The derivative is 0 on ( −− 3, 2 ) , 2 on ( −− 2, 1 ) ,0

43. The derivative is 0 on ( −− 3, 2 ) , 2 on ( −− 2, 1 ) ,0

53. = +Δ + x 1 x + 1

on ( − ) 1, 0 , − 2 on () 0,1 , 0 on () 1, 2 , 2 on () 2,3 Δ x

and 0 on () 3, 4 . The derivative is undefined at

44. The derivative is 1 except at x =− 2, 0, 2 where

x +Δ x ⎜

it is undefined. Δ

xx ( +Δ x )

xx ( +Δ x )

dy

lim −

dx Δ→ x 0 xx ( +Δ x ) x 2

46. Δ= y [3(0.1) 2 + 2(0.1) 1] – [3(0.0) + 2 + 2(0.0) 1] + ( x + 1 )( x +Δ−− x 1 )( x − 1 )( x +Δ+ x 1 ) 1

( x +Δ+ x 1 )( x + 1 )

47. Δy = 1/1.2 – 1/1 = – 0.1667 2 x +Δ−++Δ−− xx x x x 1 ⎡ x 2 +Δ−+−Δ− xx x x x 1 ⎤

⎦ = 1 2 +Δ+++Δ+ × Δ x

x Δy = 2/(0.1+1) – 2/(0+1) = – 0.1818 1

xx x x

+Δ+++Δ+ xx x x x 1 Δ x x 2 +Δ+++Δ+ xx x x x 1

= dx Δ→ x 0 x 2 +Δ+++Δ+ xx x x x 1 x 2 + 2 x +

50. Δ= y cos[2(0.573)] – cos[2(0.571)] ≈ –0.0036

Δ y ( x +Δ x )– 2 x 2 2 xx Δ+Δ ( x ) 2 2

= 2 x +Δ x ( x +Δ x ) − 1 x 2 − 1

( 2 +Δ x ) −− x ( x +Δ x ) 2 ( x − 1 ) ⎤

xx

Δ y [( x +Δ x ) – 3( 3 x +Δ x )]–( 2 x 3 –3 x 2 )

xx ( +Δ x )

() x −− x ( x +Δ−−Δ x x x x ) ⎤ ) ⎥ 1

3 x 3 xx –6–3 x Δ+Δ x ( x ) 2 Δ+Δ +Δ

lim (3 x 2 +Δ 3 xx –6–3 x Δ+Δ x ( x )) 2 dy

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1 63. The derivative is 0 at approximately t = 15 and

57. f ′ (0) ≈ – ; (2) 1 f ′ ≈

t = 201 . The greatest rate of increase occurs at

about t = 61 and it is about 0.5 degree F per day.

f ′ (5) ≈ ; (7) f ′ ≈ –3 The greatest rate of decrease occurs at about

t = 320 and it is about 0.5 degree F per day. The

58. g ′ (–1) ≈ 2; (1) g ′ ≈ 0 derivative is positive on (15,201) and negative on (0,15) and (201,365).

g (4) ≈ –2; g ′ (6) ≈ –

64. The slope of a tangent line for the dashed function is zero when x is approximately 0.3 or

1.9. The solid function is zero at both of these points. The graph indicates that the solid

function is negative when the dashed function

has a tangent line with a negative slope and positive when the dashed function has a tangent line with a positive slope. Thus, the solid function is the derivative of the dashed function.

65. The short-dash function has a tangent line with zero slope at about x = 2 . 1 , where the solid function is zero. The solid function has a tangent line with zero slope at about x = 0 . 4 , 1 . 2 and 3.5. The long-dash function is zero at these points.

The graph shows that the solid function is

61. a.

f (2) ≈ ; (2) f′ ≈

positive (negative) when the slope of the tangent

2 2 line of the short-dash function is positive

f (0.5) 1.8; (0.5) ≈ f′ ≈ –0.6 (negative). Also, the long-dash function is

positive (negative) when the slope of the tangent

b. = 0.5 line of the solid function is positive (negative).

Thus, the short-dash function is f, the solid

c. x = 5

function is f ' = g , and the dash function is g ' .

d. x = 3, 5

66. Note that since x = 0 + x, f(x) = f(0 + x) = f(0)f(x),

e. x = 1, 3, 5

hence f(0) = 1.

f. x = 0

fa ( + h )–() fa

fa ′ () = lim

g. x ≈− 0.7,

and 5 < x < 7

= lim

fafh ()()–() fa

62. The derivative fails to exist at the corners of the

fh ()–1

fh ( ) – (0) f

graph; that is, at t = 10 , 15 , 55 , 60 , 80 . The

= fa ( ) lim

= fa ( ) lim h → 0 h h → 0 h

derivative exists at all other points on the interval

= faf′ ( ) (0)

f ′ (a) exists since f ′ (0 ) exists.

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67. If f is differentiable everywhere, then it is

b. If f is an even function,

continuous everywhere, so

and b = 4 – 2m. f ( −− u ) fx ( ) For f to be differentiable everywhere,

above, then f ′− ( x 0 ) = lim

must exist.

mx +− b 4 70. Say f(–x) = –f(x). Then

fx (–)–() h = fx

x − 2 x → 2 − x − 2 –(–) fx h + fx ()

lim

= – lim

Thus m = 4 and b = 4 – 2(4) = –4 h → 0 h h → 0 h

fx ( + h )–() fx + fx ()–(–) fx h h

68. f s () x =

lim

h → 0 2 h an even function if f(x) is an odd function.

⎡ fx ( + h )–() fx fx (–)–() h fx

Say f(–x) = f(x). Then

1 h fx ( + h )–() fx 1 fx [ + (– )] – ( ) h fx → 0 h

= lim

lim

h → 0 h – h → 0 h fx (–)–() h 2 fx 2 – = lim

= fx ′ () + fx ′ () = fx ′ ( ). fx [ + (– )] – ( ) h fx

= – lim

= – fx ′ () so ′ f (x)

For the converse, let f (x) = x . Then – h → 0 – h

h h h h –– is an odd function if f(x) is an even function. – f s (0) = lim

= lim

h → 0 2 h h → 0 2 h 71.

but ′ f (0) does not exist.

a. If f is an odd function,

ft ()[ −−− f ( x )]

f ′− ( x 0 ) = lim t →− x 0 t + x 0 c. A function f (x) decreases as x increases when

0 c. A function f (x) increases as x increases when

f ′ (x) > 0.

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2.3 Concepts Review

13. D ( x x 4 + x 3 + x 2 ++ x 1)

1. the derivative of the second; second; = D x ( x 4 ) + D x ( x 3 ) + D x ( x 2 ) + D x () x + D x (1)

f (x) ′ g (x ) + g(x) ′ f ( x) = 4

2. denominator; denominator; square of the

14. D (3 x 4 gxfx 3 ()()–()() fxgx ′ x x x 2 +π+π x ′ 2 –2 –5 ) denominator;

4. kL(f); L(f) + L(g); D x = 12 x 3 –6 x 2 –10 x +π

7 5 Problem Set 2.3 –2 15. D

=π (7 x 6 ) – 2(5 x 4 ) – 5(–2 x –3 )

x ( x ) = 2(–2 x ) = –4 x –3 = 12 x − 10 x +π 10 x

) = –3 D x ( x ) = –3(–4 x ) 12 = x

7. D x ⎜⎟ ⎛⎞=π D x ( x –1 ) =π (–1 x –2 ) =π – x –2 = 3(–3 x ) (–4 + x ) = – 4 –4 x

18. D x (2 x –6 + x –1 ) = 2 D x –6 ( x ) + D x –1 ( x ) = 2(–6 x –7 ) (–1 + x –2 ) = –12 x –7 – x –2

2 = 2 2(–1 x –2 ) – (–2 x ) = – 2 + x

9. D x ⎜ 5 ⎟ = 100 D x ( x –5 ) 100(–5 = x –6 )

x ( x ) = –500 x = –

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26. D [(–3 x + 2

x ( x )– D x ⎜⎟ x

⎝ ⎠ 3 ⎝⎠ 3 = (–3 x + 2) D x (–3 x ++ 2) (–3 x + 2) D x (–3 x + 2)

2 = 2 (–1 x –2 )–0 = – = (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 12

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

24. D x 3 [3 ( xx –1)] 3 = xD x ( x 3 + x 3 2 x 2 x –1) ( –1) D x (3 ) x

2 3 3 = 5 x + 6 x 2 + 2 x = xx 3 (3 )( + x –1)(3) 12 = x –3

28. D x [( x –1)( x 1)]

(2 x + 1)(2) (2 + x + 1)(2) = 8 x + 4 = ( x –1)(2 ) ( x + x + 1)(4 x )

= 2 x 5 –2 x + 4 x 5 + 4 x 3 = 6 x 5 + 4 x 3 –2 x

29. D x [( x 2 + 17)( x 3 –3 x + 1)] = ( x 2 + 17) D ( x 3 –3 x ++ 1) ( x x 3 –3 x + 1) D x ( x 2 + 17) = ( x 2 + 17)(3 x 2 – 3) ( + x 3 –3 x + 1)(2 ) x = 3 x 4 + 48 x 2 – 51 2 + x 4 –6 x 2 + 2 x

= 5 x 4 + x 2 42 + x 2 – 51

30. D [( 4 + 2 )( x 3 x xx + 2 x 2 + 1)] = ( x 4 + 2) xD ( 3 + 2 2 ++ 1) ( x 3 2 x x 2 x x + + 1) D x ( x 4 + 2) x = ( x 4 + x 2 )(3 x 2 + 4)( x + x 3 + 2 x 2 + 1)(4 x 3 + 2) = 7 x 6 + 12 x 5 + 12 x 3 + 12 x 2 + 2

31. D x [(5 x 2 – 7)(3 x 2 –2 x + 1)] = (5 x 2 – 7) D x x 2 (3 –2 x ++ 1) (3 x 2 –2 x + 1) D x (5 x 2 – 7) = (5 x 2 – 7)(6 – 2) (3 x + x 2 –2 x + 1)(10 ) x = 60 x 3 – 30 x 2 – 32 x + 14

32. D x [(3 x 2 + xx 2 )( 4 –3 x + 1)] = (3 x 2 + 2) xD x ( x 4 –3 x ++ 1) ( x 4 –3 x + 1) D x (3 x 2 + 2) x = (3 x 2 + x 2 )(4 x 3 – 3) ( + x 4 –3 x + 1)(6 x + 2) = x 5 + x 4 x 18 2 10 – 27 –6 x + 2

(3 x 2 + 1)(0) – (6 ) x

(5 x 2 –1)(0) – 2(10 ) x

(5 x 2 –1) 2 (5 x 2 –1) 2

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(4 x 2 –3 x + 9)(0) – (8 – 3) x

(2 x 3 – 3 )(0) – 4(6 x x 2 – 3) –24 x 2 + 12

( x + 1)(1) – ( –1)(1) x

2 ⎝ x –1 ⎠ ( –1) x ( –1)(2) – (2 –1)(1) x x

(3 x + 5)(4 ) – (2 x x 2 – 1)(3)

(3 x 2 + 1)(5) – (5 – 4)(6 ) x x

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

(2 x + 1) 2 4 x 2 + 4–5 x

(2 x + 1) 2

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⎛ 5 x 2 + 2–6 x ⎞ (3 – 1) x D x (5 x 2 + x 2 – 6) – (5 x 2 + x 2 – 6) D x (3 – 1) x

42. D x ⎜

x (3 – 1)(10 x x +

(3 – 1) x 2

2) – (5 x 2 + x 2 – 6)(3)

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

( x 2 + x 2 – 3)(2 – 2) – ( x x 2 –2 x + 5)(2 x + 2)

46. a. ( – ) (3) f g ′ = f ′ (3) – g ′ (3) = 2 – (–10) 12 =

b. ( fg ⋅ ) (3) ′ = f (3) (3) g ′ + g (3) (3) f ′ = 7(–10) + 6(2) = –58

′ = = 7(–10) – 6(2) = 82 c. ( gf ) (3) –

47. D x [ ( )] fx 2 = D x [ ( ) ( )] fxfx

= fxD () x [ ( )] fx + fxD () x [ ( )] fx =⋅ 2 fx () ⋅ Dfx x ()

48. D x [ ( ) ( ) ( )] fxgxhx = fxDgxhx () x [ ( ) ( )] + gxhxDfx ()() x () = fxgxDhx ( )[ ( ) x () + hxDgx () x ( )] + gxhxDfx ()() x () = fxgxDhx ()() x () + fxhxDgx ()() x () + gxhxDfx ()() x ()

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49. D x ( x 2 –2 x + 2) = 2–2 x 54. Proof #1:

At x = 1: m tan = 2(1) – 2 = 0 D x [ fx () − gx () ] = D x [ fx ( ) ( 1) ( ) +− gx ]

Tangent line: y = 1

= D x [ fx () ] + D x [ ( 1) ( ) − gx ]

( x 2 + 4)(0) – (2 ) x

– ( + 2 2 + 2 Let Fx () = fx () − gx () 4) . Then ( x 4)

At x = 1: m tan =−

[ fx ( +− h ) gx ( + h ) − fx () − gx ()

25 Fx '( ) = lim h → 0

Tangent line: y – = – ( –1) x ⎡ fx ( +− h ) fx () gx ( +− h ) gx () ⎤

51. D 3 2 x 2 ( x – x ) = 3 x –2 x 55. a. D t 2 (–16 t + 40 t + 100) = –32 t + 40 The tangent line is horizontal when m tan = 0:

v = –32(2) + 40 = –24 ft/s

m tan = 3 x 2 –2 x = 0

b. v = –32t + 40 = 0

2 4 56. D t (4.5 t 2 + 2) t =+ 9 t ⎛ 2 ⎞

The line through (2,5) and (x 0 ,y 0 ) has slope

m tan = 4 – 2(1) = 2

Tangent line: y – 3 = 2(x – 1); y = 2x + 1 the slope of the line y = x . Solving for x gives

Set ' y equal to − 1 , the negative reciprocal of

At x = 3 : 4(3) – (3) y 2 0 0 = = 3 x =± 500 1/ 6 ≈± 2.817

m tan = 4 – 2(3) = –2

y =± 100(500) − 5/6 ≈± 0.563 Tangent line: y – 3 = –2(x – 3); y = –2x + 9 The points are (2.817,0.563) and

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58. D x ( x 2 ) = 2 x 61. The watermelon has volume πr 3 ; the volume

(, x y

The line through (4, 15) and 0 0 ) has slope

of the rind is

y 0 − 15 . 3 If (x

0 ,y ) is on the curve y = x 2 0 , then

At the end of the fifth week r = 10, so

x 2 0 –8 x 0 + 15 = 0 per cm of radius growth. Since the radius is ( x 0 – 3)( x 0 – 5) = 0 growing 2 cm per week, the volume of the rind is

At x 0 = 3 : (3) y 0 = = 9 growing at the rate of

(2) ≈ 681 cm 3 per

She should shut off the engines at (3, 9). (At

week.

x 0 = 5 she would not go to (4, 15) since she is moving left to right.)

2.4 Concepts Review

59. D x (7 – x 2 ) = –2 x

sin( x + h ) – sin( ) x

The line through (4, 0) and (, x 0 y 0 ) has

y slope 0

− 0 . If the fly is at (, x y ) when the x 0 −

7– x 2 –0

spider sees it, then m

tan = –2 x 0 =

0 3. cos x; –sin x

0 – 8x 0 +7=0 ( x 0 – 7)( x 0 –1) = 0

At x 0 = y 1: 6 0 =

Problem Set 2.4

+ 9 36 = 45 = 35 1. D x (2 sin x + 3 cos x) = 2 D x (sin x) + 3 D x ( cos x) ≈ 6. 7

d = (4 – 1) 2 + (0 – 6) 2 =

= 2 cos x – 3 sin x

They are 6.7 units apart when they see each other.

2. D (sin x 2 x ) = sin xD x (sin ) sin x + xD x (sin ) x ⎛ = sin x cos x + sin x cos x = 2 sin x cos x = sin 2x

60. P(a, b) is ⎜ a ,. ⎟ Dy x = –

so the slope of

x 2 3. D x 2 (sin x + cos 2 x ) = D x (1) = 0

1 the tangent line at P is – 2 . The tangent line is

a 4. D x (1 – cos 2 x ) = D (sin x 2 x )

(–) x a or y = – 2 (–2) x a which

= sin xD x (sin ) sin x + xD x (sin ) x

a = sin x cos x + sin x cos x

has x-intercept (2a, 0).

= 2 sin x cos x = sin 2x

dOP (,) = a 2 1 + 1 , (,) dPA = (–2) a a 2 +

a 2 a 2 5. D x (sec ) x = D x ⎜

⎟ ⎝ cos x ⎠

a 2 + 2 = dOP (,) so AOP is an isosceles

cos xD (1) – (1) D (cos ) x

cos 2 x

triangle. The height of AOP is a while the base,

OA has length 2a, so the area is (2 a)(a) = a 2 . =

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⎛ sin x + cos x ⎞

6. D x (csc ) x = D x ⎜

tan xD x (sin x + cos ) (sin x − x + cos ) xD x (tan ) x =

sin xD x (1) (1) − D x (sin ) x

– csc cot x x x x x x (sin + cos ) x

tan (cos – sin ) – sec

x (tan )

cos xD (sin ) sin x − xD (cos ) x

11. D x ( sin cos x x ) = sin xD x [ cos x ] + cos xD

x [ sin ]

8. D

x (cot ) x = D x ⎜

⎝ sin x

= sin x ( − sin x ) + cos x ( cos x ) 2 = cos x − sin 2 x

sin xD x (cos ) cos x − xD = x x (sin )

sin 2 x

12. D x ( sin tan x x ) = sin xD x [ tan x ] + tan xD x [ sin x ]

= 2 sin x sec

− sin 2 x – cos 2 x –(sin 2 x + cos 2 x )

x ) + tan x ( cos x )

sin x

sin x

( cos – x )

x ()

cos x

13 . D x ⎜

cos xD (sin x + cos ) (sin x − x + cos ) xD (cos ) x

x cos x x − sin x

cos

cos (cos – sin ) – (– sin x x x x – sin cos ) x = x

xD x ( − 1 cos x )( −− 1 cos xD )() x x

15 . D x ( x 2 cos ) x = xD 2 x (cos ) cos x + xD x ( x 2 ) =− x 2 sin x + x 2 cos x

( x 2 + 1) 2 ( x 2 + 1)(– sin x x + cos x + cos ) – 2 ( cos x xx x + sin ) x

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17. y = tan 2 x = (tan )(tan ) x x

b. D t (20 sin t) = 20 cos t

Dy

= (tan )(sec x x ) (tan )(sec + x x ) At : t = rate = 20 cos = 10 3 ≈ 17. 32 ft/s

= 2 tan sec x 2

c. The fastest rate 20 cos t can obtain is

18. y = sec 3 x = (sec 2 x )(sec ) x

20 ft/s.

Dy =

(sec x ) sec tan x x + (sec ) xD

2 x (sec x )

25 . y = tan x

3 = 2 sec x tan x + sec (sec x x ⋅ sec tan x x y ' sec = x + sec x ⋅ sec tan ) x x

When y = 0 , y = tan 0 = 0 and y ' sec 0 1 = 2 = .

= sec 3 x tan x + 2sec 3 x tan x

The tangent line at x = is y x 0 =.

= 3sec 2 x tan x

26. y = tan 2 x = (tan )(tan ) x x

19. D x (cos x) = –sin x

y ' = (tan )(sec x 2 x ) (tan )(sec + x 2 x )

At x = 1: m tan = – sin1 –0.8415 ≈ =

x 2 tan sec 2 x

y = cos 1 ≈ 0.5403 Now, sec x 2 is never 0, but tan x = at 0 Tangent line: y – 0.5403 = –0.8415(x – 1)

x = k π where k is an integer.

20. D (cot ) = – csc x 2 x x 27. y = 9sin cos x x

At : –2; x = m tan = y ' 9 sin ( sin ) cos (cos ) = [ x − x + x x ]

4 y =1

= ⎡ 2 2 ⎣ ⎤ 9 sin x − cos x ⎦

– ⎟ =− 9 [ cos 2 x ]

Tangent line: –1 –2 ⎜ x

The tangent line is horizontal when ' y = 0 or, in

21. D x sin 2 x = D x (2sin cos ) x x this case, where cos 2 x = 0 . This occurs when = ⎣ ⎡ 2 sin xD x cos x + cos xD

x sin x ⎤

where k is an integer.

=− 2sin cos x x fx '( ) = 0 when cos x = 1 ; i.e. when x = 2 k π

23. D t (30sin 2 ) t = 30 D t (2sin cos ) t t

where k is an integer.

fx '( ) = 2 when x = (2 k + 1) π where k is an

= 30 ( − 2sin 2 t + 2 cos 2 t )

integer.

60 cos 2 t

29. The curves intersect when 2 sin x =

30sin 2 t = 15

1(–1) = –1 so the curves intersect at right angles. The seat is moving to the left at the rate of 30 3

30 3 ft/sec

30. v = D t (3sin 2 ) t = 6 cos 2 t

ft/s.

At t = 0: v = 6 cm/s

24. The coordinates of the seat at time t are

t = : v= − 6 cm/s

(20 cos t, 20 sin t).

2 t = π : v = 6 cm/s

a. ⎜ 20 cos , 20sin ⎞= ⎟ (10 3, 10) ⎝

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31. D x 2 (sin x ) = lim

h → 0 h sin( x 2 + 2 xh + h 2 ) – sin x 2

= lim

= sin x 2 [cos(2 xh + h 2 ) – 1] cos + x 2 sin(2 xh + h 2 )

sin x 2 cos(2 xh + h 2 ) cos + x 2 sin(2 xh + h 2 ) – sin x 2

lim h → 0 = lim

2 cos(2 xh + h 2 )–1

sin(2 xh + h 2 ) ⎤

= lim(2 x + h ) sin

+ cos x

2 ⎥ = x 2 (sin x ⋅+ 0 cos x ⋅= 1) x 2 cos x

→ 0 2 xh + h 2

2 xh + h ⎥ ⎦

sin(5( x + h )) – sin 5 x

() = cos x − 1.25cos x + 0.225

32. D x (sin 5 ) x = lim

34. fx

h → 0 h sin(5 x + h 5 ) – sin 5 x

= lim 5sin 5 x

+ 5cos 5 x

=+ 0 5cos 5 1 5cos 5 x ⋅= x x 0 ≈ 1.95

f ′ (x 0

33. f(x) = x sin x

a.

2.5 Concepts Review

1. Dufgtgt t ; ( ( )) ( ) ′ ′

2. DwGHsHs v ; ( ( )) ′ ′ ()

3. ( ( )) ;( ( )) fx 2 fx 2

b. f(x) = 0 has 6 solutions on [,6] ππ

f ′ (x) = 0 has 5 solutions on [,6] ππ

4. x 2 cos( x 2 );6(2 x + 1) 2

c. f(x) = x sin x is a counterexample.

Consider the interval [] 0, π.

Problem Set 2.5

f () − π = f () π = 0 and fx () = 0 has

15 1. y =u and u = 1 + x

exactly two solutions in the interval (at 0 and

π ). However, f ' () x = 0 has two solutions

Dy = DyDu ⋅

= (15 u in the interval, not 1 as the conjecture 14 )(1) indicates it should have.

= 15(1 + x ) 14

d. The maximum value of fx ()– fx ′ () on

2. y =u 5 and u = 7 + x

[,6] π π is about 24.93.

Dy x = DyDu u ⋅ x

= (5u 4 )(1) = 5(7 + x ) 4

3. y =u 5 and u = 3 – 2x Dy x = DyDu u ⋅ x

= (5 u 4 )(–2) = –10(3 – 2 ) x 4

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4. y =u 7 and u =+ 42 x 2 10. y = cos u and u = 3 x 2 –2 x Dy x = DyDu u ⋅ x Dy x = DyDu u ⋅ x

= (7 u 6 )(4 ) x = x 28 (4 2 + x 26 ) = (–sin u)(6x – 2) = –(6 – 2) sin(3 x x 2 –2) x

5. y =u 11 and u = x 3 –2 x 2 + 3 x + 1 Dy x = DyDu u ⋅ x 11. y = u 3 and u = cos x

Dy x = DyDu u ⋅

= (11 u )(3 x –4 x + 3) x = (3 u 2 )(– sin ) x

= 11(3 x 2 –4 x + 3)( x 3 –2 x 2 + 3 x + 1) 10 = –3sin cos x 2 x

6. y = u –7 and u = x 2 – x + 1

12. y = u 4 , u = sin v, and v = 3 x 2

Dy = DyDu ⋅

xu

Dy x = DyDuDv u ⋅ v ⋅ x

= (–7 u –8

)(2 – 1) x

= (4 u 3 )(cos )(6 ) v x = –7(2 – 1)( x x 2 – x + 1) –8 = x 24 sin (3 3 x 2 ) cos(3 x 2 )

7. y = u –5 and u =+ x 3 x + 1

Dy x = DyDu u ⋅ x 13. y = u 3 and u =

x –1

= 5 (–5 u –6 )(1) = –5( x + 3) –6 = –

Dy x = DyDu u ⋅

Dy = DyDu ⋅ ⎛ x = + 1 ⎞ ⎛ –2 ⎞

= –9(6 x + 1)(3 x 2 + x – 3) –10 x − 2

9(6 x + 1)

x −π

(3 x 2 + x – 3) 10 Dy x = DyDu u ⋅ x

− 4 ( x −π ) D x ( x −−− 2) ( x 2) D x ( x −π = (− ) 3 u ) ⋅

9. y = sin u and u = x 2 + x ( x −π ) 2

Dy x = DyDu u ⋅ x

x −π 2

= (cos u)(2x + 1)

1) cos( x + x )

15. y = cos u and u =

( x + 2) D x (3 x 2 ) – (3 x 2 ) D x ( x + 2)

Dy x = DyDu u ⋅ x = (– sin ) u

( x + 2) 2

⎛ 3 x 2 ⎞ ( x + 2)(6 ) – (3 x x 2 )(1)

x = DyDuDv u ⋅ v ⋅ x = (3 u )( sin ) − v

(1 – ) x 2

⎛ x 2 ⎛ 2 2 2 ⎞ x ⎞ (1 – )(2 ) – ( x x x )(–1)

–3(2 – x x 2 )

Section 2.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

17. D x [(3 – 2) (3 – x 2 x 22 ) ] (3 – 2) = x 2 D x 22 (3 – x ) + (3 – x 22 ) D x (3 – 2) x 2 = (3 – 2) (2)(3 – x 2 x 2 )(–2 ) (3 – x + x 22 ) (2)(3 – 2)(3) x = 2(3 x − 2)(3 − x 2 )[(3 x − 2)( 2 ) (3 − x +− x 2 )(3)] = 2(3 x − 2)(3 − x 2 )(9 4 + x − 9 x 2 )

18. D [(2 – 3 x 24 )( x 7 + 3 3) ] (2 – 3 = x 24 ) D ( x 7 + 3) 3 + ( x 7 + 3) 3 D (2 – 3 x x 24 x x )

= (2 – 3 x 24 ) (3)( x 7 + 2 3) (7 x 6 )( + x 7 + 3 3) (4)(2 – 3 x 23 ) (–6 ) x = xx 3 (3 2 – 2) ( 3 x 7 + 2 3) (29 x 7 – 14 x 5 + 24)

⎡ ( x + 1) 2 ⎤ (3 – 4) x D x ( x + 2 1) – ( x + 1) 2 D x (3 – 4) x

(3 – 4)(2)( x x + 1)(1) – ( x + 2 1) (3) 3 x 2 – 8 – 11 x

( x + 1)(3 x − 11)

( x 2 + 2 4) (2) – (2 – 3)(2)( x x 2 + 4)(2 ) x

21. y ′ = 2 x 2 ( 2 + 4 )( x + 4 )( 2 = 2 x + 4 ) () 2 x = 4 x x 2 ( + 4 )

22. y ′ = 2 ( x + sin x )( x + sin x ′ ) = 2 ( x + sin x )( 1 + cos x )

⎛ 2 3–2 t ⎞ ( t + 5)(3) – (3 – 2)(1) t 51(3 – 2) t 2

( s + 4)(2 ) – ( s s 2 – 9)(1) s 2 + 8 s + 9

( 5) ( t + 5)(3)(3 – 2) (3) – (3 – 2) (1) t 2 t 3

(6 t + 47)(3 – 2) t 2

( t + 5) 2

26. d (sin 3 θ ) = 3sin 2 θ cos θ

⎛ sin x ⎞ (cos 2 )

(sin ) (sin )

cos 2 2 x x ⎛ 2 sin ⎞ cos cos 2 x x + 2sin sin 2 x x 3sin 2 x cos cos 2 x x + 6sin 3 x sin 2 x

3(sin 2 x ) (cos cos 2 x x + 2sin sin 2 ) x = x cos 2 4 x

Instructor’s Resource Manual

Section 2.5 117

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

d d 28. d = [sin tan( t

t 2 + 1)] sin = t ⋅ [tan( t 2 + 1)] tan( + t 2 +⋅ 1) (sin ) t

= (sin )[sec ( t 2 t 2 + 1)](2 ) tan( t + t 2 +

1) cos t = t 2 sin sec ( t 2 t 2 ++ 1) cos tan( t t 2 + 1)

30. Gt ′ () = ( t 2 + 9) 3 Dt ( 2 – 2) 4 + ( t 2 – 2) 4 Dt ( t 2 t + 9) 3 = ( t 2 + 3 9) (4)( t 2 – 2) (2 ) ( 3 t + t 2 – 2) (3)( 4 t 2 + 2 9) (2 ) t = t 2 (7 t 2 + 30)( t 2 + 2 9) ( t 2 – 2) 3

G′ (1) = –7400

31. Ft ′ ( ) [cos( = t 2 ++ 3 t 1)](2 t + 3) = (2 t + 3) cos( t 2 ++ 3 t 1) ;

F′ (1) = 5cos 5 1.4183 ≈

32. gs ′ () = (cos π sD ) (sin 2 π+ s ) (sin s 2 π sD ) s (cos π s ) = (cos π s )(2sin π s )(cos ππ+ s )( ) (sin 2 π s )(– sin ππ s )( ) =π sin π s [2 cos 2 π s – sin 2 π s ]

′ ⎛⎞ g 1 ⎜⎟ =π – ⎝⎠ 2

33. D [sin ( x 4 x 2 + x 3 )] = 4sin ( 3 x 2 + 3) xD x sin( x 2 + 3) x = 4sin ( 3 x 2 + x 3 ) cos( x 2 + 3) xD x ( x 2 + 3) x = 4sin ( 3 x 2 + x 3 ) cos( x 2 + x 3 )(2 x + 3) = 4(2 x + 3) sin ( 3 x 2 + x 3 ) cos( x 2 + 3) x

34. D t [cos (4 – 19)] 5cos (4 – 19) 5 t = 4 t D t cos(4 – 19) t = 5cos (4 – 19)[– sin(4 – 19)] 4 t t D t (4 – 19) t = –5cos (4 – 19) sin(4 – 19)(4) 4 t t = –20 cos (4 – 19) sin(4 – 19) 4 t t

35. D 3 2 t 2 [sin (cos )] 3sin (cos ) t = tD t sin(cos ) t = 3sin (cos ) cos(cos ) t tD t (cos ) t = 3sin (cos ) cos(cos )(– sin ) 2 t t t = –3sin sin (cos ) cos(cos ) t 2 t t

37. D 4 2 3 2 θ 2 [cos (sin θ )] = 4 cos (sin θ ) D θ cos(sin θ ) = 4 cos (sin 3 θ 2 )[– sin(sin θ 2 )] D θ (sin θ 2 ) = –4 cos (sin 3 θ 2 ) sin(sin θ 2 )(cos θ 2 ) D θ ( θ 2 ) = –8 cos (sin θ 3 θ 2 ) sin(sin θ 2 )(cos θ 2 )

38. Dx 2 x [ sin (2 )] x = xD sin (2 ) sin (2 ) 2 x x + 2 xDx x = x [2sin(2 ) xD

x sin(2 )] sin (2 )(1) x + x

= x [2sin(2 ) cos(2 ) x xD (2 )] sin (2 ) x + 2 x x = x [4sin(2 ) cos(2 )] sin (2 ) x x + 2 x = x 2 sin(4 ) sin (2 ) x + 2 x

39. D x {sin[cos(sin 2 )]} cos[cos(sin 2 )] x = xD x cos(sin 2 ) x = cos[cos(sin 2 )][– sin(sin 2 )] x xD x (sin 2 ) x = – cos[cos(sin 2 )]sin(sin 2 )(cos 2 ) x x xD x (2 ) x = –2 cos[cos(sin 2 )]sin(sin 2 )(cos 2 ) x x x

40. D t 2 {cos [cos(cos )]} 2 cos[cos(cos )] t = tD t cos[cos(cos )] t = 2 cos[cos(cos )]{– sin[cos(cos )]} t t D t cos(cos ) t = –2 cos[cos(cos )]sin[cos(cos )][– sin(cos )] t t tD t (cos ) t = 2 cos[cos(cos )]sin[cos(cos )]sin(cos )(– sin ) t t t t = –2sin cos[cos(cos )]sin[cos(cos )]sin(cos ) t t t t

Section 2.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

53. F ( cos x ) = F ′ ( cos x )( cos x )

≈+≈ dx

dx

=− sin xF ′ ( cos x )

42. ( f − 2 g )() 2 = f ′ ()()() 2 − 2 g 2 d 54. d cos ( Fx () ) =− sin ( Fx () ) Fx ()

= dx

f ′ () 2 − 2 g ′ () 2

dx

=− F ′ () x sin ( Fx () ) =− 120 () = 1

55. D 2

x ⎣ ⎡ tan ( F () 2 x ) ⎦ ⎤= sec ( F () 2 x ) D x ⎣ ⎡ F () 2 x ⎦ ⎤

43. ()()( fg 2 = f g ′ + g f ′ )( ) ( ) ( ) 2 = 2 0 + 1 1 = 1

= sec 2 ( F () 2 x ) × F ′ () 2 x × D x [] 2 x

g (2) (2) – (2) (2) f ′ f g ′

= 2 F ′ () 2 x sec 2 ( F () 2 x )

1 56. ⎣ ⎡ g ( tan 2 x ) ⎦ ⎤ = g ' tan 2 ( x ) ⋅ tan 2 x

( 2 ) ( sec 2 x ) ⋅ 2

45. ( f g ) (6) ′ = fg ′ ( (6)) (6) g ′

2 ' tan 2 sec 2

2 x ⎡ ⎣ () sin Fx () ⎤ ⎦

46. ( g f ) (3) ′ = gf ′ ( (3)) (3) f ′ 57. D Fx

= Fx () × D x ⎡

⎣ sin Fx () ⎦ + sin Fx () × DFx x ()

g (4) (3) f ≈ ⎜⎟ (1) =

= Fx () × 2sin Fx () × D x ⎡ ⎣ sin Fx () ⎦ ⎤

47. D x F () 2 x = F ′ ()() 2 x D x 2 x = 2 F ′ () 2 x + F ′ x sin 2 () Fx ()

2 2 2 = Fx × 2sin Fx × cos Fx × D ⎡ ′ Fx () () ( () ) x ⎣ () D ⎤

x F ( x + 1 )( = F x + 1 x x +

() x sin Fx ( () )