =

3 12 x 2

( x3 2

3x 2 − 10 x + 8 = 0

(3 x

− 4)( x − 2) = 0 3x − 4 = 0 → x =

− 1)

4 3

x −2 = 0 → x = 2

6. h( x) = 6 x5

y′ = 0

h′( x) = 30 x 4

2. f ( x ) = − 8

5x4 3 20 3 10 3 y′ = x = x 6 3

7. y =

f ′( x) = 0

3. y = x5

y′ = 5 x 4 4. f ( x) =

+ 10)( x − 2) = 0

1 64

3. 4(3) x3 + 2( 2) x = 12 x3 + 4 x = 4 x 3x 2 + 1 4.

x 2 + 8 x − 20 = 0

9.

, x = 2

−2

1 = x −6 x6

f ′( x) = − 6 x − 7 = − 5. h( x) = 3x

3

h′( x) = 9 x

3t 2 4 3 g ′(t ) = t 2

8. g (t ) =

6 x7

9. f ( x) = 4 x f ′( x) = 4

2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 90

Chapter 2

Differentiation 17. s(t ) = 4t 4 − 2t + t + 3

1 x = x 3 3 1 g ′( x) = 3

10. g ( x ) =

s′(t ) = 16t 2 − 4t + 1 18. y = 2 x3 − x 2 + 3 x − 1

11. y = 8 − x3

y′ = 6 x 2 − 2 x + 3

y = − 3x 2

19. g ( x) = x 2 3

12. y = t 2 − 6

g ′( x) =

y′ = 2t

2 −1 3 2 = 13 x 3 3x

20. h( x ) = x 5 2

13. f ( x) = 4 x 2 − 3x

h′( x) =

f ′( x) = 8 x − 3 14. g ( x) = 3x 2 + 5 x3

5 32 x 2

21. y = 4t 4 3

( 43 )t

g ′( x) = 6 x + 15 x 2 = 15 x 2 + 6 x

y′ = 4

13

15. f (t ) = − 3t 2 + 2t − 4

22. f ( x ) = 10 x1 6

f ′(t ) = − 6t + 2

f ′( x) =

16. y = 7 x3 − 9 x 2 + 8

=

16 1 3 t 3

5 −5 6 5 5 = = x 6 3 3 x5 6 3 x5

23. y = 4 x − 2 + 2 x 2

y′ = 21x 2 − 18 x

y′ = −8 x −3 + 4 x1 = −

8 + 4x x3

24. s(t ) = 8t − 4 + t

s′(t ) = 8( − 4t − 6 ) + 1 = − Function

2 25. y = 7 x4 26. y = 27. y =

28. y =

29. y =

30. y =

2 3x 2

32. y =

4 3x3

1 −3 x 64

y′ = −

3 −4 x 64

y′ = −

3 64 x 4

6

y =

y′ = −

6π −7 x 64

y′ = −

3π 32 x 7

−5

π 64

x −6

y = 128 x5

y = 4 x4

x 4

y′ = −

y =

x

3

8 7 x5

3

4x x −3

31. y = 6

y′ = −

4 y′ = − x −3 3

4

( 2 x)

Simplify

− 8 −5 y′ = x 7

2 −2 x 3

π

( 2 x)

Differentiate

2 y = x−4 7 y =

1

( 4 x)

Rewrite

y = 6 x1 2

y =

3 12 x 4

y′ = 128(5) x 4

y′ = 640 x 4

y′ = 4( 4) x3

y′ = 16 x 3

( 12 ) x

y′ = 6

y′ =

−1 2

3  1  −1 2  x 4 2 

3 x

y′ = y′ =

32 +1 t6

3 8

x

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.2 Function 33. y =

34. y =

Rewrite

1 76 x

3 2

4

35. y =

5

8x

36. y =

3

6 x2

x

3

Differentiate

1 −1 6 x 7

y′ =

1  1  −7 6 − x 7 6 

y′ = −

y =

3 −3 4 x 2

y′ =

3  3  −7 4 − x 2 4 

y′ = −

y = (8 x)

15

3

y =

6 ( x)

= 81 5 x1 5

23

1  y′ = 81 5  x − 4 5  5  y′ =

3

 2 6   x −1 3  3

y′ =

(1)

12

=

3 2

= m.

38. y = x −1

44.

y′ = x − 2 = −

(4)

−4

f ′(t ) = − 4t

−5

8

x7

5 5 1 8 8 = = 45 5 4 5x 5x 5 x

23 6 33 x

f ′( x) = − 12 −

3 2 x 2

f ′(1) = − 12 −

3 2

1 x3 2

= −2

f ( x) = 3(5 − x ) = 75 − 30 x + 3 x 2 2

f ′( x) = −30 + 6 x

1 x2

1 16  3 4 At the point  , , y′ = − 2 = − = m. 3 4 3 9  

39. f (t ) = t

9 4

f ( x) = − 12 x(1 + x 2 ) = − 12 x −

3 12 x 2 3 2

1 42 6 x 7

y′ = 81 5

43.

At the point (1, 1), y′ =

91

Simplify

y =

37. y = x 3 2 y′ =

Some Rules for Differentiation

f ′(5) = −30 + (6)(5) = 0

45. (a)

y = −2 x 4 + 5 x 2 − 3 y′ = −8 x3 + 10 x

m = y′(1) = −8 + 10 = 2

4 = − 5 t

The equation of the tangent line is

At the point 4 4 1  1 = − = −128 = m.  , 16 , f ′  = − 5 1 2   2 1     32  2

y − 0 = 2( x − 1) y = 2 x − 2. (b) and (c)

3.1

−4.7

40. f ( x) = x −1 3

4.7

1 1 f ′( x) = − x − 4 3 = − 4 3 3 3x

1 1  1 = − = m. At the point  8, , f ′(8) = − 43 48  2 3(8)

−3.1

46. (a) y = x3 + x + 4

y′ = 3 x 2 + 1

41. f ( x) = 2 x3 + 8 x 2 − x − 4

m = y′( − 2) = 3( − 2) + 1 = 13

f ′( x) = 6 x 2 + 16 x − 1

The equation of the tangent line is

At the point

(−1, 3), f ′(−1) = 6( −1) + 16( −1) − 1 = −11 = m. 2

42. f ( x) = x 4 − 2 x3 + 5 x 2 − 7 x f ′( x) = 4 x3 − 6 x 2 + 10 x − 7

At the point 3 2 (−1, 15), f ′( −1) = 4( −1) − 6(−1) + 10( −1) − 7 = − 4 − 6 − 10 − 7 = − 27 = m.

2

y − ( −6) = 13 x − ( − 2) y + 6 = 13x + 26 y = 13x + 20. (b) and (c)

50

−5

5

−50

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DOWNLOAD SOLUTIONS MANUAL 92

Chapter 2

47. (a)

Differentiation

f ( x) = f ′( x) =

3

5

x +

x = x1 3 + x1 5

50. (a)

1 −2 3 1 −4 5 1 1 + x = + 45 x 3 5 3x 2 3 5x

y = ( 2 x + 1)

2

y = 4 x2 + 4 x + 1 y′ = 8 x + 4

1 1 8 m = f ′(1) = + = 3 5 15

m = y′ = 8(0) + 4 = 4

The equation of the tangent line is

The equation of the tangent line is y − 1 = 4( x − 0)

8 ( x − 1) 15 8 22 y = x + . 15 15

y − 2 =

y = 4 x + 1. (b) and (c)

(b) and (c)

3

3.1

−4 −4.7

2

4.7 −1

51.

−3.1

48. (a)

f ( x) =

1 3

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

− x = x −2 3 − x

x2 2 f ′( x ) = − x −5 3 − 1 3 2 1 −1 = − m = f ′( −1) = 3 3

52.

53.

y − 2 = − 13 ( x + 1) y = − 13 x + 53 . 4

54.

4

−5

55.

−2

f ( x) = 6 x 2 − 5 x −2 + 7 x −3

y = 3x3 − 6

56.

y′ = 9 x 2

2 = x 2 − 2 x − 2 x −4 x4 8 f ′( x) = 2 x − 2 + 8 x −5 = 2 x − 2 + 5 x 1 = x 2 + 4 x + x −1 x 1 f ′( x) = 2 x + 4 − x −2 = 2 x + 4 − 2 x f ( x) = x 2 + 4 x +

f ( x) = x 4 5 + x

m = y′ = 9( 2) = 36 57.

The equation of the tangent line is

4 −1 5 4 x +1 = +1 5 5 x1 5

f ( x) = x1 3 − 1 f ′( x) =

2

10 21 − 4 x3 x

f ( x) = x 2 − 2 x −

f ′( x) =

2  49. (a) y = 3 x x 2 −  x 

1 −2 3 1 x = 3 3x2 3

f ( x ) = x( x 2 + 1) = x 3 + x f ′( x ) = 3 x 2 + 1

y − 18 = 36( x − 2) y = 36 x − 54. (b) and (c)

4 6 + 3 2 x x

f ′( x) = 12 x + 10 x −3 − 21x − 4 = 12 x +

The equation of the tangent line is

(b) and (c)

f ( x) = x 2 − 4 x −1 − 3 x −2

58. 50

−5

f ( x ) = ( x 2 + 2 x )( x + 1) = x 3 + 3 x 2 + 2 x f ′( x ) = 3 x 2 + 6 x + 2

5

−50

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DOWNLOAD SOLUTIONS MANUAL Section 2.2

59.

f ( x) =

f ( x) =

f ( x) =

2( x3 − 3) 6 2 x3 − 6 = = 3 3 x x x3

2 x 2 − 3x + 1 = 2 x − 3 + x −1 x

f ′( x) = 2 − x −2 = 2 − 61.

1 2x2 − 1 = x2 x2

4 x3 − 3 x 2 + 2 x + 5 = 4 x − 3 + 2 x −1 + 5 x −2 x2

f ′( x) = 4 − 2 x −2 − 10 x −3 = 4 − 62.

93

2 x3 − 4 x 2 + 3 = 2 x − 4 + 3x −2 x2

f ′( x) = 2 − 6 x −3 = 2 − 60.

Some Rules for Differentiation

2 10 4 x3 − 2 x − 10 − 3 = 2 x x x3

−6 x3 + 3 x 2 − 2 x + 1 = −6 x 2 + 3 x − 2 + x −1 x 1 f ′( x) = −12 x + 3 − x −2 = −12 x + 3 − 2 x f ( x) =

67. y = x 2 + 3

y′ = 2x Set y′ = 4.

2x = 4

63. y = x 4 − 2 x + 3

x = 2

y′ = 4 x3 − 4 x = 4 x( x 2 − 1) = 0 when x = 0, ±1

If x = 2, y = ( 2) + 3 = 7 → ( 2, 7).

If x = ±1, then y = ( ± 1) − 2( ± 1) + 3 = 2.

The graph of y = x 2 + 3 has a tangent line with slope

The function has horizontal tangent lines at the points (0, 3), (1, 2), and (−1, 2).

m = 4 at the point ( 2, 7).

4

2

64. y = x 3 + 3 x 2 y′ = 3 x 2 + 6 x = 3 x( x + 2) = 0 when x = 0, − 2.

The function has horizontal tangent lines at the points (0, 0) and (−2, 4).

2

68. y = x 2 + 2 x

y′ = 2x + 2 Set y′ = 10.

2 x + 2 = 10 x = 4 If x = 4, y = ( 4) + 2( 4) = 24 → ( 4, 24). 2

65. y =

1 x2 2

+ 5x

y′ = x + 5 = 0 when x = −5. The function has a horizontal tangent line at the point

The graph of y = x 2 + 2 x has a tangent line with slope m = 10 at the point ( 4, 24).

(−5, − 252 ).

66. y = x 2 + 2 x

y′ = 2 x + 2 = 0 when x = −1. The function has a horizontal tangent line at the point ( −1, −1).

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DOWNLOAD SOLUTIONS MANUAL 94

Chapter 2

Differentiation

69. (a)

70. (a)

y

y 5

4

g

f

4

g −4

2

3

f

−2

2

4

2

x

−2

−3 −2 −1 −4

1

−1

2

3

x

(b) f ′( x) = 2 x

(b) f ′( x) = g ′( x ) = 3x 2

f ′(1) = 2

f ′(1) = g ′(1) = 3

g ′( x) = 6 x g ′(1) = 6

(c) Tangent line to f at x = 1: f (1) = 1

(c) Tangent line to f at x = 1:

y − 1 = 3( x − 1)

f (1) = 1

y = 3x − 2

y − 1 = 2( x − 1)

Tangent line to g at x = 1:

y = 2x − 1

g (1) = 4

Tangent line to g at x = 1:

y − 4 = 3( x − 1)

g (1) = 3

y = 3x + 1

y − 3 = 6( x − 1)

y

y = 6x − 3

4

y 2

g −4

5

f

−2

2

4

g

f

4

x

3

−2

2

−4

−3 −2 −1

(d) f ′ and g′ are the same.

−1

1

2

3

x

(d) g′ is 3 times f ′. 71. If g ( x) = f ( x) + 6, then g ′( x ) = f ′( x ) because the derivative of a constant is 0, g ′( x) = f ′( x). 72. If g ( x) = 2 f ( x), then g ′( x) = 2 f ′( x) because of the Constant Multiple Rule. 73. If g ( x) = − 5 f ( x), then g ′( x) = − 5 f ′( x) because of the Constant Multiple Rule. 74. If g ( x) = 3 f ( x) − 1, then g ′( x) = 3 f ′( x) because of the Constant Multiple Rule and the derivative of a constant is 0. 75. (a)

R = − 4.1685t 3 + 175.037 2 − 1950.88t + 7265.3 R′ = −12.5055t 2 + 350.074t − 1950.88 2009: R′(9) = −12.5055(9) + 350.074(9) − 1950.88 ≈ $186.8 million per year 2

2011: R′(11) = −12.5055(11) + 350.074(1) − 1950.88 ≈ $386.8 million per year 2

(b) These results are close to the estimates in Exercise 13 in Section 2.1. (c) The slope of the graph at time t is the rate at which sales are increasing in millions of dollars per year.

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DOWNLOAD SOLUTIONS MANUAL Section 2.2 76. (a)

Some Rules for Differentiation

95

R = − 2.67538t 4 + 94.0568t 3 − 1155.203t 2 + 6002.42t − 9794.2 R′ = −10.70152t 3 + 282.1704t 2 − 2310.406t + 6002.42 2010: R′(10) = −10.70152(10) + 282.1704(10) − 2310.406(10) + 6002 ≈ $413.88 million per year 3

2

2012: R′(12) = −10.70152(12) + 282.1704(12) − 2310.406(12) + 6002 ≈ $417.86 million per year 3

2

(b) These results are close to the estimates in Exercise 14 in Section 2.1. (c) The slope of the graph at time t is the rate at which sales are increasing in millions of dollars per year. 77. (a) More men and women seem to suffer from migraines between 30 and 40 years old. More females than males suffer from migraines. Fewer people whose income is greater than or equal to $30,000 suffer from migraines than people whose income is less than $10,000.

f ′( x) = 12.3 x 2 − 24 x + 2.5 12

f′

greater at game 1. (b) The attendance rate for basketball games, f ′(t ),

3

0

(b) The derivatives are positive up to approximately 37 years old and negative after about 37 years of age. The percent of adults suffering from migraines increases up to about 37 years old, then decreases. The units of the derivative are percent of adults suffering from migraines per year. 78. (a) The attendance rate for football games, g ′(t ), is

f ( x) = 4.1x3 − 12 x 2 + 2.5 x

81.

f −12

f has horizontal tangents at (0.110, 0.135) and

(1.841, −10.486). 82.

f ( x) = x3 − 1.4 x 2 − 0.96 x + 1.44 f ′( x) = 3 x 2 − 2.8 x − 0.96 5

is greater than the rate for football games, g ′(t ),

f′

at game 3. (c) The attendance rate for basketball games, f ′(t ), is greater than the rate for football games, g ′(t ), at game 4. In addition, the attendance rate for football games is decreasing at game 4. (d) At game 5, the attendance rate for football continues to increase, while the attendance rate for basketball continues to decrease. 79. C = 7.75 x + 500

C′ = 7.75, which equals the variable cost.

−2

2

f −5

f has horizontal tangents at (1.2, 0) and ( −0.267, 1.577). 83. False. Let f ( x) = x and g ( x) = x + 1.

Then f ′( x) = g ′( x) = 1, but f ( x) ≠ g ( x). 84. True. c is a constant.

80. C = 150 x + 7000 P = R −C P = 500 x − (150 x + 7000)

P = 350 x − 7000

P′ = 350, which equals the profit on each dinner sold.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Chapter 2

96

Differentiation

Section 2.3 Rates of Change: Velocity and Marginals Skills Warm Up 1.

−63 − ( −105) 21 − 7

42 14 = 3 =

−43 − 35 − 78 2. = 6 − ( − 7) 13 = −6 3.

24 − 33 −9 = = −3 9−6 3

40 − 16 24 12 = = 4. 18 − 8 10 5 5. y = 4 x 2 − 2 x + 7

y′ = 8 x − 2 6. s = − 2t 3 + 8t 2 − 7t

s′ = − 6t 2 + 16t − 7 7. s = −16t 2 + 24t + 30

s′ = −32t + 24

8. y = −16 x 2 + 54 x + 70

y′ = −32 x + 54 9. A =

A′ =

(−2r 3 + 3r 2 + 5r ) 1 −6r 2 + 6r + 5 ) 10 ( 1 10

A′ = − 53 r 2 + 53 r + 10. y =

y′ =

1 2

(6 x3 − 18x2 + 63x − 15) 1 18 x 2 − 36 x + 63) 9( 1 9

y′ = 2 x 2 − 4 x + 7 x2 5000 2x y′ = 12 − 5000 x y′ = 12 − 2500

11. y = 12 x −

12. y = 138 + 74 x −

y′ = 74 −

x3 10,000

3x 2 10,000

120 − 63 = $9.5 billion yr 6−0

(e) 2004–2010:

408 − 305 ≈ $17.2 billion yr 30 − 24

(b) 1986–1992:

165 − 120 = $7.5 billion yr 12 − 6

(f ) 1980 –2012:

453 − 63 ≈ $12.2 billion yr 32 − 0

(c) 1992 –1998:

226 − 165 ≈ $10.2 billion yr 18 − 12

(g) 1990 –2012:

453 − 152 ≈ $13.7 billion yr 32 − 10

(d) 1998–2004:

305 − 226 ≈ $13.2 billion yr 24 − 18

(h) 2000 –2012:

453 − 269 ≈ $15.3 billion yr 32 − 20

1. (a) 1980 –1986:

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DOWNLOAD SOLUTIONS MANUAL Section 2.3 2. (a) Imports:

1980 –1990:

495 − 245 = $25 billion yr 10 − 0

(b) Exports: 1980–1990:

394 − 226 = $16.8 billion yr 10 − 0 1218 − 495 ≈ $72.3 billion yr 20 − 10

(d) Exports:

(e) Imports:

1560 − 1218 = $38.0 billion yr 29 − 20

(f ) Exports: 2000–2010:

1056 − 782 = $30.4 billion yr 29 − 20

(g) Imports: 1980 –2013:

6. f ( x) = − x 2 − 6 x − 5; [− 3, 1]

Average rate of change: f (1) − f ( −3) Δf −12 − 4 = = = −4 Δx 1 − ( −3) 4

f ′( x) = − 2 x − 6

7. f ( x) = 3 x 4 3 ; [1, 8]

Average rate of change:

782 − 394 = $38.8 billion yr 1990–2000: 20 − 10

2000–2010:

2268 − 245 ≈ $61.3 billion yr 33 − 0

f (8) − f (1) Δy 48 − 3 45 = = = Δx 8 −1 7 7 f ′( x ) = 4 x1 3

Instantaneous rates of change: f '(1) = 4, f ′(8) = 8 8. f ( x ) = x 3 2 ; [1, 4]

Average rate of change:

f ( 4) − f (1) Δy 8 −1 7 = = = Δx 4 −1 3 3 f ′( x) =

(h) Exports: 1980–2013:

97

Instantaneous rates of change: f ′( −3) = 0, f ′(1) = − 8

(c) Imports: 1990 –2000:

Rates of Change: Velocity and Marginals

1580 − 226 ≈ $41.0 billion yr 33 − 0

3. f (t ) = 3t + 5; [1, 2]

3 12 x 2

Instantaneous rates of change: f ′(1) = 9. f ( x) =

1 ; [1, 5] x

Average rate of change:

Average rate of change:

f ( 2) − f (1) Δy 11 − 8 = = = 3 Δt 2 −1 1

f (5) − f (1) Δy = = 5 −1 Δx

f ′(t ) = 3 Instantaneous rates of change: f ′(1) = 3, f ′( 2) = 3 4. h( x) = 7 − 2 x; [1, 3]

Average rate of change:

h(3) − h(1) Δh 1−5 = = = −2 Δt 3−1 2

f ′( x) = −

10. f ( x) =

=

− 54 4

= −

1 5

= −

1 6

f ′( x) =

1 25

1 ; [1, 9] x

f (9) − f (1) Δy = = 9 −1 Δx

h′( x) = 2 x − 4

3

f ′(1) = −1, f ′(5) = −

Instantaneous rates of change: h(1) = −2, h(3) = −2

h ( 2 ) − h ( −2 ) Δh −2 − 14 = = = −4 Δx 2 − ( −2 ) 4

−1

Instantaneous rates of change:

Average rate of change:

Average rate of change:

1 5

1 x2

h′(t ) = − 2

5. h( x ) = x 2 − 4 x + 2; [− 2, 2]

3 , f ′( 4) = 3 2

1 3

−1 8

=

− 23 4

1 2 x3 2

Instantaneous rates of change:

1 1 f ′(1) = − , f ′(9) = 2 54

Instantaneous rates of change: h′( −2) = −8, h′( 2) = 0 © 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 98

Chapter 2

Differentiation

11. f (t ) = t 4 − 2t 2 ; [− 2, −1]

(c) Average: [2, 3]:

s(3) − s( 2)

Average rate of change:

3−2

f ( −1) − f ( − 2) Δy −1 − 8 = = = −9 Δx −1 − ( − 2) 1 f ′(t ) = 4t 3 − 4t

Instantaneous rates of change:

v(3) = s′(3) = − 20.4 m sec (d) Average: [3, 4]:

s( 4) − s(3) 4−3

g ′( −1) = 3, g ′(1) = 3

 1   5   − 1 16. (a) H ′(v) = 3310 v −1 2  − 1 = 33   v   2  Rate of change of heat loss with respect to wind speed.  5  (b) H ′( 2) = 33 − 1 2   ≈ 83.673

0 − 1400 ≈ −467 3 The number of visitors to the park is decreasing at an average rate of 467 people per month from September to December.

13. (a) ≈

kcal sec ⋅ m3 hr kcal 1 = 83.673 3 ⋅ m 3600 = 0.023 kcal m3

Both the instantaneous rate of change at t = 8 and the average rate of change on [4, 11] are about zero.

 5  − 1 H ′(5) = 33  5 

ΔM 800 − 200 600 = = = 300 mg hr Δt 3−1 2

≈ 40.790 = 40.790

Both the instantaneous rate of change at t = 4 and the average rate of change on [2, 5] is about zero.

= 0.11 kcal m3

15. s = − 4.9t 2 + 9t + 76

(a) Average: [0, 1]:

s(1) − s(0) 1−0

80.1 − 76 = = 4.1 m sec 1

17. s = − 4.9t 2 + 170

(a) Average velocity =

(b) Average: [1, 2]:

s( 2) − s(1) 2 −1

74.4 − 80.1 = = − 5.7 m sec 1

v(1) = s′(1) = − 0.8 m sec v( 2) = s′( 2) = −10.6 m sec

s (3) − s ( 2)

3− 2 125.9 − 150.4 = 1 = − 24.5 m sec

v(0) = s′(0) = 9 m sec v(1) = s′(1) = − 0.8 m sec

kcal m 2 hr m sec

kcal sec ⋅ m3 hr kcal 1 = 40.790 3 ⋅ m 3600

(b) Answers will vary. Sample answer: [2, 5]

Instantaneous: v(t ) = s′(t ) = − 9.8t + 9

kcal m 2 hr m sec

= 83.673

(b) Answers will vary. Sample answer: [4, 11]

14. (a)

33.6 − 58.9 = − 25.3 m sec 1

v( 4) = s′( 4) = − 30.2 m sec

Average rate of change:

Instantaneous rates of change:

=

v(3) = s′(3) = − 20.4 m sec

12. g ( x ) = x 3 − 1; [−1, 1]

g ′( x ) = 3 x 2

58.9 − 74.4 = −15.5 m sec 1

v( 2) = s′( 2) = −10.6 m sec

f ′( − 2) = − 24, f ′( −1) = 0

g (1) − g ( −1) 0 − ( − 2) Δy = = =1 Δx 1 − ( −1) 2

=

(b) v = s′(t ) = − 9.8t , v( 2) = −19.6 m sec,

v(3) = − 29.4 m sec (c)

s = − 4.9t 2 + 170 = 0 4.9t 2 = 170 170 4.9 t ≈ 5.89 seconds

t2 =

(d) v(5.89) ≈ − 57.7 m sec © 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.3

Rates of Change: Velocity and Marginals

99

18. (a) s(t ) = − 4.9t 2 − 5.5t + 64

v(t ) = s′(t ) = −9.8t − 5.5 (b) [1, 2]:

s( 2) − s(1)

=

2 −1

33.4 − 53.6 = − 20.2 m sec 1

(c) v(1) = −15.3 m sec

v( 2) = − 25.1 m sec (d) Set s(t ) = 0. − 4.9t 2 − 5.5t + 64 = 0 − 49t 2 − 55t + 640 = 0 t =

− ( − 55) ±

(− 55) − 4(− 49)(640) 2( − 49) 2

=

55 ±

128,465 ≈ 3.10 sec − 98

(e) v(3.10) = − 35.88 m sec 26. R = 50( 20 x − x3 2 )

19. C = 205,000 + 9800 x

dC = 9800 dx

dR 3   = 50 20 − x1 2  = 1000 − 75 x 2  dx 

20. C = 150,000 + 7 x3

27. P = − 2 x 2 + 72 x − 145

dC = 21x 2 dx

dP = −4 x + 72 dx

21. C = 55,000 + 470 x − 0.25 x 2 , 0 ≤ x ≤ 940

dC = 470 − 0.5 x dx

(

22. C = 100 9 + 3

x

dP = −0.5 x + 2000 dx

)

29. P = 0.0013x3 + 12 x

dC  150 1  = 100 0 + 3 x −1 2  = dx x 2  

dP = −1.5 x 2 + 60 x − 164.25 dx

dR = 50 − x dx

31. C = 3.6

24. R = 30 x − x 2

x

C ′(9) = $0.60 per unit.

25. R = − 6 x3 + 8 x 2 + 200 x

dR = −18 x 2 + 16 x + 200 dx

(

x + 500

(a) C ′( x) = 1.8

dR = 30 − 2 x dx

32. R = 2 x 900 + 32 x − x 2

dP = 0.0039 x 2 + 12 dx 30. P = − 0.5 x3 + 30 x 2 − 164.25 x − 1000

23. R = 50 x − 0.5 x 2

(a)

28. P = − 0.25 x 2 + 2000 x − 1,250,000

(b) C (10) − C (9) ≈ $0.584 (c) The answers are close.

)

R = 1800 x + 64 x 2 − 2 x 3 R′( x ) = 1800 + 128 x − 6 x 2

R′(14) = $2416

(b) R(15) − R(14) = 2(15) 900 + 32(15) − 152  − 2(14) 900 + 32(14) − 14 2  = 34,650 − 32,256 = $2394 (c) The answers are close.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 100

Chapter 2

Differentiation

33. P = − 0.04 x 2 + 25 x − 1500

(a)

(b)

dP = − 0.08 x + 25 = P′( x) dx P′(150) = $13

P(151) − P(150) ΔP 1362.96 − 1350 = = = $12.96 Δx 151 − 150 1

(c) The results are close. 34. P = 36,000 + 2048

x −

1 , 150 ≤ x ≤ 275 8x2

dP 1  1 = 2048 x −1 2  − ( −2 x −3 ) dx 2   8 1024 1 = + 4 x3 x dP ≈ $83.61. (a) When x = 150, dx (d) When x = 225,

(b) When x = 175,

dP ≈ $77.41. dx

(c) When x = 200,

(e) When x = 250,

dP ≈ $64.76. dx

(f ) When x = 275,

dP ≈ $68.27. dx

35. P = 1.73t 2 + 190.6t + 16,994

36. (a)

dP ≈ $72.41. dx dP ≈ $61.75. dx

103

(a) P(0) = 16,994 thousand people

P(3) = 17,581.37 thousand people P(6) = 18,199.88 thousand people P(9) = 18,849.53 thousand people P(12) = 19,530.32 thousand people P(15) = 20,242.25 thousand people P(18) = 20,985.32 thousand people

0

(b) For t < 4, the slopes are positive, and the fever is increasing. For t > 4, the slopes are negative, and the fever is decreasing. (c) T (0) = 100.4°F

T ′( 4) = 101°F

P( 21) = 21,759.53 thousand people

T (8) = 100.4°F

The population is increasing from 1990 to 2011. (b)

dP = P′(t ) = 3.46t + 190.6 dt dP represents the population growth rate. dt

(c) P′(0) = 190.6 thousand people per year

P′(3) = 200.98 thousand people per year P′(6) = 211.36 thousand people per year P′(9) = 221.74 thousand people per year P′(12) = 232.12 thousand people per year P′(15) = 242.5 thousand people per year P′(18) = 252.88 thousand people per year P′( 21) = 263.26 thousand people per year

15 98

T (12) = 98.6°F (d)

dT = −0.075t + 0.3; the rate of change of dt temperature with respect to time

(e) T ′(0) = 0.3°F per hour

T ′( 4) = 0°F per hour T ′(8) = −0.3°F per hour T (12) = −0.6°F per hour For 0 ≤ t < 4, the rate of change of the temperature is positive; therefore, the temperature is increasing. For 4 < t ≤ 12, the rate of change of the temperature is decreasing; therefore, the temperature is decreasing back to a normal temperature of 98.6°F.

The rate of growth is increasing.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.3 37. (a) TR = −10Q 2 + 160Q

Rates of Change: Velocity and Marginals

(400, 1.75), (500, 1.50)

39. (a)

(b) (TR )′ = MR = −20Q + 160 (c)

Slope =

Q

0

2

4

6

8

10

Model

160

120

80

40

0

−40

Table

130

90

50

10

−30

1.50 − 1.75 = −0.0025 500 − 400

p − 1.75 = −0.0025( x − 400) p = −0.0025 x + 2.75 P = R − C = xp − c

= x( −0.0025 x + 2.75) − (0.1x + 25)

38. (a) R = xp = x(5 − 0.001x ) = 5 x − 0.001x 2

(b) P = R − C = (5 x − 0.001x 2 ) − (35 + 1.5 x)

101

= −0.0025 x 2 + 2.65 x − 25 (b)

800

= −0.001x 2 + 3.5 x − 35 (c)

dR = 5 − 0.002 x dx dP = 3.5 − 0.002 x dx

0

1200 0

At x = 300, P has a positive slope.

x

600

1200

1800

2400

3000

At x = 530, P has a 0 slope.

dR dx

3.8

2.6

1.4

0.2

−1.0

At x = 700, P has a negative slope.

dP dx

2.3

1.1

−0.1

−1.3

−2.5

(c) P′( x) = −0.005 x + 2.65

P

1705

2725

3025

2605

1465

P′(300) = $1.15 per unit P′(530) = $0 per unit P′(700) = −$.85 per unit

40. (36,000, 30), (32,000, 35) Slope =

35 − 30 5 1 = − = 32,000 − 36,000 4000 800

1 ( x − 36,000) 800 1 p = − x + 75 (demand function) 800

p − 30 = −

1 2  1  (a) P = R − C = xp − C = x − x + 75  − (5 x + 700,000) = − x + 70 x − 700,000 800  800 

(b)

400,000

At x = 18,000, P has a positive slope. At x = 28,000, P has a 0 slope. 0

50,000

At x = 36,000, P has a negative slope.

0

(c) P′( x) = −

1 x + 70 400

P′(18,000) = $25 per ticket P′( 28,000) = $0 per ticket P′(36,000) = − $20 per ticket

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Chapter 2

102

Differentiation

 25,000 km  1 L  0.70 dollar  41. (a) C ( x) =     yr 1L    x km 

C ( x) =

17,500 dollars x yr

dC 17,500 = − (b) dx x2

dollars yr km L

The marginal cost is the change of savings for a 1-kilometer per liter increase in fuel efficiency. (c)

x

10

15

20

25

30

35

40

C

1750

1166.67

875

700

583.33

500

437.5

dC dx

−175

− 77.78

− 43.75

− 28

−19.44

−14.29

−10.94

(d) The driver who gets 15 kilometers per liter would benefit more than the driver who gets 35 kilometers per liter. The value of dC dx is a greater savings for x = 15 than for x = 35. 42. (a) f ′(0.789) is the rate of change of the number of liters of gasoline sold when the price is $0.789 liter.

(b) In general, it should be negative. Demand tends to decrease as price increases. Answers will vary. 43. (a) Average rate of change from 2000 to 2013:

Δp 16,576.66 − 10,786.85 = ≈ $445.37 yr Δt 13 − 0

(b) Average rate of change from 2003 to 2007:

Δp 13,264.82 − 10,453.92 = ≈ $702.73 yr Δt 7−3

So, the instantaneous rate of change for 2005 is p′(5) ≈ $702.73 yr. (c) Average rate of change from 2004 to 2006:

Δp 12,463.15 − 10,783.01 = ≈ $840.07 yr Δt 6−4

So, the instantaneous rate of change for 2005 is p′(5) ≈ $840.07 yr. (d) The average rate of change from 2004 to 2006 is a better estimate because the data is closer to the years in question. 44. Answers will vary. Sample answer: The rate of growth in the lag phase is relatively slow when compared with the rapid growth in the acceleration phase. The population grows slower in the deceleration phase, and there is no growth at equilibrium. These changes could be explained by food supply or seasonal growth.

Section 2.4 The Product and Quotient Rules Skills Warm Up 1.

( x2

+ 1)( 2) + ( 2 x + 7)( 2 x) = 2 x 2 + 2 + 4 x 2 + 14 x = 6 x 2 + 14 x + 2 = 2(3 x 2 + 7 x + 1)

(

)

( )(

)

2. 2 x − x3 (8 x) + 4 x 2 2 − 3x 2 = 16 x 2 − 8 x 4 + 8 x 2 − 12 x 4

= 24 x 2 − 20 x 4 = 4 x 2 (6 − 5 x 2 ) 3. x( 4)( x 2 + 2) ( 2 x ) + ( x 2 + 4)(1) = 8 x 2 ( x 2 + 2) ( x 2 + 4) 3

3

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DOWNLOAD SOLUTIONS MANUAL Section 2.4

The Product and Quotient Rules

103

Skills Warm Up —continued— 4. x 2 ( 2)( 2 x + 1)( 2) + ( 2 x + 1) ( 2 x) = 4 x 2 ( 2 x + 1) + 2 x( 2 x + 1) 4

4

3 = 2 x( 2 x + 1) 2 x + ( 2 x + 1)   

5.

(2 x

+ 7)(5) − (5 x + 6)( 2)

(2 x

+ 7)

2

= =

6.

( x2

10 x + 35 − 10 x − 12

(2 x

− 4)

(2 x

2

+ 7)

( x2

+ 1)( 2) − ( 2 x + 1)( 2 x)

( x2

+ 1)

2

= =

=

8.

(1 − x4 )(4) − (4 x − 1)(−4 x3 ) 2 (1 − x4 )

( x2

2

− x2 − 8x − 4

( x2 ( x2

− 4)

2

+ 1)

2

−2 x 2 − 2 x + 2

( x2 + 1) −2( x 2 + x − 1) 2 ( x2 + 1) 2

=

=

( x−1 + x)(2) + (2 x − 3)(− x−2

− 4)

2x2 + 2 − 4 x2 − 2x

=

9.

2

2 x3 + x 2 − 8 x − 4 − 2 x3 − 2 x 2

= =

7.

2

23

− 4)( 2 x + 1) − ( x 2 + x)( 2 x)

( x2

+ 7)

4 − 4 x 4 + 16 x 4 − 4 x3

(1 − x4 )

2

12 x 4 − 4 x3 + 4

(1 − x4 )

2

(1 − x4 )

2

4(3 x 4 − x3 + 1)

+ 1) = 2 x −1 + 2 x + ( −2 x −1 + 2 x + 3 x −2 − 3) = 4 x + 3x −2 − 3 = 4x + =

10.

(1 − x−1)(1) − ( x − 4)( x−2 ) 2 (1 − x−1 )

3 −3 x2

4 x3 − 3x 2 + 3 x2

 1 − x −1 − x −1 + 4 x −2  x 2  =   2  −1 −2  1 − 2x + x  x 

= =

x2 − 2 x + 4 x2 − 2 x + 1 x2 − 2 x + 4

(x

− 1)

2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 104

Chapter 2

Differentiation

Skills Warm Up —continued— 11.

f ( x) = 3 x 2 − x + 4 f ′( x) = 6 x − 1

13.

f ′( 2) = 6( 2) − 1 = 12 − 1 = 11 12.

f ( x) = − x3 + x 2 + 8 x

= −

f ′( x) = −3 x 2 + 2 x + 8 f ′( 2) = −3( 22 ) + 2( 2) + 8 = −3( 4) + 4 + 8 = 0

2 2 = x −1 7x 7 2 2 f ′( x) = − x − 2 = − 2 7 7x 2 f ′( 2) = − 2 7 ( 2) f ( x) =

14.

1 14

1 x2 2 f ′( x) = 2 x + 3 x f ( x) = x 2 −

f ′( 2) = 2( 2) +

2 23

2 8 1 = 4+ 4 17 = 4 = 4+

1. f ( x) = ( 2 x − 3)(1 − 5 x) f ′( x ) = ( 2 x − 3)( − 5) + (1 − 5 x )( 2) = −10 x + 15 + 2 − 10 x = − 20 x + 17

2. g ( x) = ( 4 x − 7)(3x + 1)

g ′( x) = ( 4 x − 7)(3) + (3 x + 1)( 4) = 12 x − 21 + 12 x + 4 = 24 x − 17 3. f ( x) = (6 x − x

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

= 18 x − 3 x 2 + 24 − 8 x + 18 x − 6 x 2 = − 9 x 2 + 28 x + 24

4. f ( x) = (5 x − x3 )( 2 x + 9)

f ′( x) = (5 x − x 3 )( 2) + ( 2 x + 9)(5 − 3 x 2 ) = 10 x − 2 x 3 + 10 − 6 x 3 + 45 − 27 x 2 = − 8 x 3 − 27 x 2 + 20 x + 45

5. f ( x) = x( x 2 + 3)

f ′( x) = x( 2 x) + ( x 2 + 3)(1) = 2x2 + x2 + 3 = 3x 2 + 3

6. f ( x) = x 2 (3x3 − 1)

f ′( x ) = x 2 (9 x 2 ) + (3 x 3 − 1)( 2 x) = 9x4 + 6 x4 − 2 x = 15 x 4 − 2 x

2  7. h( x) =  − 3( x 2 + 7) = ( 2 x −1 − 3)( x 2 + 7) x  −1 h′( x) = ( 2 x − 3)( 2 x) + ( x 2 + 7)( − 2 x − 2 )

= 4 − 6 x − 2 − 14 x − 2 = − 6x + 2 −

14 x2

4  8. f ( x) = (3 − x) 2 − 5  = (3 − x)( 4 x − 2 − 5) x  f ′( x ) = (3 − x)( − 8 x − 3 ) + ( 4 x − 2 − 5)( −1) = − 24 x − 3 + 8 x − 2 − 4 x − 2 + 5 = −

24 4 + 2 + 5 x3 x

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.4 9. g ( x) = ( x 2 − 4 x + 3)( x − 2)

g ′( x ) = ( x 2 − 4 x + 3)(1) + ( x − 2)( 2 x − 4) 2

2

= x − 4x + 3 + 2x − 4x − 4x + 8

The Product and Quotient Rules

15. f (t ) =

f ′(t ) =

t +6 t2 − 8

(t 2

− 8)(1) − (t + 6)( 2t )

(t 2

2

= 3 x − 12 x + 11

=

10. g ( x) = ( x − 2 x + 1)( x − 1) 2

3

g ′( x) = ( x 2 − 2 x + 1)(3 x 2 ) + ( x3 − 1)( 2 x − 2) 4

3

2

4

3

= 3x − 6 x + 3x + 2 x − 2 x − 2 x + 2

=

= 5 x 4 − 8 x3 + 3x 2 − 2 x + 2

x 11. h( x ) = x −5 ( x − 5)(1) − x(1) = x − 5 − x = − 5 h′( x) = ( x − 5)2 ( x − 5)2 ( x − 5)2 12. h( x) = h′( x) = = =

13. f (t ) =

f ′(t ) = = =

14. f ( x) =

f ′( x) = =

16. g ( x) =

g ′( x) =

2

=

+ 3)( 2 x) − x 2 (1)

=

x x +3

(x

( x + 3)

(x

+ 3)

17. f ( x) =

2

x2 + 6 x

(x

+ 3)

f ′( x) =

2

2t 2 − 3 3t + 1

(3t

=

+ 1)( 4t ) − ( 2t 2 − 3)(3)

(3t

+ 1)

2

12t 2 + 4t − 6t 2 + 9

(3t

+ 1)

2

=

18. f ( x) =

2

6t + 4t + 9

(3t

+ 1)

f ′( x) =

2

7x + 3 4x − 9

(4 x

=

− 9)(7) − (7 x + 3)( 4)

(4 x

− 9)

2

=

− 8)

2

t 2 − 8t − 2t 2 − 12t

(t 2

− 8)

2

− t 2 − 12t − 8

(t 2

− 8)

2

4x − 5 x2 − 1

( x2

− 1)( 4) − ( 4 x − 5)( 2 x)

( x2

− 1)

2

4 x 2 − 4 − 8 x 2 + 10 x

( x2

− 1)

2

− 4 x 2 + 10 x − 4

( x2

2

2x2 + 6x − x2

105

− 1)

2

x2 + 6x + 5 2x − 1

− 1)( 2 x + 6) − ( x 2 + 6 x + 5)( 2)

(2 x

(2 x

− 1)

2

4 x 2 + 12 x − 2 x − 6 − 2 x 2 − 12 x − 10

(2 x

− 1)

2

2 x 2 − 2 x − 16

(2 x

− 1)

2

4 x2 − x + 2 3 − 4x

(3 −

4 x)(8 x − 1) − ( 4 x 2 − x + 2)( − 4)

(3 − 4 x )

2

24 x − 3 − 32 x 2 + 4 x + 16 x 2 − 4 x + 8

(3 − 4 x ) 2 −16 x 2 + 24 x + 5

(3 − 4 x )

2

28 x − 63 − 28 x − 12

= −

(x

− 1)

2

75

( 4 x − 9)2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 106

Chapter 2

19. f ( x) =

Differentiation

6 + 2 x −1 3x − 1

(3 x

f ′( x) =

20. f ( x) =

− 1)( − 2 x − 2 ) − (6 + 2 x −1 )(3)

(3 x

− 1)

f ′( x) =

2

− 6 x −1 + 2 x − 2 − 18 − 6 x −1

=

(3 x

− 1)

=

2

2 x − 2 − 12 x −1 − 18

=

(3 x

− 1)

=

2

=

Rewrite

x3 + 6 x 21. f ( x ) = 3

22. f ( x) =

x3 + 2 x 2 10

+ 2)( 2 x − 3 ) − (5 − x − 2 )(1)

2x

−2

x + 2 + 4x

(x

1 3 1 2 x + x 10 5

−3

− 5 + x− 2

+ 2)

2

4 x −3 + 3x − 2 − 5

(x

+ 2)

2

4 + 2 x − 5 x3 x 3 ( x + 2)

Differentiate

1 f ( x) = x 3 + 2 x 3 f ( x) =

(x

4 3 + 2 −5 3 x x = 2 ( x + 2)

2 12 − − 18 2 2 − 12 x − 18 x 2 x x = = 2 x 2 (3 x − 1) (3x − 1)2

Function

5 − x−2 x + 2

f ′( x) = x 2 + 2

f ′( x) =

3 2 2 x + x 10 5

2

Simplify f ′( x) = x 2 + 2

f ′( x) =

3 2 2 x + x 10 5

23. y =

7 x2 5

y =

7 2 x 5

y' =

7 ⋅ 2x 5

y′ =

14 x 5

24. y =

2x4 9

y =

2 4 x 9

y′ =

2 ⋅ 4 x3 9

y′ =

8 3 x 9

25. y =

7 3x3

y =

7 −3 x 3

y ' = −7 x −4

y′ = −

7 x4

26. y =

4 5x2

y =

4 −2 x 5

8 y′ = − x −3 5

y′ = −

8 5 x3

27. y =

4 x 2 − 3x 8 x

y =

1 32 3 12 x − x , x ≠ 0 2 8

y′ =

3 12 3 x − x −1 2 4 16

y′ =

3 4

y =

5 53 5 53 x + x ,x ≠ 0 2 3

y′ =

25 2 3 10 −1 3 x + x ,x ≠ 0 6 9

y′ =

25 3 2 10 x + 3 6 9 x

28. y =

5(3 x 2 + 2 x) 63 x

x −

3 16 x

29. y =

x2 − 4 x + 3 2( x − 1)

y =

1 ( x − 3), x ≠ 1 2

y′ =

1 (1), x ≠ 1 2

y′ =

1 ,x ≠ 1 2

30. y =

x2 − 4 4( x + 2)

y =

1 ( x − 2), x ≠ − 2 4

y′ =

1 (1), x ≠ − 2 4

y′ =

1 , x ≠ −2 4

31. f ′( x ) = ( x 3 − 3 x)( 4 x + 3) + (3 x 2 − 3)( 2 x 2 + 3 x + 5) = 4 x 4 + 3 x 3 − 12 x 2 − 9 x + 6 x 4 + 9 x3 + 9 x 2 − 9 x − 15 = 10 x 4 + 12 x 3 − 3 x 2 − 18 x − 15

Product Rule and Simple Power Rule

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.4 32. h′(t ) = (t 5 − 1)(8t − 7) + (5t 4 )( 4t 2 − 7t − 3) = 8t 6 − 7t 5 − 8t + 7 + 20t 6 − 35t 5 − 15t 4 = 28t 6 − 42t 5 − 15t 4 − 8t + 7

Product Rule and Simple Power Rule 33. h(t ) = h′(t ) =

(6t − 4) 1 6 = 2 3( ) 1 3

f ( x) = f ′( x) =

35. f ′( x) =

= =

Constant Multiple and Simple Power Rules 34.

The Product and Quotient Rules

(3x − 8) 1 3 = 3 2( ) 2 1 2

( x 2 − 1)(3x 2 + 3) − ( x3 + 3x + 2)(2 x) 2 ( x 2 − 1) 3x 4 − 3 − 2 x 4 − 6 x 2 − 4 x

( x 2 − 1)

2

x4 − 6 x2 − 4 x − 3

( x 2 − 1)

2

Quotient Rule and Simple Power Rule 36. f ( x) =

Constant Multiple and Simple Power Rules

107

f ′( x) = = =

2 x3 − 4 x 2 − 9 x3 − 5

( x3 − 5)(3x 2 − 8x) − (2 x3 − 4 x 2 − 9)(3x2 ) 2 ( x3 − 5) 3 x5 − 8 x 4 − 15 x 2 + 40 x − 6 x5 + 12 x 4 − 27 x 2

( x3 − 5)

2

− 3x5 + 4 x 4 − 42 x 2 + 40 x

( x3 − 5)

2

Quotient Rule and Simple Power Rule 37.

f ( x) =

( x − 5)( x + 4) = x − 5, x ≠ − 4 x 2 − x − 20 = x + 4 ( x + 4)

f ′( x) = 1 Simple Power Rule

(3t + 1)(1 + 7) = 3t + 1, t ≠ − 7 3t 2 + 22t + 7 = t +7 t+7 h′(t ) = 3, t ≠ −7

38. h(t ) =

Simple Power Rule 39. g (t ) = ( 2t 3 − 1) = ( 2t 3 − 1)( 2t 3 − 1) 2

g ′(t ) = ( 2t 3 − 1)(6t 2 ) + ( 2t 3 − 1)(6t 2 ) = 12t 2 ( 2t 3 − 1)

Product Rule and Simple Power Rule 40. f ( x ) = ( 4 x 3 − 2 x − 3) = ( 4 x3 − 2 x − 3)( 4 x 3 − 2 x − 3) 2

f ′( x) = ( 4 x 3 − 2 x − 3)(12 x 2 − 2) + ( 4 x3 − 2 x − 3)(12 x 2 − 2) = 48 x5 − 24 x 3 − 36 x 2 − 8 x 3 + 4 x + 6 + 48 x5 − 24 x3 − 36 x 2 − 8 x3 + 4 x + 6 = 96 x 5 − 48 x 3 − 72 x 2 − 16 x3 + 8 x + 12

Product Rule and Simple Power Rule

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 108

Chapter 2

41. g ( s ) =

g ′( s ) = = =

Differentiation 45. f ( x ) = ( x + 4)( 2 x + 9)( x − 3)

s 2 − 2s + 5 s 2 − 2s + 5 = s1 2 s 12

s

(2s

(

− 2) − ( s − 2s + 5) 2

1 s −1 2 2

)

f ′( x) = ( 2 x 2 + 17 x + 36)(1) + ( x − 3)( 4 x + 17)

s

= ( 2 x 2 + 17 x + 36) + ( 4 x 2 + 5 x − 51)

2s 3 2 − 2 s1 2 − 12 s 3 2 + s1 2 − 52 s −1 2 s

= 6 x 2 + 22 x − 15 2

3 12 5 3s − 2s − 5 s − s −1 2 − s −3 2 = 2 2 2s3 2

Quotient Rule and Simple Power Rule 42. f ( x) =

= ( 2 x 2 + 17 x + 36)( x − 3)

3

2

3

2

x − 5x − 6x x − 5x − 6x = x1 2 x

Product Rule and Simple Power Rule 46.

f ( x) = (3 x 3 + 4 x)( x − 5)( x + 1) = (3 x 3 + 4 x)( x 2 − 4 x − 5)

f ′( x) = (3 x 3 + 4 x)( 2 x − 4) + ( x 2 − 4 x − 5)(9 x 2 + 4)

= x5 2 − 5 x3 2 − 6 x1 2

= (6 x 4 − 12 x 3 + 8 x 2 − 16 x)

5 3 2 15 1 2 x − x − 3 x −1 2 2 2 5 15 1 2 3 x − 12 = x3 2 − 2 2 x

= 15 x 4 − 48 x 3 − 33 x 2 − 32 x − 20

+ (9 x 4 − 36 x3 − 41x 2 − 16 x − 20)

f ′( x) =

=

Product Rule and Simple Power Rule 47. f ( x) = (5 x + 2)( x 2 + x)

5 x 2 − 15 x − 6 2 x1 2

Constant Multiple and Simple Power Rules 43. f ( x) = f ′( x ) = = = = =

(x

− 2)(3x + 1)

3x 2 − 5 x − 2 = 4x + 2

4x + 2

(4 x

+ 2)(6 x − 5) − (3x 2 − 5 x − 2)( 4)

(4 x

+ 2)

2

f ′( x) = = =

= 10 x 2 + 9 x + 2 + 5 x 2 + 5 x

m = f ′( −1) = 3

+ 2)

y = 3x + 3

2

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

2(6 x 2 + 6 x − 1)

= 3x 4 − 6 x 2 + 3 + 2 x 4 − 6 x 2

2

= 5 x 4 − 12 x 2 + 3

6 x2 + 6 x − 1 2( 2 x + 1)

(x

−10

48. f ( x) = ( x 2 − 1)( x3 − 3 x)

12 x + 12 x − 2

2( 2 x + 1)

5

2

2

4( 2 x + 1)

(−1, 0)

−5

y − 0 = 3( x − ( −1))

24 x − 8 x − 10 − 12 x 2 + 20 x + 8

(4 x

10

= 15 x 2 + 14 x + 2

2

m = f ′( − 2) = 5( − 2) − 12( − 2) + 3 = 35 4

2

Quotient Rule and Simple Power Rule 44. f ( x) =

f ′( x) = (5 x + 2)( 2 x + 1) + ( x 2 + x )(5)

+ 1)( 2 x − 7) 2x + 1

(2 x + 1)

y + 6 = 35 x + 70

+ 1)

y = 35 x + 64 2 −6

6

2

8 x 2 − 6 x − 5 − 4 x 2 + 10 x + 14

(2 x

y − ( − 6) = 35( x − ( − 2))

2 x2 − 5x − 7 = 2x + 1

(2 x + 1)(4 x − 5) − (2 x 2 − 5 x − 7)(2)

2

2

(−2, −6) −8

2

4x + 4x + 9

(2 x

+ 1)

2

Quotient Rule and Simple Power Rule

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.4 49. f ( x) = x3 ( x 2 − 4)

50. f ( x ) =

f ′( x) = x1 2 (1) + ( x − 3) 12 x −1 2

= 2 x 4 + 3 x 4 − 12 x 2 = 5 x − 12 x

=

m = f ′(1) = − 7

−5

(1, −3)

y − ( − 3) = − 7( x − 1) y = −7x + 4

1 x1 2 2

= x1 2 +

10

2

5

109

x ( x − 3) = x1 2 ( x − 3)

(

f ′( x) = x 2 ( 2 x ) + ( x 2 − 4)(3 x 2 ) 4

The Product and Quotient Rules

3 12 x 2

m = f ′(9) =

3 2

)

3 −1 2 x 2

3 −1 2 x 2

(9 )

12

3 2

(9 )

−1 2

9 2

=

1 2

= 4

y − 18 = 4( x − 9)

−10

y = 4 x − 18 10

(9, 18)

0

20 0

51. f ( x) =

3x − 2 x +1

f ′( x) =

( x + 1)(3) − (3x − 2)(1) 2 ( x + 1)

f ′( 4) =

1 5

1 ( x − 4) 5 1 4 y −2 = x− 5 5 1 6 y = x+ 5 5

(x

− 1)

( x + 1)

2

=

5

( x + 1)

2

(4, 2) −2

10

−4

− 1)2 − ( 2 x + 1)

(x

3x + 3 − 3 x + 2

4

y −2 =

52. f ′( x) =

=

2

f ′( 2) = −3

=

−3

(x

− 1)

2

(3x − 2)(6 x + 5)

53. f ( x) =

2x − 3

f ′( x) =

y − 5 = −3( x − 2) y = −3x + 11

=

(2 x

= 8

−7 −2

18 x 2 + 3x − 10 2x − 3

− 3)(36 x + 3) − (18 x 2 + 3 x − 10)( 2)

(2 x

− 3)

2

72 x 2 − 102 x − 9 − 36 x 2 − 6 x − 20

(2 x

8

(2, 5)

=

− 3)

2

36 x 2 − 108 x + 11

(2 x

m = f ′( −1) =

− 3)

2

36( −1) − 108( −1) + 11 2

(2(−1) − 3)

31 ( x − (−1)) 5 31 26 y = x+ 5 5

2

y − ( −1) =

=

31 5 10

−5

5

(−1, −1)

−10

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 110

Chapter 2

54. f ( x ) =

f ′( x) = = =

(x

Differentiation

+ 2)( x 2 + x)

x3 + 3 x 2 + 2 x x − 4

=

x − 4

− 4)(3 x 2 + 6 x + 2) − ( x3 + 3x 2 + 2 x)(1)

(x

(x

− 4)

2

3x3 − 12 x 2 + 6 x 2 − 24 x + 2 x − 8 − x3 − 3 x 2 − 2 x

(x

− 4)

2

2 x3 − 9 x 2 − 24 x − 8

(x

m = f ′(1) =

− 4)

2

2(1) − 9(1) − 24(1) − 8 3

2

(1 − 4)

= −

2

13 3

13 ( x − 1) 3 13 7 y = x + 3 3

y − ( − 2) = −

55. f ′( x) =

(x

−8

− 1)

4

(1, −2) −4

− 1)( 2 x) − x 2 (1)

(x

4

2

=

x2 − 2 x

(x

− 1)

59.

2

f ( x) = x( x + 1) = x 2 + x f ′( x) = 2 x + 1

f ′( x) = 0 when x 2 − 2 x = x( x − 2) = 0, which

6

implies that x = 0 or x = 2. Thus, the horizontal tangent lines occur at (0, 0) and ( 2, 4). 56. f ′( x) =

( x 2 + 1)(2 x) − ( x 2 )(2 x) 2 ( x 2 + 1)

=

f −2

2x

( x 2 + 1)

−3

2

60.

f ′( x) = 0 when 2 x = 0, which implies that x = 0.

3

f′

( x3 + 1)(4 x3 ) − x 4 (3x 2 ) = x6 + 4 x3 2 2 ( x3 + 1) ( x3 + 1) f ′( x) = 0 when x 6 + 4 x3 = x 3 ( x3 + 4) = 0, which

57. f ′( x) =

3

−4. Thus, the horizontal

tangent lines occur at (0, 0) and 2

+ 1)( 4 x

3

3

f

−2

2

−1

61.

f ( x) = x( x + 1)( x − 1)

11

= x3 − x

)

−4, − 2.117 .

) − ( x + 3)(2 x) 2 ( x + 1) 2 x( x 2 + 3)( x 2 − 1) = 2 ( x 2 + 1) f ′( x) = 0 when 2 x( x 2 + 3)( x 2 − 1) = 0, which

58. f ′( x) =

(x

(

f ( x) = x 2 ( x + 1) = x3 + x 2 f ′( x) = 3 x 2 + 2 x = x(3 x + 2)

Thus, the horizontal tangent line occurs at (0, 0).

implies that x = 0 or x =

2

f′

f′

f ′( x) = 3x 2 − 1

−2

2

f

4

−6

2

implies that x = 0 or x = ±1. Thus, the horizontal tangent lines occur at (0, 3), (1, 2), and (−1, 2).

62.

f ( x) = x 2 ( x + 1)( x − 1) = x 4 − x 2 f ′( x) = 4 x 3 − 2 x = 2 x( 2 x 2 − 1) 3

f′

−2

f

2

−1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.4

63.

 3p  x = 2751 −  5 p + 1 

66.

=

 3  dx = −275  ≈ −1.87 units 2 dp  ( 21) 

= =

per dollar. 64.

dx ( p + 1)(2) − (2 p)(1) = 0 −1− 2 dp ( p + 1) = −1 −

2

( p + 1)

2

50 (t + 2) − (t + 1750) 2 2500(t + 2) −1748 50(t + 2) 25(t + 2)

(a) When t = 1,

=

+ 1)

2

− p2 − 2 p − 3

(p

+ 1)

When p = 3,

2

dx −9 − 6 − 3 = ≈ −1.13 units dp 16

per dollar.

 50 + t 2 ( 4) − ( 4t )( 2t )   2 ( )  = 500  200 − 4t  65. P′ = 500  2    50 + t 2 2  (50 + t 2 ) )   (  184  When t = 2, P′ = 500   ≈ 31.55 bacteria hour. 2  (54) 

2

dP −874 = ≈ −3.88 percent day. dt 225

(b) When t = 10,

−( p + 1) − 2

(p

2

−874

2

=

111

50(t + 2)(1) − (t + 1750)(50) dP = 2 dt 50(t + 2)

 (5 p + 1)3 − (3 p )(5)    dx 3  = −275  = −275 2 2 dp    (5 p + 1)  (5 p + 1)

When p = 4,

The Product and Quotient Rules

67. P =

P′ =

dP −874 = dt 3600 437 = − 1800 ≈ −0.24 percent day.

t2 − t + 1 t2 + 1

(t 2 + 1)(2t − 1) − (t 2 − t + 1)(2t ) = t 2 − 1 2 2 (t 2 + 1) (t 2 + 1)

(a) P′(0.5) = − 0.480 week (b) P′( 2) = 0.120 week (c) P′(8) = 0.015 week Each rate in parts (a), (b), and (c) is the rate at which the level of oxygen in the pond is changing at that particular time.

 4t 2 + 16t + 75  68. T = 10 2   t + 4t + 10 

Initial temperature: T (0) = 75°F

 t 2 + 4t + 10 (8t + 16) − 4t 2 + 16t + 75 ( 2t + 4)  ( ) ( )  = − 700(t + 2) T ′(t ) = 10 2 2   2 (t + 4t + 10) (t 2 + 4t + 10)   (a) T ′(1) ≈ − 9.33°F hr (b) T ′(3) ≈ − 3.64°F hr (c) T ′(5) ≈ −1.62°F hr (d) T ′(10) ≈ − 0.37°F hr Each rate in parts (a), (b), (c), and (d) is the rate at which the temperature of the food in the refrigerator is changing at that particular time.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 112

Chapter 2

Differentiation 72. (a) P = ax 2 + bx + c

69. C = x3 − 15 x 2 + 87 x − 73, 4 ≤ x ≤ 9

Marginal cost: Average cost: (a)

When x = 10, P = 50: 50 = 100a + 10b + c.

dC = 3x 2 − 30 x + 87 dx

When x = 12, P = 60: 60 = 144a + 12b + c. When x = 14, P = 65: 65 = 196a + 14b + c.

C 73 = x 2 − 15 x + 87 − x x

Solving this system, we have

60

a = − 58 , b =

dC dx

Thus, P = C x

4

(b)

75 , and 4

− 85 x 2

+

c = −75.

75 x 4

− 75.

80

9 0

(b) Point of intersection: 3 x 2 − 30 x + 87 = x 2 − 15 x + 87 − 2 x 2 − 15 x +

73 x

0

73 = 0 x

(c) Marginal profit: P′ = − 54 x +

x ≈ 6.683

C dC = ≈ 20.50. x dx

73.

= 0  x = 15

70. (a) As time passes, the percent of people aware of the product approaches approximately 95%.

f ( x ) = 2 g ( x ) + h( x ) f ′( x) = 2 g ′( x) + h′( x) f ′( 2) = 2( −2) + 4 = 0

Thus, the point of intersection is (6.683, 20.50). At this point average cost is at a minimum. 74.

f ( x) = 3 − g ( x) f ′( x) = − g ′( x) f ′( 2) = −( −2) = 2

(b) As time passes, the rate of change of the percent of people aware of the product approaches zero.

75. f ( x) = g ( x)h( x)

x   200 71. C = 100 2 + , x ≥ 1 x + 30   x

 ( x + 30) − C ′ = 100 −2( 200 x −3 ) + 2  ( x + 30)

75 4

This is the maximum point on the graph of P, so selling 15 units will maximize the profit.

2 x 3 − 15 x 2 + 73 = 0

When x = 6.683,

20

0

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

x  

 400  30  = 100 − 3 + 2  x ( x + 30)  30   400 (a) C ′(10) = 100 − 3 + 2  = −38.125 40   10 (b) C ′(15) ≈ −10.37 (c) C ′( 20) ≈ −3.8 Increasing the order size reduces the cost per item. An order size of 2000 should be chosen since the cost per item is the smallest at x = 20.

= (3)( 4) + ( −1)( − 2) = 14

76.

f ( x) = f ′( x) = f ′( 2) =

g ( x) h( x )

h( x) g ′( x) − g ( x )h′( x)

h( x)

2

( −1)(−2) − (3)( 4) 2 ( −1)

= −10

77. Answers will vary.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Chapter 2

Quiz Yourself

113

Chapter 2 Quiz Yourself 1. f ( x) = 5 x + 3

f ′( x) = lim

2. f ( x) =

f ( x + Δ x) − f ( x)

x + Δx + 3 − x + 3 Δx x + Δx + 3 − ( x + 3)

= lim

Δx → 0

= lim

5 x + 5Δ x + 3 − 5 x − 3 Δx

= lim

= lim

5Δ x Δx

= lim

Δx → 0

Δx

Δx → 0

5( x + Δ x) + 3 − (5 x + 3) = lim  Δx → 0 Δx Δx → 0

f ( x + Δx ) − f ( x )

f ′( x) = lim

Δx

Δx → 0

x +3

Δx → 0

Δx

Δx → 0

= lim 5 = 5

=

Δx → 0

At ( − 2, − 7): m = 5

2

(

x + Δx + 3 +

1 x + Δx + 3 +

x +3

)

x +3

1 x +3

At (1, 2): m =

1 1 = 4 2 1+3

3. f ( x) = 3x − x 2 f ′( x) = lim

Δx → 0

f ( x + Δ x) − f ( x) Δx

3( x + Δ x) − ( x + Δ x)2  − (3 x − x 2 )  = lim  Δx → 0 Δx = lim

(

3x + 3Δ x − x 2 + 2 x( Δ x) + ( Δ x) Δx

Δx → 0

2

) − (3 x − x )

= lim

3x + 3Δ x − x − 2 x( Δ x) − ( Δ x) − 3 x + x 2 Δx

= lim

3 x − 2 x( Δ x ) − ( Δ x ) Δx

= lim

Δ x (3 − 2 x − Δ x ) Δx

2

2

Δx → 0

Δx → 0

Δx → 0

2

2

= lim (3 − 2 x − Δ ) = 3 − 2 x Δx → 0

At ( 4, − 4): m = 3 − 2( 4) = 3 − 8 = − 5 4. f ( x) = 12

8. f ( x) = 4 x − 2

f ′( x) = 0

f ′( x) = −8 x −3 = −

5. f ( x) = 19 x + 9

9. f ( x) = 10 x −1 5 + x −3

f ′( x) = 19 6. f ( x) = x − 3x − 5 x + 8 4

8 x3

3

2

f ′( x) = −2 x − 6 5 − 3 x − 4 = −

2 x

−6 5

3 x4

f ′( x) = 4 x3 − 9 x 2 − 10 x 7. f ( x) = 12 x1 4

f ′( x) = 3x −3 4 =

3 x3 4

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 114

Chapter 2

10. f ( x ) =

f ′( x) = =

Differentiation

2x + 3 3x + 2

(3 x

16. f ( x) =

+ 2)( 2) − ( 2 x + 3)(3)

(3 x

+ 2)

= −

+ 2)

2

5

(3 x

+ 2)

( ) f ′( x) = ( x 2 + 1)( −2) + ( −2 x + 4)( 2 x) = −6 x 2 + 8 x − 2

) + 3 x + 4)(5) + (5 x − 2)( 2 x + 3)

12. f ( x) = x 2 + 3x + 4 (5 x − 2)

Instantaneous rates of change: 1 1 f ′( − 2) = − , f ′( − 5) = − 12 75 17. f ( x) =

3

x ; [8, 27]

Average rate of change: f ( 27) − f (8) Δy 3− 2 1 = = = 27 − 8 19 19 Δx f ( x) =

3

= 15 x 2 + 26 x + 14

f ′( x) =

1 −2 3 1 x = 3 3x 2 3

f ′( x) = = =

=

4x x2 + 3

1  1 3 − −  − 1 6  15  = 30 = − 3 3 30

1 3x 2

= 5 x 2 + 15 x + 20 + 10 x 2 + 11x − 6

13. f ( x) =

f ( − 2) − f (5) Δy = = Δx − 2 − ( − 5) f ′( x) = −

2

11. f ( x) = x 2 + 1 ( − 2 x + 4)

( f ′( x) = ( x 2

Average rate of change:

2

6x + 4 − 6x − 9

(3 x

1 ; [− 5, − 2] 3x

x = x1 3

Instantaneous rates of change: f ′(8) =

( x 2 + 3)(4) − 4 x(2 x) 2 ( x 2 + 3) 4 x 2 + 12 − 8 x 2

( x 2 + 3)

f ′( 27) =

(a)

− 4 x 2 + 12 2

14. f ( x) = x 2 − 3x + 1; [0, 3]

Average rate of change: f (3) − f (0) 1−1 Δy = = = 0 3−0 3 Δx f ′( x) = 2 x − 3 Instantaneous rates of change: f ′(0) = −3, f ′(3) = 3

1 27

18. P = −0.0125 x 2 + 16 x − 600

2

( x 2 + 3) − 4( x 2 − 3) 2 ( x 2 + 3)

dP = −0.025 x + 16 dx

When x = 175,

f (1) − f ( −1) 6− 4 Δy = = =1 1 − ( −1) 2 Δx

f ′( x) = 6 x 2 + 2 x − 1

dP = $11.625. dx

(b) P(176) − P(175) = 1828.8 − 1817.1875 = $11.6125 (c) The results are approximately equal.

19. f ( x) = 5 x 2 + 6 x − 1

f ′( x) = 10 x + 6 At ( −1, − 2), m = −4. y + 2 = −4( x + 1) y = −4 x − 6 2

15. f ( x) = 2 x3 + x3 − x + 4; [−1, 1]

Average rate of change:

1 , 12

−5

4

(− 1, −2)

−4

Instantaneous rates of change: f ′( −1) = 3, f ′(1) = 7

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.5

8

f ( x) =

20.

x

3

m = f ′( 4) = −

( 4)

= −

4

3 ( x − 4) 8 3 3 y −1 = − x + 8 2 3 5 y = − x+ 8 2 y −1 = −

22. f ( x) = f ′( x) =

( ) f ′( x) = ( x 2 + 1)( 4) + ( 4 x − 3)( 2 x)

12 12 = − 2 x5 2 x x

12 2

115

21. f ( x) = x 2 + 1 ( 4 x − 3)

= 8 x −3 2

f ′( x) = −12 x −5 2 = −

The Chain Rule

= 4x2 + 4 + 8x2 − 6 x = 12 x 2 − 6 x + 4

3 8

m = f ′(1) = 12(1) − 6(1) + 4 = 10 2

y − 2 = 10( x − 1)

5

10

y = 10 x − 8

(1, 2) −5

(4, 1) −1

5

8

−10

−1

5x + 4 2 − 3x

(2

− 3 x)(5) − (5 x + 4)( − 3)

(2

m = f ′(1) =

− 3x)

22

(2 − 3(1))

2

=

10 − 15 x + 15 x + 12

(2

− 3 x)

2

=

22

(2

− 3 x)

2

1

= 22

2

0

8

y − ( − 9) = 22( x − 1)

(1, −9)

y + 9 = 22 x − 22 y = 22 x − 31

−15

23. S = − 0.01722t 3 + 0.7333t 2 − 7.657t + 45.47, 7 ≤ t ≤ 13

(a)

dS = S ′(t ) = − 0.051666t 2 + 1.4666t − 7.657 dt

(b) 2008: S ′(8) ≈ $0.77 yr 2011: S ′(11) ≈ $2.22 yr 2012: S ′(12) ≈ $2.50 yr

Section 2.5 The Chain Rule Skills Warm Up 1.

5

2.

4

2

(2 x

3

4x2 + 1

1 6

3

= (1 − 5 x)

25

− 1) = ( 2 x − 1)

1

3.

4.

(1 − 5 x)

2x + 9

5.

x −1 3 = x1 2 (1 − 2 x) 1 − 2x

34

6.

= ( 4 x 2 + 1)

3

−1 2

= ( 2 x3 + 9)

(3 − 7 x ) 2x

3

=

(3 − 7 x ) 2x

32

= ( 2 x)

−1

(3 − 7 x )3 2

7. 3x 3 − 6 x 2 + 5 x − 10 = 3 x 2 ( x − 2) + 5( x − 2) −1 6

= (3 x 2 + 5)( x − 2)

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 116

Chapter 2

Differentiation

Skills Warm Up —continued—

(

x − x −5

8. 5 x

x +1 = x5

) (

x −1 −15

(

= ( x − 1) 5

)

x −1

)

10. − x 5 + 6 x 3 + 7 x 2 − 42 = − x 3 ( x 2 − 6) + 7( x 2 − 6) = ( − x 3 + 7)( x 2 − 6)

x −1

= − ( x 3 − 7)( x 2 − 6)

9. 4( x 2 + 1) − x( x 2 + 1) = ( x 2 + 1) 4 − x( x 2 + 1) 2

3

2

= ( x 2 + 1) ( 4 − x 3 − x) 2

y = f ( g ( x))

u = g ( x)

y = f (u )

u = 6x − 5

y = u4

u = x2 − 2 x + 3

y = u3

5x − 2

u = 5x − 2

y =

9 − x2

u = 9 − x2

y =

1. y = (6 x − 5)

4

2. y = ( x 2 − 2 x + 3) 3. y = 4. y =

3

5. y = (3x + 1)

3

−1

6. y = ( x 2 − 3)

−1 2

7. y = ( 4 x + 7)

−1 3  4 y′ =  (5 x 4 − 2 x) (10 x3 − 1)  3

u

u = 3x + 1

y = u −1

u = x2 − 3

y = u −1 2

y′ =

4(10 x3 − 1)

3(5 x 4 − 2 x)

y′ = 2( 4 x + 7) ( 4) 1

12. y = ( x3 + 2 x 2 )

y′ = −

y′ = 8( 4 x + 7) = 32 x + 56

13. f ( x) =

3

y′ = 3(3 x 2 − 2) (6 x) 2

y′ = 18 x(3 x 2 − 2)

9. y =

14. f ( x ) =

2

3 − x 2 = (3 − x 2 )

12

−1 2 1 y′ = (3 − x 2 ) ( − 2 x) 2

y′ = − x(3 − x y′ = −

13

2

)

−1 2

= −

x

(3 − x 2 )

12

x 3 − x2

10. y = 4 4 6 x + 5 = 4(6 x + 5)

14

−5 4 1 y′ = 4 (6 x + 5) (6)  4

y′ = 6(6 x + 5)

−5 4

=

+ 5)

40 x3 − 4 3

3 5x4 − 2x

−2

(3 x 2 + 4 x )

3x 2 + 4 x

( x3 + 2 x 2 )

2

−1 2 = 2(1 − x3 ) ; (c) General Power Rule 3 1− x

7

−3

(1 − x)3

= 7(1 − x) ; (c) General Power Rule

15. f ( x) =

3

82 ; (b) Constant Rule

16. f ( x) =

3

x 2 = x 2 3 ; (a) Simple Power Rule

17. f ( x) =

x2 + 9 ; (d) Quotient Rule x + 4x2 − 6

18. f ( x) =

x1 2 ; (d) Quotient Rule x + 2x − 5

3

3

19. y′ = 3( 2 x − 7) ( 2) = 6( 2 x − 7) 2

20. y = (3 − 5 x)

2

4

y′ = 4(3 − 5 x) ( − 5) = − 20(3 − 5 x) 3

6

(6 x

=

−1

y′ = ( −1)( x3 + 2 x 2 )

2

8. y = (3 x 2 − 2)

23

−1 3  2 y′ =  (5 x 4 − 2 x) ( 20 x3 − 2) 3   −1 3  2 y′ =  (5 x 4 − 2 x) ( 2)(10 x3 − 1)  3

u 3

11. y = (5 x 4 − 2 x)

3

54

(

)(

)

(

)(

21. h′( x) = 2 6 x − x3 6 − 3x 2 = 6 x 6 − x 2 2 − x 2

)

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.5 22. f ( x) = ( 2 x 3 − 6 x )

43

13  4 f ′( x ) =  ( 2 x 3 − 6 x) (6 x 2 − 6) 3  

f ′( x ) = 8( 2 x 3 − 6 x)

13

f ′(t ) =

f ′( x) = − 7 x( x 2 + 11)

( x2 − 1)

f ′( x) = −

t + 1 = (t + 1)

12

5 − 3 x = (5 − 3 x )

1 3 −1 2 (5 − 3x) ( −3) = − 2 2 5 − 3x

−1 2 1 4t + 5 s′(t ) = ( 2t 2 + 5t + 2) ( 4t + 5) = 2 2 2t 2 + 5t + 2

(7 x

y − 54 = 216( x − 2)

(2, 54)

−2

4

3

3

f ( 2) = 3(14) = 115,248 4

y − 115,248 = 296,352( x − 2)

−3

y = 296,352 x − 477,456

+ 6 x)

5

−4

( − 9)

−6

g ′( x) = −15(7 x 2 + 6 x)

(14 x

15(14 x + 6)

(7 x 2 + 6 x)

=

6

−6

(14 x

200,000

54

(2

= 3(7 x 2 + 6 x)

g ′( x) = 3( − 5)(7 x 2 + 6 x)

g ′( x) = −

200

3

3 2

2

f ′( 2) = 12(14) (9) = 296,352

f ′( x) = 2( − 3)( 2 − 9 x)

28. g ( x) =

2

32. f ′( x) = 12(9 x − 4) (9) = 108(9 x − 4)

−2 3

23

27. f ( x) = 2( 2 − 9 x)

) (2 x) = 12 x( x2 − 1)

−400

24 x + 3)

9

y = 216 x − 378

−2 3 1 y′ = 9 ( 4 x 2 + 3) (8 x)  3

(4 x

+ 11)

f ( 2) = 54

13

2

(x

2

f ′( 2) = 24(32 ) = 216

26. y = 9 3 4 x 2 + 3 = 9( 4 x 2 + 3)

y′ =

7x

= −

−4 3

(

12

y′ = 24 x( 4 x + 3)

+ 11)

31. f ′( x) = 2(3) x 2 − 1

2t 2 + 5t + 2 = ( 2t 2 + 5t + 2)

2

(x

92

−7 3 4x2  4 y′ =  − ( 4 − x3 ) ( − 3x 2 ) = 73  3 3( 4 − x 2 )

12

25. s(t ) =

−7 2

−9 2

7x 2

30. y = ( 4 − x3 )

1 1 (t + 1)−1 2 (1) = 2 2 t +1

24. g ( x) = g ′( x) =

( x 2 + 11)

= ( x 2 + 11)

7

117

−9 2  7 f ′( x) =  − ( x 2 + 11) ( 2 x) 2  

13  4 f ′( x ) =  ( 2 x 3 − 6 x) (6)( x 2 − 1)  3

23. f (t ) =

1

29. f ( x) =

The Chain Rule

− 9 x)

(2, 115,248)

−5

+ 6)

+ 6)

4

−1

5

0

4 x 2 − 7 = ( 4 x 2 − 7)

12

33. f ( x) =

f ′( x) =

−1 2 1 2 (4 x − 7) (8x) = 2

8 3 f ( 2) = 3

f ′( 2) =

8 ( x − 2) 3 8 7 y = x − 3 3

y −3 =

4x 4x2 − 7 10

(2, 3) −1

4

−4

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 118

34.

Chapter 2

Differentiation

f ( x) = x

x 2 + 5 = x( x 2 + 5)

12

38. f ′( x) =

12 −1 2 1  f ′( x) = x  ( x 2 + 5) ( 2 x) + ( x 2 + 5) (1) 2 

= x ( x + 5) 2

2

= ( x 2 + 5) =

2x + 5 x2 + 5

f

f ′ is never 0.

−5

4

f′

10

−3

2 − 5x 2 x

40. f ′( x) = −1

4

2

4

f has a horizontal tangent when f ′ = 0.

−1

−1

4

f f′ −4

x 2 − 2 x + 1 = ( x 2 − 2 x + 1)

41. y = ( 4 − x 2 )

−1 2 1 f ′( x) = ( x 2 − 2 x + 1) ( 2 x − 2) 2 x −1 = x2 − 2x + 1 x −1 = x −1

f ′( 2) = 1

=

−2

−3

−2 3

=

8 1 = − 4 −32

=

1 4

−1 1 = (t 2 + 3t − 1) t + 3t − 1

y′ =

( 2 , 14 ( 4

y′ = −2

1 − 3x 2 − 4 x3 2 x ( x 2 + 1)

43. y = −

−2

y′ =

2

2

f has a horizontal tangent when f ′ = 0.

( 2t

+ 3)

2t + 3

(t

2

+ 3t − 1)

2

(t

5t

(t + 8)

2

+ 8) (5) − (5t )( 2)(t + 8)(1) 2

((t + 8) )

2 2

5(t + 8) (t + 8) − 2t 

(t 2 + 8)

4

5(t + 8)( − t + 8)

(t + 8)

4

=

5(t − 8)

(t

+ 8)

3

Quotient Rule and Chain Rule

f −1

−2

General Power Rule

2

1 1 = − ( x − 2) 4 4 1 3 y = − x+ 4 4

2

2

2

= −

−5 3 2 4x 36. f ′( x) = − ( 4 − 3x 2 ) ( −6 x) = 53 3 (4 − 3x 2 )

(−8)

(4 − x 2 )

s′(t ) = −1(t 2 + 3t − 1)

y = x −1

53

( − 2 x)

2x

42. s(t ) =

4

y − 1 = 1( x − 2)

4( 2)

−2

General Power Rule (2, 1)

f ( 2) = 1

−1

y′ = ( −1)( 4 − x 2 )

3

f ( 2) = ( −8)

3

4

12

f ( x) =

37. f ′( x) =

( x + 1) x 2 x( x + 1)

39. f ′( x) = −

2

13 ( x − 2) 3 13 8 y = x − 3 3

y −

f′

−1 −1

 x 2 + ( x 2 + 5)  

y −6 =

f ′( 2) =

f

f ′ is never 0.

12

2

13 f ′( 2) = 3 f ( 2) = 6

35.

x ( x + 1)

2

32

+ ( x + 5)

−1 2

−1 2

3

2

5

f′

−2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.5 44. f ( x) = 3 x( x3 − 4)

−2

50.

f ′( x) = 3 x ( − 2)( x3 − 4)  = −18 x ( x − 4) 3

3

−3

−3

f ( x) = x3 ( x − 4)

+ 3( x − 4)

−2

f ′( x) = 5 x 4 − 32 x3 + 48 x 2 = x 2 (5 x 2 − 32 x + 48)

−3

− 3(5 x3 + 4)

( x 3 − 4)

3

51. y = x 2 x + 3 = x( 2 x + 3)

12

(

) f ′( x) = ( 2 x − 1)( − 6 x) + (9 − 3 x 2 )( 2)

45. f ( x) = ( 2 x − 1) 9 − 3x 2

= −12 x 2 + 6 x + 18 − 6 x 2 = − 6(3 x 2 − 2 x − 3)

( ) y′ = (7 x + 4)(3 x 2 − 4 x) + ( x3 − 2 x 2 )(7) = 21x3 − 16 x 2 − 16 x + 7 x3 − 14 x 2 = 28 x3 − 30 x 2 − 16 x

Product Rule and Simple Power Rule

t + 6 = 2t (t + 6)

12

= t ( t + 6)

−1 2

+ 2(t + 6)

12

= ( t + 6)

−1 2

t + 2(t + 6)

= ( t + 6)

−1 2

(3t

+ 12)

3(t + 4) 3t + 12 = t + 6 t + 6

53. y = t 2

−1 3

−4 3 3x 2  1 g ′( x ) = 3 − ( x 3 − 1) (3x 2 ) = − 43  3 ( x3 − 1)

General Power Rule 49. f ( x) = x(3x − 9)

2x + 3

Product and General Power Rule

General Power Rule

x −1

 x + ( 2 x + 3)

−1 2 12 1  y′ = 2t  (t + 6) (1) + (t + 6) ( 2) 2  

=

1 −1 2 47. y = = ( x + 2) x+ 2 1 1 −3 2 = − y ′ = − ( x + 2) 32 2 2( x + 2)

= 3( x 3 − 1)

−1 2

3( x + 1)

52. y = 2t

46. y = (7 x + 4) x3 − 2 x 2

3

= ( 2 x + 3)

Product and General Power Rule

Product Rule and Simple Power Rule

3

12 −1 2 1  y′ = x  ( 2 x + 3) ( 2) + ( 2 x + 3) 2 

=

= 18 + 6 x − 18 x 2

3

= x 2 (5 x − 12)( x − 4) Simple Power Rule

Product Rule and Chain Rule

48. g ( x ) =

2

= x5 − 8 x 4 + 16 x3

= − 3( x3 − 4) 6 x3 − ( x3 − 4) =

119

= x3 ( x 2 − 8 x + 16)

−2 (3x 2 ) + ( x3 − 4) (3) 3

The Chain Rule

12

−1 2 12 1  y′ = t 2  (t − 2) (1) + 2t (t − 2) 2  1 −1 2 = (t − 2) t 2 + 4t (t − 2) 2 t 2 + 4t (t − 2) = 2 t −2

=

3

t − 2 = t 2 (t − 2)

t (5t − 8) 2 t −2

Product and General Power Rule

f ′( x) = x(3)(3 x − 9) (3) + (3 x − 9) (1) 2

3

= (3 x − 9) 9 x + (3 x − 9) 2

= 9( x − 3) (12 x − 9) 2

= 27( x − 3) ( 4 x − 3) 2

Product and General Power Rule

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 120

Chapter 2

Differentiation

x ( x − 2) = x1 2 ( x − 2) 2

54. y =

58. f ( x) = (3 x3 + 4 x)

15

2

1 2 1  y′ = x1 2 2( x − 2) (1) + ( x − 2)  x −1 2    2 

= 2 = = =

x ( x − 2) +

(x

− 2)

2

2

(x

− 2)(5 x − 2)

59.

2

f (t ) =

f ′(0) =

2(6 − 5 x)(5 x 2 − 12 x + 5) − 1)

 4x2 − 5  y′ = 3   2− x 

 ( 2 − x)(8 x) − ( 4 x − 5)( −1)    2  ( 2 − x) 

 4x2 − 5  = 3   2− x 

2

16 x − 8 x 2 + 4 x 2 − 5    2   ( 2 − x)

2

= s′(3) =

 4 x − 5   − 4 x 2 + 16 x − 5  = 3    2  ( 2 − x)  2 − x  

y −

3( 4 x 2 − 5) ( − 4 x 2 + 16 x − 5) 2

( 2 − x )3

Quotient and General Power Rule 57. y = ( x3 − 2 x 2 − x + 1)

= 36(3 − t ) −3

(−1)

=

−2

72

(3 − t )3 12

(0, 4) −4

4 −2

1 2

x − 3x + 4

= ( x 2 − 3x + 4)

−1 2

−3 2 1 2 ( x − 3x + 4) (2 x − 3) 2 3 − 2x

2( x 2 − 3 x + 4) 3−6 2( 4)

32

= −

32

3 16

1 3 = − ( x − 3) 2 16 3 17 y = − x+ 16 16 3

2

( 3, 12 (

y′ = 2( x3 − 2 x 2 − x + 1)(3 x 2 − 4 x − 1)

)(

)

m = y′(1) = 2 (1) − 2(1) − (1) + 1 3(1) − 4(1) − 1 3

2

3 0

72 8 = 27 3

s′( x ) = −

2

(

(3 − t )

60. s( x) = 2

=

36

−3

8 (t − 0 ) 3 8 y = t + 4 3

3

2

(2, 2)

y −4 =

3

Quotient and General Power Rule  4x2 − 5  56. y =    2− x 

4

f ′(t ) = −72(3 − t )

 2   6 − 5 x   ( x − 1)( −5) − (6 − 5 x)( 2 x)  y′ = 2 2  2   x − 1  ( x 2 − 1)  

(x

1 2

1 ( x − 2) 2 1 y = x +1 2

x

2

45

y − 2 =

Product and General Power Rule

=

5(3 x3 + 4 x)

m = f ′( 2) =

x

− 2) 4 x + ( x − 2) 2 x

 6 − 5x  55. y =  2   x − 1

9x2 + 4

=

2

(x

2

2

x

4 x( x − 2) + ( x − 2)

−4 5 1 3 (3x + 4 x) (9 x2 + 4) 5

f ′( x) =

2

= 2( −1)( − 2) = 4

2

−1

5

−1

4

y − 1 = 4( x − 1) y = 4x − 3

−4

(1, 1)

5

−2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.5 61.

f (t ) = (t 2 − 9) t + 2 = (t 2 − 9)(t + 2)

12

63.

−1 2  12 1 f ′(t ) = (t 2 − 9)  (t + 2)  + (t + 2) ( 2t ) 2  1 2 −1 2 12 = (t − 9)(t + 2) + 2t (t + 2) 2 −1 2  1  = (t + 2)  (t 2 − 9) + 2t (t + 2) 2 

= (t + 2) = (t + 2) =

5t2 2

(2)

32 6

(2, 3)

8

−4

−2

25 + x

= x( 25 + x 2 )

2

y′ = x − 12 ( 25 + x 2 )  = − x 2 ( 25 + x 2 )

2x 2x = − 12 1− x (1 − x)

= ( 25 + x 2 )

  12 −1 2 1  (1 − x) ( 2) −  2 (1 − x) ( −1)( 2 x)     y′ = −  12 2   − x 1 ( )  

=

)

−3 2

−3 2

(2 x)

−1 2

+ ( 25 + x 2 )

+ ( 25 + x 2 )

−3 2

 (1 − x) ( 2(1 − x) + x)   = − 1− x      (1 − x)−1 2 ( 2 − 2 x + x)   = − 1− x  

65.

(25 + x 2 )

f ( x) = f ′( x) =

 (1 − x)−1 2 ( 2 − x)   = − 1− x  

f ′( x) =

 ( 2 − x)   = − 32  (1 − x) 

3

(1 − x)3 2

−1 2

32

1

−3

3

−1

x 2 + 4 = ( x 2 + 4)

13

−2 3 1 2 x + 4) ( 2 x ) ( 3 2x

3( x 2 + 4)

Set f ′( x) =

x − 2

(1)

25

1 5 1 y − 0 = ( x − 0) 5 1 y = x 5

−1 2

−1 2

− x 2 + ( 25 + x 2 )  

y′(0) =

 2(1 − x)1 2 + x(1 − x) −1 2   = − 1− x  

5 y − 3 = − ( x − ( − 3)) 8 5 15 y −3 = − x − 8 8 5 9 y = − x + 8 8

− 3)

1−3 = −2 1

x

64. y = −16

(− 3) − 2 32 (1 − (− 3))

−1 2

4

(−1, −8)

(

( 12 )(2 x − 3)

− ( x + 1)

y = −2 x + 7

2 −4

y − ( −8) = −6 t − ( −1) y = −6t − 14

y′( − 3) =

(1)

y − 3 = −2( x − 2)

9 2

f ′( −1) = −6

=

(2 x − 3)

f ′( 2) =

t + 2

62. y = −

12

x − 4

=

9  t + 4t −  2 2 

− 3)

(2 x (2 x − 3) − ( x + 1) (2 x − 3)3 2

=

−1 2  5 2

+ 4t −

(2 x

f ′( x) =

9  2  2 t − 2 + 2t + 4t   

121

x +1 x +1 = 12 2x − 3 (2 x − 3)

f ( x) =

−1 2  1 2

The Chain Rule

23

2x 3( x 2 + 4)

23

= 0.

2x = 0 =

x = 0 → y = f (0) =

−5 5 = − 43 2 8

(

Horizontal tangent at: 0,

3

4

3

4

)

6

(−3, 3) −7

1 −1

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 122

66.

Chapter 2

5 x 2 + x − 3 = (5 x 2 + x − 3)

12

f ( x) = f ′( x) = f ′( x) =

Differentiation

−1 2 1 2 5 x + x − 3) (10 x + 1) ( 2 10 x + 1

10 x + 1

2(5 x + x − 3)

12

= 0.

10 x + 1 = 0 x = −

1  1 → y = f −  = 10  10 

61 20

61 is not a real number, there is no point 20 of horizontal tangency. −

Because

67.

x x = 2x − 1 (2 x − 1)1 2

f ( x) =

f ′( x) =

f ′( x) = f ′( x) = f ′( x) =

(2 x

− 1)

(2 x

− 1)

(2 x

− 1)

12

−1 2 1  x ( 2 x − 1) ( 2)  2  2 ( 2 x + 1)1 2   

(1) −

− x( 2 x − 1) (2 x + 1)

12

−1 2

−1 2

( 2 x − 1) − x ( 2 x − 1)

x −1

(2 x

Set f ′( x) =

− 1)

(3 x

− 2)

12

(5) − 5 x (3x 1

2

− 2)

−1 2

(3) 

12 2

5 (3x − 2)−1 2 2(3 − x 2) − 3x 2 f ′( x) = (2 x − 1) f ′( x) =

5(3 x − 4)

2(3 x − 2)

Set f ′( x) =

32

5(3x − 4)

2(3 x − 2)

32

= 0.

5(3x − 4) = 0 3x − 4 = 0 x =

4 20  4 → y = f  = 3 3 2  3

 4 10  Horizontal tangent at:  ,  3 3 2 59

r  1 r    69. A′ = 1000(60)1 +    = 50001 +  12 12 12      

59

59

32

x −1

(2 x

5x 5x = 12 3x − 2 ( 3 x − 2)

(3x − 2)    15 −1 2 12 5(3 x − 2) − x(3 x − 2) 2 ′ f ( x) = (2 x + 1)

2(5 x 2 + x − 3) 2

f ( x) =

f ′( x) =

12

Set f ′( x) =

68.

− 1)

32

= 0.

x −1 = 0 x = 1 → y = f (1) =

Horizontal tangent at: (1, 1)

1 =1 1

0.08   (a) A′(0.08) = 501 +  12   ≈ $74.00 per percentage point 59

0.10   (b) A′(0.10) = 501 +  12   ≈ $81.59 per percentage point 59

0.12   (c) A′(0.12) = 501 +  12   ≈ $89.94 per percentage point

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.6 −2 70. N = 400 1 − 3(t 2 + 2)   

71. V =

−3 dN = N ′(t ) = 400 ( − 3)( − 2)(t 2 + 2) ( 2t )   dt 4800t = 3 (t 2 + 2)

Higher-Order Derivatives

123

k t +1

When t = 0, V = 10,000. k  k = 10,000 0+1 10,000 V = t +1

10,000 =

(a) N ′(0) = 0 bacteria day

V = 10,000(t + 1)

(b) N ′(1) ≈ 177.8 bacteria day

−1 2

dV 5000 −3 2 = −5000(t + 1) (1) = − 32 dt (t + 1)

(c) N ′( 2) ≈ 44.4 bacteria day (d) N ′(3) ≈ 10.8 bacteria day

When t = 1, dV 5000 2500 = − 32 = − ≈ −$1767.77 per year. dt 2 ( 2)

(e) N ′( 4) ≈ 3.3 bacteria day (f ) The rate of change of the population is decreasing as time passes.

When t = 3,

dV 5000 = − 3 2 = −$625.00 per year. dt ( 4)

72. (a) From the graph, the tangent line at t = 4 is steeper than the tangent line at t = 1. So, the rate of change after 4 hours is greater. (b) The cost function is a composite function of x units, which is a function of the number of hours, which is not a linear function. 73. (a) r = (0.3017t 4 − 9.657t 3 + 97.35t 2 − 266.8t − 242)

12

−1 2 dr 1 = r ′(t ) = (0.3017t 4 − 9.657t 3 + 97.35t 2 − 266.8t − 242) ⋅ (1.2068t 3 − 28.971t 2 + 194.7t − 266.8) dt 2 1.2068t 3 − 28.971t 2 + 194.7t − 266.8 = 2 0.3017t 4 − 9.657t 3 + 97.35t 2 − 266.8t − 242

Chain Rule 4

(b)

8

13

−2

(c) The rate of change appears to be the greatest when t = 8 or 2008. The rate of change appears to be the least when t ≈ 9.60, or 2009, and when t ≈ 12.57, or 2012.

Section 2.6 Higher-Order Derivatives Skills Warm Up 1. −16t 2 + 292 = 0

−16t 2 = − 292 73 t2 = 4 73 t = ± 2

2. −16t 2 + 88t = 0 −8t ( 2t − 11) = 0 −8t = 0 → t = 0 2t − 11 = 0 → t =

11 2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Chapter 2

124

Differentiation

Skills Warm Up —continued— 3. −16t 2 + 128t + 320 = 0 −16(t 2 − 8t − 20) = 0

7.

y =

(2 x + 7)(2 x) − ( x 2 )(2) dy = 2 dx ( 2 x + 7)

−16(t − 10)(t + 2) = 0 t − 10 = 0 → t = 10 t + 2 = 0 → t = −2

=

4. −16t 2 + 9t + 1440 = 0

t =

−9 ±

92 − 4( −16)(1440)

5.

=

2( −16)

=

−9 ±

92241 −32 9 ± 3 10249 = 32 t ≈ − 9.21 and t ≈ 9.77 =

x2 2x + 7

8.

y =

4 x 2 + 14 x − 2 x 2

(2 x

+ 7)

2

2 x 2 + 14 x

( 2 x + 7) 2 x( x + 7) 2 ( 2 x + 7) 2

x 2 + 3x 2x2 − 5

(2 x 2 − 5)(2 x + 3) − ( x 2 + 3x)(4 x) dy = 2 dx (2 x 2 − 5)

y = x 2 ( 2 x + 7) dy = x 2 ( 2) + 2 x ( 2 x + 7 ) dx

=

4 x3 + 6 x 2 − 10 x − 15 − 4 x3 − 12 x 2

(2 x2

= 2 x 2 + 4 x 2 + 14 x = 6 x 2 + 14 x

6. y = ( x 2 + 3x)( 2 x 2 − 5)

dy = ( x 2 + 3x)( 4 x) + ( 2 x + 3)( 2 x 2 − 5) dx

=

− 5)

2

−6 x 2 − 10 x − 15

(2 x 2

− 5)

2

9. f ( x) = x 2 − 4

Domain: ( −∞, ∞)

= 4 x3 + 12 x 2 + 4 x3 − 10 x + 6 x 2 − 15

Range: [− 4, ∞)

= 8 x3 + 18 x 2 − 10 x − 15

10. f ( x) =

x −7

Domain: [7, ∞) Range: [0, ∞)

1.

2.

f ( x) = 9 − 2 x

f ( x) = 3 x 2 + 4 x

f ′( x) = −2

f ′( x) = 6 x + 4

f ′′( x) = 0

f ′′( x) = 6

f ( x) = 4 x + 15 f ′( x) = 4 f ′′( x) = 0

3.

4.

f ( x) = x 2 + 7 x − 4 f ′( x) = 2 x + 7 f ′′( x) = 2

5.

g (t ) = 13 t 3 − 4t 2 + 2t g ′(t ) = t 2 − 8t + 2 g ′′(t ) = 2t − 8

6. f ( x) = − 54 x 4 + 3 x 2 − 6 x f ′( x) = − 5 x3 + 6 x − 6 f ′′( x) = −15 x 2 + 6

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.6 7. f (t ) =

2 = 2t − 3 t3

13.

f ′′(t ) = 24t

f ′′′( x) = 60 x 2 − 72 x

5 5 = t −4 6t 4 6 10 − 5 g ′(t ) = − t 3 50 − 6 50 g ′′(t ) = t = 6 3 3t

f ( x) = 3( 2 − x 2 )

14.

f ′′( x) = 12 x 2 − 12 x f ′′′( x) = 24 x − 12 = 12( 2 x − 1)

3

15.

2

f ′( x) = 20 x3 + 180 x 2 + 480 x + 320

= 18( 2 − x 2 ) 4 x 2 − ( 2 − x 2 ) = 18( 2 − x

y = 4( x 2 + 5 x)

)(5 x

f ′′( x) = 60 x 2 + 360 x + 480

− 2)

f ′′′( x) = 120 x + 360

3

16.

y′ = 4(3)( x 2 + 5 x) ( 2 x + 5) 2

= 12 x11 − 216 x8 + 1296 x5 − 2592 x 2 f ′′( x) = 132 x10 − 1728 x 7 + 6480 x 4 − 5184 x f ′′′( x) = 1320 x9 − 12,096 x 6 + 25,920 x3 − 5184

y′′ = 120 x 4 + 1200 x3 + 3600 x 2 + 3000 x

f ′( x) =

x +1 x −1 ( x − 1)(1) − ( x + 1)(1)

17.

( x − 1)2

= −

2

( x − 1)

f ′′( x) = 4( x − 1)

= −2( x − 1)

2

−3

(1)

=

−2

− 1)

3

18.

1 − 4x 12. g ( x) = x−3 ( x − 3)(− 4) − (1 − 4 x)(1) g ′( x) = ( x − 3)2 = =

11

(x

− 3)

2

− 3)

f ′( x) = f ′′( x) =

2

= 11( x − 3)

g ′′( x) = − 22( x − 3)

−3

(1)

= −

−2

45 x7

2 25 x5 2 − x −5 25 2 −6 x 5 12 − x−7 5 84 −8 84 x = 5 5 x8

f ( x) = − f ( x) =

− 4 x + 12 − 1 + 4 x

(x

3 3 = x−4 8x4 8 3 f ′( x) = − x − 5 2 15 − 6 f ′′( x) = x 2 f ( x) =

f ′′′( x) = − 45 x − 7 = −

4

(x

4 3

= 24 x5 + 300 x 4 + 1200 x3 + 1500 x 2

f ( x) =

f ( x) = ( x3 − 6)

f ′( x) = 4( x3 − 6) (3 x 2 )

= ( 24 x + 60)( x 4 + 10 x3 + 25 x 2 )

11.

3

= 5 x 4 + 60 x 3 + 240 x 2 + 320 x

2

2

f ( x ) = 5 x ( x + 4)

= 5 x( x3 + 12 x 2 + 48 x + 64)

2

f ′′( x) = ( −18 x)2( 2 − x 2 )( −2 x) + ( 2 − x 2 ) ( −18)

2

f ( x) = x 4 − 2 x 3 f ′( x) = 4 x3 − 6 x 2

f ′( x) = 9( 2 − x 2 ) ( −2 x) = −18 x( 2 − x 2 )

10.

f ( x) = x5 − 3 x 4 f ′′( x) = 20 x3 − 36 x 2

24 = 5 t

8. g (t ) =

9.

125

f ′( x) = 5 x 4 − 12 x 3

f ′(t ) = − 6t − 4 −5

Higher-Order Derivatives

f ′′′( x) =

22

(x

− 3)

3

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DOWNLOAD SOLUTIONS MANUAL 126 19.

Chapter 2

Differentiation

g (t ) = 5t 4 + 10t 2 + 3

26. f ′′( x) = 20 x3 − 36 x 2

g ′(t ) = 20t 3 + 20t

f ′′′( x) = 60 x 2 − 72 x = 12 x(5 x − 6)

g ′′(t ) = 60t 2 + 20

27. f ′′′( x) = 4 x − 4

g ′′( 2) = 60( 4) + 20 = 260

f (4) ( x) = −16 x − 5

f ( x) = 9 − x 2

20.

80 f (5) ( x) = 80 x − 6 = 6 x

f ′( x) = −2 x

(

f ′′( x) = −2

)

28.

f ′′ − 5 = −2 21.

f ( x) =

4 − x = ( 4 − x)

1 −1 2 ( 4 − x) 2 1 −3 2 f ′′( x) = − ( 4 − x) 4 3 −3 −5 2 f ′′′( x) = − ( 4 − x) = 52 8 8( 4 − x)

22.

−3 8(9)

f (t ) =

= −

52

x − 2 = 4( x − 2)

12

−1 2 −1 2 1 f ′′′( x) = 4 ( x − 2) (1) = 2( x − 2) 2  

12

−3 2 −3 2  1 f (4) ( x) = 2 − ( x − 2) (1) = − ( x − 2)  2 3 3 −5 2 f (5) ( x) = ( x − 2) (1) = 52 2 2( x − 2)

f ′( x) = −

f ′′′( −5) =

f ′′( x) = 4

29. f (5) ( x) = 2( x 2 + 1)( 2 x) = 4 x3 + 4 x

1 648

f (6) ( x) = 12 x 2 + 4

2t + 3 = ( 2t + 3)

12

30.

1 (2t + 3)−1 2 ( 2) = ( 2t + 3)−1 2 2 1 −3 2 −3 2 f ′′(t ) = − ( 2t + 3) ( 2) = −( 2t + 3) 2 3 3 −5 2 f ′′′(t ) = ( 2t + 3) ( 2) = 2 (2t + 3)5 2

f ′′′( x) = 4 x + 7 f (4) ( x) = 4

f ′(t ) =

f (5) ( x) = 0

31.

f ′( x) = 3 x 2 − 18 x + 27 f ′′( x) = 6 x − 18

3 1 f ′′′  = 32  2

f ′′( x) = 0  6 x = 18 x = 3

23. f ( x) = ( x3 − 2 x) = x9 − 6 x 7 + 12 x5 − 8 x3 3

f ′( x) = 9 x − 42 x + 60 x − 24 x 8

6

4

2

f ′′( x) = 72 x 7 − 252 x5 + 240 x3 − 48 x f ′′(1) = 12 24. g ( x) = ( x 2 + 3x) = x8 + 12 x 7 + 54 x 6 + 108 x5 + 81x 4 4

g ′( x) = 8 x 7 + 84 x 6 + 324 x5 + 540 x 4 + 324 x3 g ′′( x) = 56 x 6 + 504 x5 + 1620 x 4 + 2160 x3 + 972 x 2

32.

f ( x) = ( x + 2)( x − 2)( x + 3)( x − 3) = ( x 2 − 4)( x 2 − 9) = x 4 − 13 x 2 + 36 f ′( x) = 4 x3 − 26 x f ′′( x) = 12 x 2 − 26 f ′′( x) = 0  12 x 2 = 26 x = ±

13 78 = ± 6 6

g ′′( −1) = −16 25. f ′( x) = 2 x 2

f ′′( x) = 4 x

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.6

33.

f ( x) = x f ′( x) = x

= =

127

x 2 − 1 = x( x 2 − 1)

12

12 12 −1 2 1 2 x2 + ( x 2 − 1) x − 1) ( 2 x) + ( x 2 − 1) = ( 12 2 ( x 2 − 1)

( x 2 − 1) (2 x) − x 2  12 ( x 2 − 1) (2 x) −1 2

12

f ′′( x) =

Higher-Order Derivatives

x2 − 1

( x 2 − 1)(2 x) − x3 + x 32 12 ( x 2 − 1) ( x 2 − 1)

+

−1 2 1 2 ( x − 1) (2 x) 2

x2 − 1 x2 − 1

2 x3 − 3x

( x 2 − 1)

32

f ′′( x) = 0  2 x 3 − 3 x = x( 2 x 2 − 3) = 0 x = ±

3 6 = ± 2 2

x = 0 is not in the domain of f. 34.

( x2 + 3)(1) − ( x)(2 x) = 3 − x2 = 3 − x2 x2 + 3 −2 ( )( ) 2 2 ( x2 + 3) ( x2 + 3) −3 −2 f ′′( x) = (3 − x 2 ) −2( x 2 + 3) ( 2 x) + ( x 2 + 3) ( −2 x)   f ′( x) =

= −2 x( x 2 + 3) 2(3 − x 2 ) + ( x 2 + 3) −3

=

−2 x(9 − x 2 )

( x2 + 3) 2 x( x 2 − 9) = 3 ( x2 + 3) f ′′( x) = 0  2 x( x 2 − 9) 3

= 0

x = 0, ± 3

35. (a) s(t ) = − 4.9t 2 + 44.1t

v(t ) = s′(t ) = − 9.8t + 44.1 a(t ) = v′(t ) = s′′(t ) = − 9.8 (b) s(3) = 88.2 m v(3) = 14.7 m sec a(3) = − 9.8 m sec 2 (c)

v (t ) = 0 − 9.8t + 44.1 = 0 − 9.8t = − 44.1 t = 4.5 sec

s( 4.5) = 99.225 m

s (t ) = 0

(d) 2

− 4.9t + 44.1t = 0 − 4.9t (t − 9) = 0 t = 0 sec

t = 9 sec

v(9) = − 9.8(9) + 44.1 = − 44.1 m sec This is the same speed as the initial velocity. 36. (a) s(t ) = − 4.9t 2 + 381 v(t ) = s′(t ) = − 9.8t a(t ) = v′(t ) = − 9.8

(b) s(t ) = 0 when 4.9t 2 = 381, or t =

381 ≈ 8.8 sec. 4.9

(c) v(8.8) = − 86.42 m sec

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 128

37.

Chapter 2

Differentiation

(t + 10)(27.5) − (27.5t )(1) = 275 dv = 2 2 dt (t + 10) (t + 10) t

0

10

20

30

40

50

60

v

0

13.75

18.33

20.63

22

22.92

23.57

dv dt

2.75

0.69

0.31

0.17

0.11

0.08

0.06

As time increases, the acceleration decreases. After 1 minute, the automobile is traveling at about 23.57 meters per second. 38. s(t ) = − 2.5t 2 + 20t v(t ) = s′(t ) = − 5t + 20 a(t ) = s′′(t ) = − 5

t

0

1

2

3

4

s(t)

0

17.5

30

37.5

40

v(t)

20

15

10

5

0

a(t)

−5

−5

−5

−5

−5

It takes 4 seconds for the car to stop, at which time it has traveled 40 meters. 39. f ( x) = x 2 − 6 x + 6

41. (a) y(t ) = − 21.944t 3 + 701.75t 2 − 6969.4t + 27,164

f ′( x) = 2 x − 6

(b) y′(t ) = − 65.832t 2 + 1403.5t − 6969.4

y′′(t ) = −131.664t + 1403.5

f ′′( x) = 2

(c) Over the interval 8 ≤ t ≤ 13, y′(t ) > 0; therefore,

7

(a)

y is increasing over 8 ≤ t ≤ 13, or from 2008 to 20013.

f′

f″ −5

10

−131.664t + 1403.5 = 0

−3

−131.664t = −1403.5

(b) The degree decreased by 1 for each successive derivative. (c) f ( x) = 3x 2 − 9 x

f ′( x) = 6 x − 9 f ′′( x) = 6

y′′(t ) = 0

(d)

f

t ≈ 10.66 or 2010

10

f′

42. Let y = xf ( x).

f″

−4

4

Then, y′ = xf ′( x) + f ( x)

y′′ = xf ′′( x) + f ′( x) + f ′( x) = xf ′′( x) + 2 f ′( x)

f − 10

(d) The degree decreases by 1 for each successive derivative. 40. Graph A is the position function. Graph B is the velocity function. Graph C is the acceleration function. Explanations will vary. Sample explanation: The position function appears to be a third-degree function, while the velocity is a second-degree function, and the acceleration is a linear function.

y′′′ = xf ′′′( x) + f ′′( x) + 2 f ′′( x) = xf ′′′( x) + 3 f ′′( x).

In general y (n) =  xf ( x)

( n)

= xf (n) ( x) + nf (n −1) ( x).

43. True. If y = ( x + 1)( x + 2)( x + 3)( x + 4), then y is a

fourth-degree polynomial function and its fifth derivative d5y equals 0. dx 5

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.7 44. True. The second derivative represents the rate of change of the first derivative, the same way that the first derivative represents the rate of change of the function.

Implicit Differentiation

129

45. Answers will vary.

Section 2.7 Implicit Differentiation Skills Warm Up 1. x −

y = 2 x

x2 = 6 − y2

x2 − y = 2 x

x2 − 6 = − y 2

− y = 2 x − x2

6 − x2 = y2

y = x2 − 2x

2.

± 6 − x2 = y

4 1 = x −3 y

7.

4y = x − 3 y =

3(1

2

x x

8.

x 6 ≠ −6

7 + 4 y = 3x 2 + x 2 y

)

=

3( 4) − 4 3

=

8 3

x2 − 2 , (0, − 3) 1− y 02 − 2 −2 1 = = − 1 − ( − 3) 4 2

9.

4 y − x 2 y = 3x 2 − 7

y(4 − x 2 ) = 3x 2 − 7 y =

3x 2 − 4 , ( 2, 1) 3y2 3( 2 2 ) − 4

x −3 4

3. xy − x + 6 y = 6 xy + 6 y = 6 + y ( x + 6) = 6 + 6+ y = x + y = 1, x 4.

6 − y2

x = ±

6.

7x  1  ,  − , − 2 4 y 2 + 13 y + 3  7   1 7 −  1 −1 −1  7 = = = 2 7 4( − 2) + 13( − 2) + 3 16 − 26 + 3 − 7

3x 2 − 7 , x ≠ ±2 4 − x2

5. x 2 + y 2 = 5 y 2 = 5 − x2 y = ±

5 − x2

x3 y = 6

1. x3

2. 3 x 2 − y = 8 x dy = 8 dx dy − = 8 − 6x dx dy = 6x − 8 dx

dy + 3x 2 y = 0 dx dy x3 = − 3x 2 y dx

6x −

dy 3x 2 3 = − 3 y = − y dx x x

y2 = 1 − x2

3.

2y

dy = −2 x dx dy x = − dx y

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 130

Chapter 2

Differentiation

y 3 = 5 x3 + 8 x

4.

3y2

10.

dy = 15 x 2 + 8 dx

xy − y 2 = y − x

2

y ( x − y ) = −( x − y )

dy 15 x + 8 = dx 3y2

y = −1 dy = 0 dx

y4 − y2 + 7 y − 6x = 9

5.

4 y3

dy dy dy − 2y +7 −6 = 0 dx dx dx = 6 (4 y 3 − 2 y + 7) dy dx dy 6 = 3 dx 4y − 2y + 7

11.

12 y 2

2 y = 4 xy 2 + 12 x dy dy = 8 xy + 4 y 2 + 12 dx dx dy dy 2 − 8 xy = 4 y 2 + 12 dx dx 2

dy dy dy + 10 y − − 9x2 = 8 dx dx dx = 8 + 9x2 (12 y 2 + 10 y − 1) dy dx 8 + 9x2 dy = 12 y 2 + 10 y − 1 dx

dy 4 y 2 + 12 = dx 2 − 8 xy 4 y2 = x2 y −9

12.

xy 2 + 4 xy = 10

7.

2y = 4x y +3 2

2 y = 4 x( y 2 + 3)

4 y 3 + 5 y 2 − y − 3 x3 = 8 x

6.

xy − y 2 =1 y − x

2

dy   2 ( y 2 − 9)8 y dy  − 4y 2y  dx   dx 

dy dy + 4 y + 4x = 0 y 2 + 2 xy dx dx dy (2 xy + 4 x) = − y 2 − 4 y dx

( y 2 − 9) 8y

dy y2 + 4 y = − dx 2 xy + 4 x

2 y 3 + 6 xy 2

dy dx = 2 x 2 2 ( y − 9) − 72 y

dy dy − 2 xy − x 2 = 0 dx dx = 2 xy − 2 y 3 (6 xy 2 − x 2 ) dy dx

2 x( y 2 − 9) dy = dx − 72 y

2x + y =1 x − 5y 2x + y = x − 5 y 6 y = −x 1 y = − x 6 1 dy = − 6 dx

2

x( y 2 − 9) dy = − dx 36 y

dy 2 xy − 2 y 3 = dx 6 xy 2 − x 2 9.

= 2x

dy 2 ( y − 9 − y2 ) dx = 2x 2 ( y 2 − 9)

2 xy 3 − x 2 y = 2

8.

2

13.

2

x 2 + y 2 = 16 dy = 0 dx dy 2y = −2 x dx dy x = − dx y

2x + 2 y

At (0, 4),

dy 0 = − = 0. dx 4

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.7 x 2 − y 2 = 25

14.

dy = 0 dx dy −2 y = −2 x dx dy x = dx y

At (5, 0),

x2

dy dy + 2 xy + y 3 + 3 y 2 x = 0 dx dx dy 2 ( x + 2 xy) = −2 xy − y 2 dx − ( 2 xy + y 3 ) dy = dx x 2 + 3 xy 2

dy is undefined. dx

y + xy = 4

15.

At (2, −1),

dy dy + x + y = 0 dx dx dy (1 + x) = − y dx dy y = − dx x +1 At ( −5, − 1),

x

At (7, 2) ,

x3 + y 3 dy 3x 2 + 3 y 2 dx dy 3x 2 + 3 y 2 dx dy 2 (3 y − 6 x) dx dy dx

dy 2 2 = − = . dx 7 − 6( 2) 5

dy  dy  + y + 2y = 0 2x −  x dx  dx  dy dy − y + 2y = 0 2x − x dx dx dy ( 2 y − x) = y − 2 x dx

= 6 xy  dy  = 6 x + y dx   dy = 6x + 6y dx = 6 y − 3x 2 =

6 y − 3x 2 3 y2 − 6x

3( 2 y − x 2 ) dy = dx 3( y 2 − 2 x) 2 y − x2 dy = 2 dx y − 2x

dy y − 3x 2 = dx 2y − x

4  4 8  dy At  ,  , = . 5  3 3  dx

(−1) − 3(− 2) = − 7 ; dy is dy At ( − 2, −1), = dx 2( −1) − ( − 2) 0 dx 2

undefined.

) = − −5 = 1. ) 10 2

2

20.

x 2 − xy + y 2 = 4

17.

(−1) 2(2) + (−1) dy = − 2 dx (2) (2) + 3(−1)

xy − x = y dy dy x + y −1 = dx dx dy dy x − =1− y dx dx dy ( x − 1) = 1 − y dx dy y −1 1− y = = − dx x −1 x −1 3−1 2  3  dy = − = − = − 4. At  , 3, 3 1  2  dx −1 2 2

dy 1 = − . dx 4

dy dy + y − 6y = 0 dx dx dy ( x − 6 y) = − y dx dy y = − dx x − 6y

( (

y(2 x + y 2 ) dy = dx x( x + 3 y 2 )

19.

xy − 3 y 2 = 2

16.

131

x2 y + y3 x = − 6

18.

2x − 2 y

Implicit Differentiation

21.

x1 2 + y1 2 = 9 1 −1 2 1 −1 2 dy x + y = 0 dx 2 2 dy x −1 2 + y −1 2 = 0 dx dy − x −1 2 = −1 2 = − dx y dy 5 At (16, 25), = − . dx 4

y x

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 132 22.

Chapter 2

Differentiation

x2 3 + y 2 3 = 5 2 −1 3 2 −1 3 dy + y = 0 x 3 3 dx

dy    dy  y 2  2 x + 2 y  + ( x 2 + y 2 ) 2 y  = 4 x dx    dx  dy dy dy 2 xy 2 + 2 y 3 + 2 x2 y + 2 y3 = 4x dx dx dx dy (4 y3 + 2 x2 y) = 4 x − 2 xy 2 dx 2 x( 2 − y 2 ) dy = dx 2 y(2 y 2 + x 2 )

− x −1 3 dy y1 3 y = −1 3 = − 1 3 = − 3 dx y x x

At (8, 1),

1 dy = − . dx 2

xy = x − 2 y

23.

x dy  1 x  y −1 2  + dx  2

y = x − 2y

x( 2 − y 2 ) dy = dx y(2 y 2 + x 2 )

dy 1  y  x −1 2  = 1 − 2 dx 2  At (1, 1) ,

y x dy dy +2 = 1− dx 2 y dx 2 x y dy x ⋅2 2 = dx x 2 +2 2 y 1−

= = =

2

x

y

x

y

xy − y

x+4

( x + y)

3

3

xy

2( x − 2 y) − y

dy dx dy 3x 2 + 3 y 2 dx

3x 2 + 3 y 2

x 2 − ( x + y)

x( 4 y − x 2 − y 2 ) dy = 2 dx x y + y3 − 2 x2 dy = 0. dx

27. 3x 2 − 2 y + 5 = 0 6x − 2

2

x 2 − ( x 2 + 2 xy + y 2 )

= 8x2 y

4( 4 xy − x3 − xy 2 ) dy = dx 4( x 2 y + y 3 − 2 x 2 )

At ( 2, 2) ,

3

2

dy  dy  2( x 2 + y 2 ) 2 x + 2 y  = 8 x 2 + y(16 x) dx dx   dy dy dy 4 x3 + 4 x 2 y + 4 xy 2 + 4 y 3 = 8x2 + 16 xy dx dx dx dy (4 x 2 y + 4 y3 − 8x 2 ) = 16 xy − 4 x3 − 4 xy 2 dx

2x − 5 y 5x − 8 y

−(2 xy + y 2 ) y (2 x + y) dy = = − 2 dx x + 2 xy x( x + 2 y )

dy At ( −1, 1), = −1. dx

( x2 + y2 )

x + 4( x − 2 y )

= x + y

dy  2 3( x + y ) 1 +  = dx   2 2 dy 3( x + y ) + 3( x + y ) = dx dy 2 dy ( x + y) − y 2 = dx dx dy  2 ( x + y) − y 2  = dx 

dy 1 = . dx 3

26.

1 dy At ( 4, 1), = . dx 4

24.

y 2 ( x2 + y2 ) = 2 x2

25.

At (1, 4),

dy = 0 dx dy = 3x dx

dy = 3. dx

28. 4 x 2 + 2 y − 1 = 0 8x + 2

dy = 0 dx 8x dy = − = −4 x dx 2

dy (−1) = −4(−1) = 4 dx

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.7 x2 + y 2 = 4

29.

2x + 2 y

At

(

33. Implicitly: 1 − 2 y

dy = 0 dx dy x = − dx y

)

3, 1 ,

dy 3 = − = − dx 1

12

dy 1 −1 2 = ± ( x − 1) (1) dx 2 1 = ± 2 x −1 y 1 = 2 2± x −1

dy = 0 dx dy 4x = − dx 9y

x2 − y3 = 0 2x − 3y2

dy = 0 dx dy 2x = dx 3y2

At ( −1, 1),

dy 2 = − . dx 3

(

34. Implicitly: 8 y

2y

( 4 − x)(3x 2 ) − x3 (−1) dy = dx ( 4 − x)2

2y

dy 12 x 2 − 3 x3 + x3 = 2 dx ( 4 − x)

2y

dy 12 x 2 − 2 x3 = dx ( 4 − x)2 2 x 2 ( x − 6) dy = − 2 dx 2 y( 4 − x) x 2 ( x − 6) dy = − 2 dx y ( 4 − x)

At ( 2, 7) ,

dy = 2. dx

1

−1 −2

3

(2, −1)

4

y=− x−1

dy − 2x = 0 dx dy x = dx 4y 1 x2 + 7 2 12 1 ± ( x 2 + 7) 2 −1 2 1 ± ( x 2 + 7) ( 2 x) 4 x ± 2 x2 + 7 x  1  x2 + 7  4 ±  2  x 4y

y = ±

Explicitly:

= dy = dx = =

=

At (3, 2),

y= x−1

2

dy 1 = − . dx 2

At ( 2, −1),

x3 4− x

)

1 2y

=

32. ( 4 − x) y 2 = x3

y2 =

x −1

= ±( x − 1)

3.

4  dy 4 5 5  At  5, , = − = − . 3  dx 9( 4 3) 3  31.

133

dy = 0 dx dy 1 = dx 2y

y = ±

Explicitly:

30. 4 x 2 + 9 y 2 = 36

8 x + 18

Implicit Differentiation

3 dy = . dx 8 y

y=

x2 + 7 4 2 2

−4

−2

(3, 2)

2

4

x

−2

y=−

x2 + 7 2

−4

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

x


DOWNLOAD SOLUTIONS MANUAL 134

Chapter 2

Differentiation

x 2 + y 2 = 100

35.

2x + 2 y

y 2 = 5 x3

37.

dy = 0 dx dy x = − dx y

2y

dy 15 x 2 = dx 2y

(

At (8, 6):

)

At 1,

4 m = − 3 4 y − 6 = − ( x − 8) 3 4 50 y = − x+ 3 3

y −

At ( −6, 8):

5 : m =

15 3 5 = 2 2 5

5 =

3 5 ( x − 1) 2

y =

3 5 5 x − 2 2

(

)

At 1, −

3 4

m =

dy = 15 x 2 dx

m =

3 y − 8 = ( x + 6) 4 3 25 y = x + 4 2

y +

16

(−6, 8)

5 :

−15 3 5 = − 2 2 5 5 = −

3 5 ( x − 1) 2

y = −

3 5 5 x + 2 2

(8, 6)

30

−24

24

(1, 5 ) −5

5

−16

(1, − 5 ) −30

x2 + y 2 = 9

36.

2x + 2 y

dy = 0 dx dy x = − dx y

4x

At (0, 3): m = 0

)

5: m = −

y −

2 2 5 = − 5 5

5 = −

2 5 ( x − 2) 5

y = −

2 5 9 5 x+ 5 5

6

(0, 3) −9

( 2, 2

3 2 3 y − 1 = − ( x − 1) 2 3 5 y = − x+ 2 2 m = −

y = 3

(

dy + 4 y + 2x = 0 dx dy 4 y + 2x 2y + x = − = − dx 4x 2x

At (1, 1):

y − 3 = 0( x − 0)

At 2,

4 xy + x 2 = 5

38.

5)

9

At (5, −1): 3 10 3 y + 1 = − ( x − 5) 10 3 1 y = − x + 10 2

2

m = −

(1, 1) 8

0

(5, − 1)

−2

−6

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.7 x3 + y 3 = 8

39.

dy = 0 dx dy = −3x 2 3y2 dx

3x 2 + 3 y 2

2y

dy x = − 2 dx y

dy = 0 dx y − 2 = 0( x − 0) m =

−8

(2, − 2) −3

(2, 0)

y − 2 = 2( x − 2) y = 2x − 2

8

At ( 2, − 2):

dy is undefined. dx

−5

m = −2 y + 2 = −2( x − 2)

The tangent line is x = 2.

y = −2 x + 2

x2 y − 8 = − 4 y x2 y + 4 y = 8

x + y 3 = 6 xy 3 − 1

42.

y ( x + 4) = 8 2

y 3 − 6 xy 3 = −1 − x

−1 8 = 8( x 2 + 4) x + 4

y 3 (1 − 6 x) = − (1 + x)

2

y3 =

−2 dy = 8( −1)( x 2 + 4) ( 2 x) dx dy 16 x = − 2 2 dx ( x + 4)

At ( − 2, 1): m =

4

m = 2

(0, 2)

At ( 2, 0):

y =

(2, 2) 0

At ( 2, 2): 5

y = 2

40.

3

x 2 (6 − x ) dy = 2 dx y (4 − x)

At (0, 2):

m =

(4 − x)(3x 2 ) − ( x3 )(−1) dy = 2 dx (4 − x)

2 x 2 (6 − x) dy 2y = 2 dx (4 − x)

2

135

x3 4− x

y2 =

41.

Implicit Differentiation

16( − 2) dy = − dx ( − 2)2 + 4

(

)

2

=

3y2

dy = dx

3y2

dy 6x − 1 − 6x − 6 = dx (6 x − 1)2

32 1 = 64 2

(6 x

− 1)

2

dy 7 = − 2 dx 3 y 2 (6 x − 1)

1 ( x − (− 2)) 2 1 y = x + 2 2

At ( −1, 0):

y −1 =

 1 At  6, :  5 16(6) dy 96 3 m = = − = − = − 2 2 dx 1600 50 (6) + 4   5 1 3 y − = − ( x − 6) 5 50 (−2, 1) 1 3 9 −8 y − = − x + 5 50 25 3 14 y = − x + 50 25 −5

x +1 6x − 1 (6 x − 1)(1) − ( x + 1)(6)

m =

dy is undefined. The tangent line is x = −1. dx

At (0, −1): m =

dy 7 = − dx 3 7 ( x − 0) 3 7 y = − x −1 3

y − ( −1) = −

(6, 15 )

8

2

−3

3

(−1, 0) (0, −1) −2

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 136

Chapter 2

43. p =

Differentiation

2 , x ≥ 0 0.00001x 3 + 0.1x

0.00001x3 + 0.1x = 0.00003x 2

= −

2 p2

dx 2 = − 2 dp p (0.00003 x 2 + 0.1) 44. p =

 dx  dx 2 x( 2 p ) + p 2  2  = − dp  dp  dx (2 p 2 + 1) = −4 xp dp dx 4 xp = − 2 dp 2p + 1 100 x 0.75 y 0.25 = 135,540

47. (a)

dy   100 x 0.75  0.25 y −0.75  + y 0.25 (75 x −0.25 ) = 0 dx  

4 , x ≥ 0 0.000001x + 0.05 x + 1 2

25 x 0.75 dy 75 y 0.25 ⋅ = − 0.25 0.75 y dx x

4 0.000001x + 0.05 x + 1 = p 2

0.000002 x

500 − x , 0 < x ≤ 500 2x

2 xp 2 = 500 − x

2 p

dx dx 2 + 0.1 = − 2 dp dp p

dx (0.00003x2 + 0.1) dp

p =

46.

dy 3y = − dx x

dx dx 4 + 0.05 = − 2 dp dp p

(0.000002 x

+ 0.05)

dx 4 = − 2 dp p dx 4 = − 2 dp p (0.000002 x + 0.05)

p =

45.

When x = 1500 and y = 1000, (b)

200 − x , 0 < x ≤ 200 2x

2 xp = 200 − x dx  dx 2 x( 2 p ) + p  2  = − dp  dp 

0

2,000

If more labor is used, then less capital is available.

2

dx (2 p 2 + 1) dp

100,000

0

2

dy = −2. dx

If more capital is used, then less labor is available. 48. (a) As price increases, the demand decreases.

= −4 xp

dx 4 xp = − 2 dp 2p + 1

(b) For x > 0, the rate of change of demand, x, with respect to the price, p, is always decreasing; that is, for x > 0,

dx is never increasing. dp

49. (a) y 2 − 35,892.5 = − 27.0021t 3 + 888.789t 2 − 9753.25t y 2 = − 27.0021t 3 + 888.789t 2 − 9753.25t + 35,892.5 y = ±

− 27.0021t 3 + 888.789t 2 − 9753.25t + 35,892.5

40

8

0

12

The numbers of cases of Chickenpox decreases from 2008 to 2012. (b) It appears that the number of reported cases was decreasing at the greatest rate during 2008, t = 8.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.8

Related Rates

137

(c) y 2 − 35,892.5 = − 27.0021t 3 + 888.789t 2 − 9753.25t 2y

dy = − 81.0063t 2 + 1777.578t − 9753.25t dt dy − 81.0063t 2 + 1777.578t − 9753.25 = dt 2y

y′ =

t

8

9

10

11

12

y

30.40

20.51

15.39

14.51

13.40

y′

−11.79

− 7.71

− 2.54

− 0.06

− 3.26

The table of values for y′ agrees with the answer in part (b) when the greatest value of y′ is −11.79 thousand cases per year.

Section 2.8 Related Rates Skills Warm Up 1. A = π r 2

9.

d 2  x + 2 y + xy dx  dy dy + y + x 2x + 2 dx dx dy dy + x 2 dx dx dy ( 2 + x) dx dy dx

2. V = 43 π r 3 3. SA = 6 s 2 4. V = s 3 5. V =

1π r 2h 3

6. A =

1 bh 2

7.

x2 + y 2 = 9 d 2  x + y 2  dx  dy 2x + 2 y dx dy 2y dx dy dx

d = [9] dx = 0 = −2 x −2 x = 2y =

8.

x 2 + 2 y + xy = 12

−x y

10.

=

d (12) dx

= 0 = − y − 2x = − y − 2x =

− y − 2x 2 + x

x + xy 2 − y 2 = xy d  x + xy 2 − y 2  dx  dy dy − 2y 1 + y 2 + 2 xy dx dx dy dy dy − 2y − x 2 xy dx dx dx dy (2 xy − 2 y − x) dx

d [ xy] dx dy = y + x dx

=

= y − y2 − 1 = y − y2 − 1

dy y − y2 − 1 = dx 2 xy − 2 y − x

3xy − x 2 = 6 d 3 xy − x 2  dx  dy − 2x 3 y + 3x dx dy 3x dx dy dx

=

d [6] dx

= 0 = 2x − 3y =

2x − 3y 3x

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 138

Chapter 2

1. y =

x,

Differentiation

dy dx 1 1 dx dx , = x −1 2 = = 2 dt dt 2 2 x dt dt

x

dy dt

dy 3 dx  1  =  = 3, (3) = . dt 4 dt 2 4

(a) When x = 4 and (b) When x = 25 and

dy dx = 2, = 2 25 ( 2) = 20. dt dt

dy dy dx dx dy dx dx 2. y = 3 x − 5 x, = 6x −5 , = (6 x − 5) , dt = dt dt dt dt dt 6 x − 5 dt 2

(a) When x = 3 and

dx dy = 2, = (6(3) − 5( 2)) = 26. dt dt

(b) When x = 2 and

dx 4 4 dy = = . = 4, 6( 2) − 5 7 dt dt

3. xy = 4, x

 x  dy dy dy dx  y  dx dx = −  , + y = 0, = −  dt dt dt  x  dt dt  y  dt

(a) When x = 8, y =

dx 1 dy 12 5 = 10, , and = − (10) = − . dt 2 dt 8 8

(b) When x = 1, y = 4, and 4. x 2 + y 2 = 25, 2 x

dy dx 1 3 = − 6, = − ( − 6) = . dt dt 4 2

dy x dx dx dx dy y dy = − , + 2y = 0, = − dt y dt dt dt dt x dt

(a) When x = 3, y = 4, and

dx dy 3 = 8, = − (8) = − 6. dt dt 4

(b) When x = 4, y = 3, and

dy dx 3 3 = −2, = − ( −2) = . dt dt 4 2

5. A = π r 2 ,

dr dA dr = 3, = 2π r = 6π r dt dt dt dA = 2π (6)(3) = 36π cm 2 min. dt

(a) When r = 6, (b) When r = 24, 6. V =

dA = 2π ( 24)(3) = 144π cm 2 min. dt

dV dr 4 3 dr πr , = 3, = 4π r 2 = 12π r 2 dt dt dt 3

(a) When r = 9, (b) When r = 16, 7. A = π r 2 ,

dV 2 = 12π (9) = 972π cm3 min. dt dV 2 = 12π (16) = 3072π cm3 min. dt

dA dr = 2π r dt dt

8. V =

dr 4 3 dV πr , = 4π r 2 dt dt 3

dr dV is constant, is not constant since it is dt dt proportional to the square of r.

If

dV dr 4 3 dV πr , = 10, = 4π r 2 , dt dt dt 3 dr  1  dV =  2 dt  4π r  dt

9. V =

(a) When r = 1,

dr 1 5 10 = m min. = 2( ) dt 2π 4π (1)

(b) When r = 2,

dr 1 5 10 = m min. = 2( ) dt 8π 4π ( 2)

dr dA dA is constant, then is not constant; is dt dt dt proportional to r.

If

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Section 2.8

10. V =

1 2 1 π r h = π r 2 (3r ) = π r 3 3 3

dV dr = 3π r 2 = 6π r 2 dt dt (a) When r = 6, (b) When r = 24,

dV 2 = 6π (6) = 216π cm3 min. dt dV 2 = 6π ( 24) = 3456π cm 3 min. dt

dC dx 11. (a) = 0.75 = 0.75(150) dt dt = 112.5 dollars per week

(b)

(c)

dR dx 1 dx = 250 − x dt dt 5 dt 1 = 250(150) − (1000)(150) 5 = 7500 dollars per week P = R −C dP dR dC = − = 7500 − 112.5 dt dt dt = 7387.5 dollars per week

12. (a)

(b)

(c)

dR dx dx = 510 − 0.6 x , dt dt dt dR dx = (510 − 0.6 x) dt dt

(a) When

dx = 9 units day and x = 400 units, dt

dR = 510 − 0.6( 400)(9) = $2430 per day. dt (b) When

dx = 9 units day and x = 600 units, dt

dR = 510 − 0.6(600)(9) = $1350 per day. dt 15. V = x3 ,

dx dV dx = 6, = 3x 2 dt dt dt

(a) When x = 2, (b) When x = 10, 16. A = 6 x 2 ,

 2(5000)  dR 2 x  dx  =  500 − =  500 − ( 250)  dt 25  dt 25    = 25,000 dollars week

(b) When x = 10,

dR dx dx = 1200 − 2x , dt dt dt dR dx = (1200 − 2 x) dt dt

13. R = 1200 x − x 2 ,

(a) When

dx = 23 units day and x = 300 units, dt

dR = 1200 − 2(300)( 23) = $13,800 per day. dt (b) When

dV 2 = 3( 2) (6) = 72 cm3 sec. dt dV 2 = 3(10) (6) = 1800 cm3 sec. dt

dx dA dx = 6, = 12 x dt dt dt

(a) When x = 2,

P = R −C

139

14. R = 510 x − 0.3 x 2 ,

dC dx = 1.05 = 1.05(250) = 262.5 dollars week dt dt

dP dR dC = − = 25,000 − 262.5 dt dt dt = 24,737.5 dollars week

Related Rates

dA = 12( 2)(6) = 144 cm 2 sec. dt dA = 12(10)(6) = 720 cm 2 sec. dt

17. Let x be the distance from the boat to the dock and y be the length of the rope.

32 + x 2 = y 2 dy = −1 dt dx dy = 2y 2x dt dt dx y dy = dt x dt When y = 5, x = 4 and As x → 0,

dx 5 = ( −1) = −1.25 m sec. dt 4

dx increases. dt

dx = 23 units day and x = 450 units, dt

dR = 1200 − 2( 450)( 23) = $6900 per day. dt

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 140

Chapter 2

Differentiation

18.

dx dy = − 720, = − 960, and dt dt x( dx dt ) + y ( dy dt ) dL = dt L

21. (a) L 2 = x 2 + y 2 , 5m

When x = 240 and y = 320, L = 400 and

2m

240( − 720) + 320( − 960) dL = = 1200 km h. dt 400

x y

y

5 y (a)  5 y − 5x = 2 y = 2 y − x

400 300

3 y = 5x y =

y

5 x 3

200

x 100

dx dy = 1.5 m sec and x = 3 m. when Find dt dt dy 5 dx = dt 3 dt dy 5 = (1.5) = 2.5 m sec dt 3

(b) Find

dx d = 1.5 m sec and ( y − x) when dt dt

dy = 2.5 m sec when x = 3 m. dt d dy dx − ( y − x) = dt dt dt = 2.5 − 1.55 = 1 m sec

19. x 2 + 102 = s 2 2x

dx ds = 2s dt dt dx s ds = dt x dt

When s = 26, x = 24 and

200

300

400

x

(b) t = 22.

400 1 = hr = 20 min 1200 3

S = 2250 + 50 x + 0.35 x 2 dS dx dx = 50 + 0.70 x dt dt dt dS = 50(125) + 0.70(1500)(125) dt = $137,500 per week

23. V = π r 2 h, h = 0.025, V = 0.025π r 2 , dV dr = 0.05π r dt dt

When r = 50 and

dr 1 = , dt 4

dV 1 = 0.05π (50)  = 0.625π ≈ 1.96 m3 min. dt  4 ds = 384: dt

dx 26 = (384) = 416 km h. dt 24 dx 20. s 2 = 902 + x 2 , x = 26, = −30 dt 2s

L

100

ds dx = 2x dt dt ds x dx = dt s dt

When x = 26, ds 26 = (−30) ≈ −8.33 ft sec. 2 dt 90 + 262

24. P = R − C

= xp − C = x(50 − 0.01x) − ( 4000 + 40 x − 0.02 x 2 ) = 50 x − 0.01x 2 − 4000 − 40 x + 0.02 x 2 = 0.01x 2 + 10 x − 4000 dP dx dx = 0.02 x + 10 dt dt dt When x = 800 and

dx = 25, dt

dP = 0.02(800)( 25) + (10)( 25) = $650 week. dt

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2

141

25. P = R − C = xp − C = x(6000 − 25 x) − ( 2400 x + 5200)

= −25 x 2 + 3600 x − 5200 dP dx dx = −50 x + 3600 dt dt dt 1 dx dP = 3600 − 50 x dt dt

When x = 44 and

26. (a) For supply, if

(b) For supply, if

dx 1 dP = = 5600, (5600) = 4 units per week. dt dt 3600 − 50( 44) dx dp dx dp is negative, then is negative. For demand, if is negative, then is positive. dt dt dt dt dp dx dp dx is positive, then is positive. For demand, if is positive, then is negative. dt dt dt dt

Review Exercises for Chapter 2 1. Slope ≈

−4 = −2 2

2. Slope ≈

4 = 2 2

3. Slope ≈ 0 4. Slope ≈

−2 1 = − 4 2

5. Answers will vary. Sample answer: t = 8; slope ≈ $225 million yr; Revenue was

increasing by about $225 million per year in 2008. t = 10; slope ≈ $350 million yr; Revenue was increasing by about $350 million per year in 2010. 6. Answers will vary. Sample answer: t = 10; slope ≈ − 20 thousand year; The number of

farms was decreasing by about 20 thousand per year in 2010. t = 12; slope ≈ −10 thousand year; The number of farms was decreasing by about 10 thousand per year in 2012. 7. Answers will vary. Sample answer: t = 1: m ≈ 65 hundred thousand visitors month; The

number of visitors to the national park is increasing at about 65,000,000 per month in January. t = 8: m ≈ 0 visitors month; The number of visitors to

the national park is neither increasing nor decreasing in August. t = 12: m ≈ −1000 hundred thousand month; The

number of visitors to the national park is decreasing at about 1,000,000,000 visitors per month in December.

8. (a) At t 1 , the slope of g (t ) is greater than the slope of

f (t ), so the rafter whose progress is given by g (t ) is traveling faster. (b) At t 2 , the slope of f (t ) is greater than the slope of g (t ), so the rafter whose progress is given by f (t ) is traveling faster. (c) At t 3 , the slope of f (t ) is greater than the slope of g (t ), so the rafter whose progress is given by f (t ) is traveling faster. (d) The rafter whose progress is given by f (t ) finishes first. The value of t where f (t ) = 9 is smaller than the value of t where g (t ) = 9. 9. f ( x) = − 3 x − 5; ( − 2, 1) f ′( x) = lim

Δx → 0

= lim

f ( x + Δx ) − f ( x ) Δx −3( x + Δx) − 5 − ( −3 x − 5) Δx

Δx → 0

−3Δx = lim = −3 Δx → 0 Δx

f ′( −2) = −3 10. f ( x) = 7 x + 3; ( −1, − 4) f ′( x) = lim

Δx → 0

= lim

f ( x + Δx ) − f ( x ) Δx 7( x + Δx ) + 3 − (7 x + 3)

Δx → 0

Δx

7 Δx = lim = 7 Δx → 0 Δx

f ′( −1) = 7

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 142

Chapter 2

Differentiation

11. f ( x) = x 2 + 9; (3, 18)

f ′( x) = lim

Δx → 0

= lim

12. f ( x) = x 2 − 7 x; (1, − 6)

f ( x + Δx) − f ( x) Δx

(x

2

= lim

Δx

x + 2 xΔx + ( Δx) + 9 − x − 9 Δx 2

( x + Δx )

2

− 7( x + Δx) − ( x 2 − 7 x) Δx

x + 2 xΔx + ( Δx) − 7 x − 7 Δx − x 2 + 7 x Δx 2

2

= lim

Δx → 0

2 xΔx + ( Δx) Δx → 0 Δx Δx( 2 x + Δx) = lim Δx → 0 Δx = lim ( 2 x + Δx) = 2 x

2 xΔx + ( Δx) − 7 Δx Δx → 0 Δx Δx( 2 x + Δx − 7) = lim Δx → 0 Δx = lim ( 2 x + Δx − 7) = 2 x − 7

2

2

= lim

= lim

Δx → 0

Δx → 0

f ′(3) = 2(3) = 6 13. f ( x) =

f ( x + Δx) − f ( x) Δx

Δx → 0

2

2

Δx → 0

Δx → 0

+ Δx) + 9 − ( x 2 + 9)

Δx → 0

= lim

f ′( x) = lim

f ′(1) = 2(1) − 7 = − 5

x + 9; ( − 5, 2)

f ′( x) = lim

f ( x + Δx) − f ( x) Δx

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

x + Δx + 9 − Δx

(x

x +9

x + Δx + 9 + x + Δx + 9 +

x +9 x +9

+ Δx + 9) − ( x + 9)

Δx  x + Δx + 9 + 1 x + Δx + 9 +

x + 9  =

x +9

2

1 x +9

1 f ′( − 5) = 4

14. f ( x) =

x − 1; (10, 3)

f ′( x) = lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

f ′(10) =

f ( x + Δx) − f ( x) Δx x + Δx − 1 − Δx

(x

x −1

x + Δx − 1 + x + Δx − 1 +

x −1 x −1

+ Δx − 1) − ( x − 1)

Δx  x + Δx − 1 +

x − 1

1 x + Δx − 1 +

=

x −1

2

1 x −1

1 6

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2 17. f ( x) = 9 x + 1

1 ; (6, 1) x−5 f ( x + Δx) − f ( x) f ′( x) = lim Δx → 0 Δx 1 1 − 5 x + Δ x − x −5 = lim Δx → 0 Δx ( x − 5) − ( x + Δx − 5) = lim Δx → 0 Δx( x + Δx − 5)( x − 5)

15. f ( x) =

f ′( x) = lim

Δx

9( x + Δ x) + 1 − (9 x + 1) = lim  Δx → 0 Δx = lim

9 x + 9Δ x + 1 − 9 x − 1 Δx

= lim

9Δ x Δx

Δx → 0

Δx → 0

= lim 9 = 9 Δx → 0

f ′(6) = −1

18. f ( x) = 1 − 4 x

1 x+6

f ′( x) = lim

Δx → 0

f ( x + Δx ) − f ( x ) Δx 1 1 − x + Δ x + x + 6 6 = lim Δx → 0 Δx ( x + 6) − ( x + Δx + 6) = lim Δx → 0 Δx ( x + Δx + 6)( x + 6)  

f ( x + Δ x) − f ( x) Δx

1 − 4( x + Δ x) − (1 − 4 x) = lim  Δx → 0 Δx

f ′( x) = lim

Δx → 0

= lim

1 − 4 x − 4Δ x − 1 + 4 x Δx

= lim

− 4Δ x Δx

Δx → 0

Δx → 0

= lim − 4 = − 4

−1 Δx → 0 ( x + Δx + 6)( x + 6)

Δx → 0

= lim = −

f ( x + Δ x) − f ( x)

Δx → 0

1 −1 = lim = − 2 Δx → 0 ( x + Δx − 5)( x − 5) ( x − 5)

16. f ( x) =

143

1

(x

f ′( −3) = −

+ 6)

2

1

( − 2 + 6) 2

= −

1 16

1 19. f ( x) = − x 2 + 2 x 2

f ′( x) = lim

f ( x + Δ x) − f ( x) Δx

Δx → 0

= lim

− 1  2

Δx → 0

= lim

(x

(

+ Δ x) + 2( x + Δ x) − − 12 x 2 + 2 x  Δx 2

− 12 x 2 − x( Δ x) −

1 2

( Δ x)

+ 2 x + 2Δ x +

1 x2 2

− 2x

Δx

Δx → 0

= lim

2

)

(

Δx − x −

Δx → 0

(

1 Δx 2

)

+ 2

Δx

)

= lim − x − 12 Δ x + 2 = − x + 2 Δx → 0

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 144

Chapter 2

20. f ( x) = 3x 2 − f ′( x) = lim

Δx → 0

Differentiation 1 x 4 f ( x + Δ x) − f ( x) Δx

1 2    2 1  3( x + Δ x) − 4 ( x + Δ x) −  3 x − 4 x  = lim Δx → 0 Δx 1 1 1 x − Δ x − 3x 2 + x 4 4 4

3x 2 + 6 x( Δ x) + 3( Δ x) − 2

= lim

Δx

Δx → 0

= lim

6 x( Δ x) + 3( Δ x)

2

Δx

Δx → 0

1 − Δx 4

1  Δ x 6 x + 3( Δ x) −  4  = lim Δx → 0 Δx 1 1  = lim  6 x + 3( Δ x) −  = 6 x − Δx → 0  4 4

21. f ( x) =

x −5

f ′( x) = lim

Δx → 0

f ( x + Δ x) − f ( x) Δx x + Δx − 5 − Δx

= lim

Δx → 0

= lim

Δx → 0

(x

Δx

Δx → 0

+ Δ x − 5) − ( x − 5) x + Δx − 5 +

1 x + Δx − 5 +

= lim 22. f ( x) =

(

x−5

x −5

x −5

x + Δx − 5 +

x −5

x + Δx − 5 +

x −5

)

=

2

1 x −5

x +3

f ′( x) = lim

f ( x + Δ x) − f ( x)

 x + Δ x + 3 −  = lim  Δx → 0 Δx Δx → 0

= lim

Δx → 0

= lim

Δx → 0

x + Δx − Δx Δx

(

(x

x

(

+ Δ x) − x x + Δx +

1 x + Δx +

x

x =

Δx → 0

)

x +3

x + Δx +

x

x + Δx +

x

) x

f ( x + Δ x) − f ( x) Δx

5 5 − x + Δx x = lim Δx → 0 Δx 5 x − 5( x + Δ x) x( x + Δ x ) = lim Δx → 0 Δx = lim

1 2

5 x

f ′( x) = lim

Δx

Δx → 0

= lim

23. f ( x) =

Δx → 0

− 5Δ x Δ x  x( x + Δ x)

= lim − Δx → 0

5 5 = − 2 x( x + Δ x ) x

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2 24. f ( x) =

1 x + 4

f ′( x) = lim

Δx → 0

f ( x + Δ x) − f ( x) Δx

( x + 4) − ( x + Δ x + 4) ( x + 4)( x + Δ x + 4)

Δx → 0

Δx

−Δ x Δ x ( x + 4)( x + Δ x + 4) 1 1 = lim − = − 2 Δx → 0 ( x + 4)( x + Δ x + 4) ( x + 4) = lim

Δx → 0

25. y is not differentiable at x = −1. At ( −1, 0), the graph

has a vertical tangent line. 26. y is not differentiable at x = 0. At (0, 3), the graph has

a node.

28. y is not differentiable at x = −1. At ( −1, 0), the graph

has a cusp. 29. y = − 6

y′ = 2 x −1 3 y′ =

2 x1 3

37. g ( x) = 2 x 4 + 3x 2

g ′( x) = 8 x3 + 6 x 38. f ( x) = 6 x 2 − 4 x

f ′( x) = 12 x − 4 39. y = x 2 + 6 x − 7 y′ = 2 x + 6

40. y = 2 x 4 − 3 x3 + x

41. f ( x) = 2 x −1 2 ; ( 4, 1) f ′( x ) = − x − 3 2 f ′( 4) = − ( 4)

−3 2

= − 0.125

3 1  + 3;  , 6  2x 2  3 y = x −1 + 3 2 3 3 y′ = − x − 2 = − 2 2 2x 3 1 = 6 y′  = 2  2 2 12

42. y =

y′ = 0 30. f ( x) = 5

f ′( x) = 0 31. f ( x) = x 7

f ′( x) = 7 x 6 1 x−4

h( x ) = x − 4 h′( x) = − 4 x − 5 h′( x) =

36. y = 3 x 2 3

y′ = 8 x 3 − 9 x 2 + 1

27. y is not differentiable at x = 0. The function is discontinuous at x = 0.

32. h( x) =

5 x3 4 15 x 2 f ′( x) = 4

35. f ( x) =

1 1 − x + Δx + 4 x + 4 = lim Δx → 0 Δx

= lim

145

−4 x5

33. f ( x) = 4 x 2

f ′( x) = 8 x 34. g (t ) = 8t 6

g ′(t ) = 48t 5

()

43. g ( x) = x 3 − 4 x 2 − 6 x + 8; ( −1, 9) g ′( x ) = 3 x 2 − 8 x − 6 g ′( −1) = 3( −1) − 8( −1) − 6 = 5 2

44. y = 2 x 4 − 5 x 3 + 6 x 2 − x ; (1, 2) y′ = 8 x 3 − 15 x 2 + 12 x − 1 y′(1) = 8(1) − 15(1) + 12(1) − 1 = 4 3

2

45. f ′( x) = 4 x − 3

10

f ′( 2) = 5 y − 3 = 5( x − 2) y = 5x − 7

(2, 3) −3

5 −2

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DOWNLOAD SOLUTIONS MANUAL 146

Chapter 2

Differentiation

y′ = 44 x3 − 10 x

46.

50.

20

y′( −1) = −34 y − 7 = −34( x + 1)

dN = N ′(t ) = 0.8332t 3 − 23.862t 2 + 23.92t − 706.5 dt

(−1, 7)

y = −34 x − 27

−2

(a) 2010: N ′(10) = m ≈ − 2020.33

0

2012: N ′(12) = m ≈ − 2415.85

−2

1 = x1 2 − x −1 2 47. f ( x) = x − x 1 1 f ′( x) = x −1 2 + x −3 2 2 2 1 1 = + 2 2 x3 2 2 x

(b) Results should be similar. (c) The slope shows the rate at which the number of farms was increasing or decreasing in that particular year, or value of t.

f ′(1) = 1 −1

y − 0 = 1( x − 1)

(1, 0)

3

−2

f ( x) =

48.

In 2010, the number of farms was decreasing about 2020.33 thousand per year, and in 2012, the number of farms was decreasing about 2415.85 thousand per year. 51. f (t ) = 4t + 3; [− 3, 1]

y = x −1 3

N = 0.2083t 4 − 7.954t 3 + 11.96t 2 − 706.5t + 3891

13

x − x = x

− x

1 −2 3 1 −1 = −1 x 3 3 3 x2 1 1 11 −1 = −1 = − f ′( − 8) = 2 12 12 3 3 ( − 8) f ′( x) =

Average rate of change: f (1) − f ( − 3) 7 − ( − 9) = = 4 1 − ( − 3) 4 Instantaneous rate of change: f ′(t ) = 4 f ′(1) = 4 f ′( − 3) = 4

11 ( x + 8) 12 11 22 y −6 = − x − 12 3 11 4 y = − x − 12 3 y −6 = −

52. f ( x) = x 2 + 3x − 4; [0, 1]

Average rate of change:

f (1) − f (0) 1−0

=

0 − ( − 4) 1

= 4

Instantaneous rate of change:

8

f ′( x) = 2 x + 3

(−8, 6)

f ′(1) = 5 −10

3

−4

49.

R = − 0.5972t 3 + 51.187t 2 − 485.54t + 2199.0 dR = R′(t ) = − 1.7916t 2 + 102.374t − 485.54 dt

(a) 2008: R′(8) = m ≈ 218.8 2010: R′(10) = m ≈ 359.0 (b) Results should be similar. (c) The slope shows the rate at which sales were increasing or decreasing in that particular year, or value of t.

f ′(0) = 3 53. f ( x) = x 2 3 ; [1, 8]

Average rate of change:

f (8) − f (1) 8−1

=

4 −1 3 = 7 7

Instantaneous rate of change: 2 3 x1 3 1 f ′(8) = 3 2 f ′(1) = 3 f ′( x ) =

In 2008, the revenue was increasing about $218.8 million per year, and in 2010, revenue was increasing about $359.0 million per year.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2

147

54. f ( x) = x 3 − x 2 + 3; [− 2, 2] f ( 2) − f ( − 2) 7 − ( − 9) = = 4 2 − ( − 2) 4

Average rate of change:

Instantaneous rate of change: f ′( x) = 3 x 2 − 2 x f ′( 2) = 8 f ′( − 2) = 16

55. s (t ) = − 4.9t 2 − 10t + 200 s(3) − s(1)

(a) Average velocity =

3−1

=

125.9 − 185.1 = − 29.6 m sec 2

(b) v(t ) = s′(t ) = − 9.8t − 10 v(1) = −19.8 m sec v(3) = − 39.4 m sec s (t ) = 0

(c) 2

− 4.9t − 10t + 200 = 0 4.9t 2 + 10t − 200 = 0 t =

−10 ±

102 − 4( 4.9)( − 200) 2( 4.9)

=

−10 ± 4020 9.8

t ≈ 5.45 sec (d) v(t ) = s′(5.45) = − 9.8(5.45) − 10 ≈ − 63.4 m sec 56. (a) s(t ) = − 4.9t 2 + 84

58.

v(t ) = s′(t ) = − 9.8t (b) Average velocity =

dC = 450 − 2 x dx

s ( 2) − s ( 0)

2 − 0 64.4 − 84 = 2 = − 9.8 m sec

(c) v(t ) = − 9.8t

59.

60.

61.

4.9t 2 = 84 84 4.9 t ≈ 4.14 sec

62. R = 150 x −

dC = 320 dx

3 2 x 4

3 dR = 150 − x dx 2

(e) v( 4.14) = − 40.6 m sec 57. C = 2500 + 320 x

R = 150 x − 0.6 x 2 dR = 150 − 1.2 x dx

− 4.9t 2 + 84 = 0

t2 =

C = 475 + 5.25 x 2 3 dC 3.5 2  = 5.25 x −1 3  = 3 dx 3 x  

v(3) = − 29.4 m sec s (t ) = 0

C = 370 + 2.55 x = 370 + 2.25 x1 2 1 1.275 dC = ( 2.55)( x −1 2 ) = 2 dx x

v( 2) = −19.6 m sec (d)

C = 24,000 + 450 x − x 2 , 0 ≤ x ≤ 225

63.

R = − 4 x 3 + 2 x 2 + 100 x dR = −12 x 2 + 4 x + 100 dx

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 148 64.

Chapter 2

Differentiation

R = 4 x + 10 x1 2

69.

f ′( x) = 15 x 2 − 15 x 4 = 15 x 2 (1 − x 2 )

dR 5 = 4 + 12 dx x

Simple Power Rule

65. P = − 0.0002 x 3 + 6 x 2 − x − 2000 dP = −0.0006 x 2 + 12 x − 1 dx 1 3 x + 4000 x 2 − 120 x − 144,000 15 dP 1 = − x 2 + 8000 x − 120 dx 5

66. P = −

2

67. P = − 0.05 x + 20 x − 1000

(a) Find

dP when x = 100. dx

dP = − 0.1x + 20 = P′( x) dx

When x = 100, (b) Find

f ( x ) = x 3 (5 − 3 x 2 ) = 5 x 3 − 3 x 5

dP = P′(100) = $10. dx

ΔP for 100 ≤ x ≤ 101. Δx

P(101) − P(100) 101 − 100

= 509.95 − 500 = $9.95

(c) Parts (a) and (b) differ by only $0.05. 68. P = − 0.021t 2 + 2.77t + 148.9

(a) P(0) = 148.9 P( 4) = 159.644 P(8) = 169.716 P(12) = 179.116 P(16) = 187.844 P( 20) = 195.9

70.

f ( x) = 4 x 2 ( 2 x 2 − 5) = 8 x 4 − 20 x 2 f ′( x) = 32 x3 − 40 x = 8 x( 4 x 2 − 5) Simple Power Rule

( y′ = ( 4 x − 3)(3x

− 2 x2 )

Product Rule and Simple Power Rule 1  72. s =  4 − 2 (t 2 − 3t ) = ( 4 − t −2 )(t 2 − 3t ) t   s′ = ( 4 − t −2 )( 2t − 3) + (t 2 − 3t )( 2t −3 ) = 8t − 12 − 2t −1 + 3t −2 + 2t −1 − 6t −2 = 8t − 12 − 3t −2

Product Rule and Simple Power Rule 73. g ( x ) = g ′( x) = g ′( x) =

x x +3 ( x + 3)(1) − x(1)

(x

+ 3)

2

3

(x

+ 3)

2

Quotient Rule and Simple Power Rule 74. f ( x ) =

These values are the populations in millions for Brazil from 1990 to 2013.

f ′( x) =

P′( 4) = 2.602

3

= 16 x3 − 33 x 2 + 12 x

f ′( x) =

(c) P′(0) = 2.77

2

= 12 x3 − 25 x 2 + 12 x + 4 x3 − 8 x 2

P( 23) = 201.501

dP (b) = − 0.042t + 2.77 = P′(t ) dt

) − 4 x) + 4( x

71. y = ( 4 x − 3) x3 − 2 x 2

2 − 5x 3x + 1

(3 x

+ 1)( − 5) − ( 2 − 5 x)(3)

(3 x

+ 1)

2

−15 x − 5 − 6 + 15 x

f ′( x) = −

(3 x

+ 1)

2

11

(3 x

+ 1)

2

Quotient Rule and Simple Power Rule

P′(8) = 2.434 P′(12) = 2.266 P′(16) = 2.098 P′( 20) = 1.93 P′( 23) = 1.804

These are the rates at which the population of Brazil is changing in millions per year from 1990 to 2013. © 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2 75. f ( x ) =

f ′( x) = =

=

6x − 5 x 2 + 1`

(x

2

80. g ( x ) =

+ 1)(6) − (6 x − 5)( 2 x)

( x + 1) 2

+ 1)

= −

2

+ 1)

= =

( x 2 − 1) ( x2

( x − 1) 2

= −

2

=

2

− 1)

2

82. g (t ) =

2

x +1

( x 2 − 1)

2

g ′(t ) =

Quotient Rule and Simple Power Rule 77. f ( x) = (5 x 2 + 2)

f ′( x) = 3(5 x 2 + 2) (10 x) 2

=

2

f ( x) = f ′( x) =

3

x 2 − 1 = ( x 2 − 1)

13

1

(x

+ 1)

32

General Power Rule

 x 2 + ( x 2 + 1)  

2x2 + 1 x2 + 1

t

(1 − t )

3

(1 − t ) (1)

− t (3)(1 − t ) ( −1)

3

2

(1 − t )

6

(1 − t ) + 3t (1 − t ) 6 (1 − t ) (1 − t ) + 3t = 2t + 1 4 4 (1 − t ) (1 − t ) 2

2

= −16 x 2 (1 − 4 x 2 ) + (1 − 4 x 2 )

2

2

= (1 − 4 x 2 ) −16 x 2 + (1 − 4 x 2 ) = (1 − 4 x 2 )(1 − 20 x 2 )

2 −1 2 = 2( x + 1) x +1

−3 2  1 h′( x) = 2 − ( x + 1)  2

−1 2

f ′( x) = x( 2)(1 − 4 x 2 )( −8 x) + (1 − 4 x 2 )

General Power Rule

= −

12

83. f ( x ) = x(1 − 4 x 2 )

−2 3 1 2 ( x − 1) (2 x) = 2 2 x 2 3 3 3( x − 1)

79. h( x) =

x 2 + 1 = x( x 2 + 1)

Quotient Rule and General Power Rule

General Power Rule 78.

5

3

=

3

= 30 x(5 x 2 + 2)

− 5)

Product and General Power Rule

2 x3 + x 2 − 2 x − 1 − 2 x3 − 2 x 2 + 2 x −x − 1

(3x 2 − 5 x)

= ( x 2 + 1)

− 1)( 2 x + 1) − ( x 2 + x − 1)( 2 x)

2

(6 x

12 −1 2 1  g ′( x) = x  ( x 2 + 1) ( 2 x) + (1)( x 2 + 1) 2 

x2 + x − 1 x2 − 1

( x2 f ′( x) =

−5

24(6 x − 5)

81. g ( x) = x

2

Quotient Rule and Simple Power Rule 76. f ( x) =

− 5 x)

−4

General Power Rule

2(3 + 5 x − 3 x 2 )

( x2

(3x

= 6(3x 2 − 5 x)

4

g ′( x ) = 6( − 4)(3 x 2 − 5 x)

2

6 + 10 x − 6 x 2

( x2

6 2

149

Product and General Power Rule 5

84.

5 1  f ( x) =  x 2 +  = ( x 2 + x −1 ) x 

f ′( x) = 5( x 2 + x −1 ) ( 2 x − x −2 ) 4

4

1  1  = 5 x 2 +   2 x − 2  x  x  

General Power Rule

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL 150

Chapter 2

Differentiation 3

85. h( x) =  x 2 ( 2 x + 3) = x 6 ( 2 x + 3)

3

87.

4

4

2

= x( x − 1) (7 x − 2) 4

= 18 x5 ( 2 x + 3) ( x + 1) 2

Product and General Power Rule

Product and General Power Rule 88.

2

= 2( x − 2)( x + 4)( 2 x + 2) = 4( x − 2)( x + 4)( x + 1)

Product and General Power Rule

= =

=

(1 − 3t )2

2

− 1)

32

(2s)

+ 3s 2 ( s 2 − 1)

= s 2 ( s 2 − 1)

32

5s 2 + 3( s 2 − 1)  

= s 2 ( s 2 − 1)

32

(8s 2

52

− 3)

Product and General Power Rule

(3t + 1) (1 − 3t )2

(1 − 3t ) (1 2)(3t 2

+ 1)

( 52 )( s

52

12

3t + 1

(3t

f ( s) = s 3 ( s 2 − 1) f ′( s) = s 3

f ′( x ) = 2 ( x − 2)( x + 4) ( x − 2) + ( x + 4)

h′(t ) =

5

= x( x − 1) 5 x + 2( x − 1)

= 6 x5 ( 2 x + 3)  x + ( 2 x + 3)

89. h(t ) =

5

f ′( x) = 5 x 2 ( x − 1) + 2 x( x − 1)

2 3 h′( x) = x 6 3( 2 x + 3) ( 2) + 6 x5 ( 2 x + 3)  

86. f ( x ) = ( x − 2)( x + 4)

f ( x) = x 2 ( x − 1)

−1 2

+ 1)

−1 2

(3)

− (3t + 1)

12

(1 − 3t )

4

(2)(1 − 3t )(−3)

(1 − 3t )(3 2) + (3t + 1)6 3 (1 − 3t )

3(9t + 5)

2 3t + 1(1 − 3t )

3

Quotient and General Power Rule

(3x + 1)  3x + 1  90. g ( x) =  2  = 2 2  x + 1 ( x + 1) 2

g ′( x) =

( x2

2

+ 1) ( 2)(3 x + 1)(3) − (3x + 1) 2( x 2 + 1)2 x 2

2

( x2

+ 1)

6( x 2 + 1) (3 x + 1) − 4 x(3 x + 1) ( x 2 + 1) 2

=

=

=

=

4

2

( x 2 + 1) 2(3 x + 1)( x 2 + 1) 3( x 2 + 1) − 2 x(3 x + 1) 4 ( x 2 + 1) 2(3 x + 1)( x 2 + 1)( − 3 x 2 − 2 x + 3) 4 ( x 2 + 1) 2(3 x + 1)( − 3x 2 − 2 x + 3) 3 ( x 2 + 1) 4

Quotient and General Power Rule.

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2

T =

91.

−1 1300 = 1300(t 2 + 2t + 25) t + 2t + 25 2

T ′(t ) = −1300(t 2 + 2t + 25) = −

2600(t + 1)

(t 2

+ 2t + 25)

(a) T ′(1) = −

−2

(2t

3 = − 3x − 4 x4

f (5) ( x) = −60 x −6

2

f (6) ( x) = 360 x − 7 =

T ′(3) = −

13 ≈ − 6.5°F hr 2

T ′(5) = −

13 ≈ − 4.33°F hr 3

(b)

f ′′′( x) = −

f (4) ( x) = 12 x −5

+ 2)

325 ≈ − 6.63°F hr 49

T ′(10) = −

95.

151

96.

f ( x) =

360 x7

x = x1 2

1 −1 2 x 2 1 f ′′( x) = − x −3 2 4 3 −5 2 f ′′′( x) = x 8 15 15 4) ( f ( x) = − x −7 2 = − 16 16 x 7 2 f ′( x) =

1144 ≈ −1.36°F hr 841

60

97. f ′( x ) = 8 x 5 2 f ′′( x) = 20 x 3 2 0

f ′′′( x ) = 30 x1 2

24

0

f (4) ( x) = 15 x −1 2 =

The rate of decrease is approaching zero. 92. When L = 12, L 12 3 2 2 2 ( D − 4) = ( D − 4) = ( D − 4) 16 16 4 dV 3 = ( D − 4). dD 2 V =

(a) When D = 8,

98. f ′′( x ) = 9 3 x = 9 x1 3 f ′′′( x) = 3 x − 2 3 f (4) ( x) = − 2 x − 5 3 f (5) ( x) =

dV  3 =  (8 − 4) = 6 board ft in. dD  2

(b) When D = 16,

99.

dV  3 =  (16 − 4) = 18 board ft in. dD  2

(d) When D = 36, dV  3 =  (36 − 4) = 48 board ft in. dD  2 93.

f ( x) = 3x 2 + 7 x + 1

10 −8 3 10 x = 3 3 x8 3

f ( x) = x 2 +

3 = x 2 + 3 x −1 x

f ′( x ) = 2 x − 3x −2 f ′′( x ) = 2 + 6 x −3 = 2 +

(c) When D = 24, dV  3 =  ( 24 − 4) = 30 board ft in. dD  2

15 x1 2

100.

f ′′′( x) = 20 x 4 −

6 x3

2 = 20 x 4 − 2 x −3 x3

f (4) ( x) = 80 x3 + 6 x −4 f (5) ( x) = 240 x 2 − 24 x −5 = 240 x 2 −

24 x5

f ′( x) = 6 x + 7 f ′′( x) = 6

94.

f ′( x) = 5 x 4 − 6 x 2 + 2 x f ′′( x) = 20 x3 − 12 x + 2 f ′′′( x) = 60 x 2 − 12 = 12(5 x 2 − 1)

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DOWNLOAD SOLUTIONS MANUAL 152

Chapter 2

Differentiation

101. (a) s (t ) = − 4.9t 2 + 1.5t + 10 v(t ) = s′(t ) = − 9.8t + 1.5

dy dx dy dy 2y + dx dx dy (2 y + 1) dx dy dx 2y

a(t ) = v′(t ) = s′′(t ) = − 9.8

(b) s (t ) = 0 = − 4.9t 2 + 1.5t + 10 Using the Quadratic Formula, t ≈ 1.59 seconds. v(t ) = s′(t ) = − 9.8t + 1.5

(c)

v(1.59) ≈ −14.08 m sec

(d) a(t ) = v′(t ) = − 9.8 m sec 2 102. s(t ) =

At ( 2, 1),

1 −2 = (t + 1) t + 2t + 1

a(t ) = v′(t ) = 6(t + 1) 2

−3

−4

= − =

+ 1)

3

+ 1)

4

=

1 2y + 1

dy 1 = . dx 3

dy 4 y1 2 = − 23 dx 9x At (8, 4),

dy dy + 2y = 0 dx dx dy (9 x + 2 y ) = −2 x − 9 y dx dy −2 x − 9 y 2x + 9 y = = − dx 9x + 2 y 9x + 2 y

2x + 9 y + 9x

105. y 2 − x 2 + 8 x − 9 y − 1 = 0

dy dy − 2x + 8 − 9 = 0 dx dx dy (2 y − 9) = 2 x − 8 dx dy 2x − 8 = dx 2y − 9

dy 2 = − . dx 9

2 ( x − 8) 9 2 52 y = − x + 9 9

y − 4 = −

x 2 + 9 xy + y 2 = 0

2y

=1

2 −2 3 3 dy x + y −1 2 = 0 3 2 dx

3

dy dy 2 x + 3x + 3y + 3y2 = 0 dx dx dy (3x + 3 y 2 ) = −2 x − 3 y dx 2x + 3y −2 x − 3 y dy = = − 3x + 3 y 2 dx 3( x + y 2 )

104.

=1

2 x1 3 + 3 y1 2 = 10

108.

x + 3xy + y = 10

103.

dy dx

1 ( x − 2) 3 1 1 y = x + 3 3

2

(t 6

(t

=1−

y −1 =

2

v(t ) = s′(t ) = −2(t + 1)

y2 = x − y

107.

109.

y 2 − 2 x = xy dy dy − 2 = x + y dx dx dy ( 2 y − x) = y + 2 dx dy y + 2 = dx 2y − x 2y

At (1, 2),

dy 4 = . dx 3

4 ( x − 1) 3 4 2 y = x + 3 3

y − 2 =

106. y 2 + x 2 − 6 y − 2 x − 5 = 0

2y

dy dy + 2x − 6 − 2 = 0 dx dx dy (2 y − 6) = 2 − 2 x dx dy 2 − 2x 1− x = = dx 2y − 6 y −3

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DOWNLOAD SOLUTIONS MANUAL Review Exercises for Chapter 2

153

y 3 − 2 x 2 y + 3xy 2 = −1

110.

3y2

dy dy dy − 2x2 − 4 xy + 6 xy + 3y2 = 0 dx dx dx dy (3 y 2 − 2 x2 + 6 xy) = 4 xy − 3 y 2 dx dy 4 xy − 3 y 2 = dx 3 y 2 − 2 x 2 + 6 xy

At (0, −1),

dy = −1. dx

y + 1 = −1( x − 0) y = −x − 1

111.

113. Let b be the horizontal distance of the water and h be the depth of the water at the deep end.

A = πr2 dA dr = 2π r dt dt

(a) Find

dr dA = 2 cm min. when r = 3 cm and dt dt

dA = 2π (3)( 2) = 12π cm 2 min dt ≈ 37.7 cm 2 min (b) Find

dr dA = 2 cm min. when r = 10 cm and dt dt

dA = 2π (10)( 2) = 40π cm 2 min dt ≈ 125.7 cm 2 min 112.

P = 375 x − 1.5 x

2

dP dx dx = 375 − 3.0 x dt dt dt dP dx = (375 − 3.0 x ) dt dt

dP = 375 − 3.0(50)( 2) = $450 day dt dP (b) = 375 − 3.0(100)( 2) = $150 day dt (a)

Then b = 6h for 0 ≤ h ≤ 2. 1 V = bh(6) = 3bh = 3(6h)h = 18h 2 2 dV dh = 36h dt dt dh 1 dV 1 1 = = ( 2) = dt 36h dt 36h 18h When h = 1,

dh 1 1 = = m min. dt 18(1) 18

114. P = R − C

= xp − C = x( 211 − 0.002 x) − (30 x + 1,500,000) = 181x − 0.002 x 2 − 1,500,000 dP dx dx = 181 − 0.004 x dt dt dt dP dx = (181 − 0.004 x) dt dt dP = 181 − 0.004(1600) (15) = $2619 week dt

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DOWNLOAD SOLUTIONS MANUAL 154

Chapter 2

Differentiation

Chapter 2 Test Yourself 7. f ( x) = ( x + 3)( x 2 + 2 x)

1. f ( x) = x 2 + 3; (3, 12)

f ( x + Δx) − f ( x) Δx

f ′( x) = lim

Δx → 0

(x

= lim

f ( x) = x3 + 5 x 2 + 6 x f ′( x) = 3 x 2 + 10 x + 6

+ Δx) + 3 − ( x + 3) 2

2

(Or use the Product Rule.)

Δx

Δx → 0

x 2 + 2 xΔx + ( Δx) + 3 − ( x 2 + 3) 2

= lim

Δx

Δx → 0

= lim

Δx → 0

8. f ( x) =

2 xΔx + ( Δx) Δx

f ′( x) =

2

Δx 2 x + ( Δx) = lim Δx → 0 Δx = lim 2 x + Δx

9.

= 2x

Δx x + Δx − 2 − Δx

Δx → 0

= lim

Δx → 0

= lim

Δx → 0

x + Δx − Δx 1 x + Δx +

(

x − 2

x

1 2 4

=

x

f ( x) =

=

1 4

1 − 2 x = (1 − 2 x)

12

(5 x

− 1)

3

x x(3)(5 x − 1) (5) − (5 x − 1) 2

x

(5 x

− 1) 15 x − (5 x − 1) x2

(5 x

− 1) (10 x + 1)

2

2

x2

1 x 1 f ′( x ) = 1 + 2 x 1 f ′(1) = 1 + 2 = 2 1

f ′(t ) = 3t 2 + 2 4. f ( x) = 4 x 2 − 8 x + 1

f ′( x) = 8 x − 8

3

2

12. f ( x) = x −

3. f (t ) = t + 2t

y − 0 = 2( x − 1)

4

−6

(1, 0)

6

−4

y = 2x − 2

5. f ( x) = x3 2 + 6 x1 2 3 12 3 x x + 3 x −1 2 = + 2 2

11.

=

3

f ′( x) =

)

f ′( x ) =

x

At ( 4, 0): m =

f ( x) =

1 −1 2 (1 − 2 x) (−2) 2 1 = − 1 − 2x

1 2

2

f ′( x) =

f ( x + Δx) − f ( x)

= lim

=

10.

x − 2; ( 4, 0) Δx → 0

f ( x) = (3x 2 + 4)

= 36 x3 + 48 x

At (3, 12): m = 2(3) = 6

f ′( x ) = lim

5 −1 2 3 5 3 x + x1 2 = + x 2 2 2 2 x

f ′( x) = 2(3x 2 + 4)(6 x)

Δx → 0

2. f ( x) =

x (5 + x) = 5 x1 2 + x3 2

3 x

3 = 5 x 2 − 3x − 3 x3 9 f ′( x) = 10 + 9 x − 4 = 10 x + 4 x

6. f ( x ) = 5 x 2 −

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DOWNLOAD SOLUTIONS MANUAL Chapter 2 13. S = − 2.1083t 3 + 70.811t 2 − 777.05t + 2893.6

(a)

17.

ΔS for 10 ≤ t ≤ 12 Δt

f ( x) = f ′( x ) =

S (12) − S (10) 122.6416 − 95.9 = 12 − 10 2 = $13.3708 billion yr

=

(2 x

− 1)

2

4

(2 x

− 1)

2 −2

−3

2010: S ′(10) = $6.68 billion yr

f ′′( x) = 8( 2 x − 1)

2012: S ′(12) = $11.6284 billion yr

f ′′′( x) = −48( 2 x − 1)

(c) The annual sales of CVS Caremark from 2010 to 2012 increased by an average of about $13.37 billion per year, and the instantaneous rates of change for 2010 and 2012 are $6.68 billion per year and $11.63 billion per year, respectively.

( 2) −4

= 16( 2 x − 1)

( 2)

= −

a( 2) = − 9.8 m sec 2 x + xy = 6

19.

P = R −C

1+ x

P = (1700 x − 0.016 x 2 ) − (715,000 + 240 x) P = − 0.016 x 2 + 1460 x − 715,000

2y

f ′( x ) = 4 x + 3 f ′′( x ) = 4 f ′′′( x) = 0

3 − x = (3 − x )

12

1 1 (3 − x)−1 2 ( −1) = − (3 − x)−1 2 2 2 1 1  1 −3 2 −3 2 f ′′( x) = −  − (3 − x) ( −1) = − (3 − x) 2 2  4 f ′( x) =

1 3  −5 2 f ′′′( x) = −  − (3 − x) ( −1) 4 2  3 −5 2 = − (3 − x ) 8 3 = − 52 8(3 − x)

dy + y = 0 dx dy x = −y − 1 dx dy y +1 = − dx x

20. y 2 + 2 x − 2 y + 1 = 0

15. f ( x) = 2 x 2 + 3 x + 1

16. f ( x) =

4

At t = 2: s( 2) = 25.4 m

(a) Revenue: R = xp

dP = − 0.032 x + 1460 = P′( x) dx P′(700) = $1437.60

− 1)

a(t ) = v′(t ) = s′′(t ) = − 9.8

v( 2) = − 9.6 m sec

(b)

96

(2 x

v(t ) = s′(t ) = − 9.8t + 10

Profit = Revenue − Cost

R = 1700 x − 0.016 x 2

−3

18. s (t ) = − 4.9t 2 + 10t + 25

14. P = 1700 − 0.016 x, C = 715,000 + 240 x

R = x(1700 − 0.016 x)

155

2x + 1 2x − 1 (2 x − 1)(2) − (2 x + 1)(2)

= −4( 2 x − 1)

(b) S ′(t ) = − 6.3249t 2 + 141.622t − 777.05

Test Yourself

21.

dy dy + 2− 2 = 0 dx dx dy ( 2 y − 2) = −2 dx dy 1 = − dx y −1 4 x 2 − 3 y 2 + x3 y = 5

8x − 6 y

dy dy + x3 3x 2 y = 0 dx dx dy dy −6y + x3 = − 8 x − 3x 2 y dx dx = − (8 x + 3x 2 y ) ( x3 − 6 y) dy dx dy 8 x + 3x 2 y = − 3 dx x − 6y x(8 + 3 xy ) dy 8 x + 3x 2 y = = 3 dx 6y − x 6 y − x3

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DOWNLOAD SOLUTIONS MANUAL 156 22.

Chapter 2

Differentiation

V = π r 2 h = 20π r 3 dV dr = 60π r 2 dt dt

(a)

dV 2 = 60π (0.5) (0.25) = 3.75π cm3 /min dt

(b)

dV 2 = 60π (1) (0.25) = 15π cm3 /min dt

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Calculus An Applied Approach Brief International Metric Solutions Manual 10th Larson  

Calculus An Applied Approach Brief International Metric Edition Solutions Manual 10th Edition Larson. SOLUTIONS MANUAL for Calculus An Appli...

Calculus An Applied Approach Brief International Metric Solutions Manual 10th Larson  

Calculus An Applied Approach Brief International Metric Edition Solutions Manual 10th Edition Larson. SOLUTIONS MANUAL for Calculus An Appli...

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