Solution manual calculus 8th edition varberg, purcell, rigdon ch02

2

CHAPTER

2.1 Concepts Review

The Derivative
4.

1. tangent line
2. secant line
3.

f (c + h ) − f ( c )
h

4. average velocity

Problem Set 2.1
1. Slope =

2. Slope =

5–3
2 – 32

=4

6–4
= –2
4–6

Slope ≈ 1.5
5.

3.

Slope ≈

Slope ≈ −2

5
2

6.

Slope ≈ –

94

Section 2.1

3
2

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7. y = x 2 + 1

[(2.01)3 − 1.0] − 7
2.01 − 2
0.120601
=
0.01
= 12.0601

d.

msec =

e.

mtan = lim

a., b.

f (2 + h) – f (2)
h
h →0

[(2 + h)3 – 1] – (23 − 1)
h
h →0

= lim

12h + 6h 2 + h3

h
h→0

= lim

c.

m tan = 2

d.

msec =

e.

(1.01)2 + 1.0 − 2
1.01 − 1
0.0201
=
.01
= 2.01

f (1 + h) – f (1)
h
h →0

mtan = lim

[(1 + h)2 + 1] – (12 + 1)
h

h →0

= lim

2 + 2h + h 2 − 2
h
h →0
h(2 + h)
= lim
h
h →0
= lim (2 + h) = 2
= lim

h(12 + 6h + h 2 )
h
h→0
= 12
= lim

9. f (x) = x 2 – 1
f (c + h ) – f (c )
mtan = lim
h
h→0
[(c + h)2 – 1] – (c 2 – 1)
h
h→0

= lim

c 2 + 2ch + h 2 – 1 – c 2 + 1
h
h→0
h(2c + h)
= lim
= 2c
h
h→0
At x = –2, m tan = –4
x = –1, m tan = –2
x = 1, m tan = 2
x = 2, m tan = 4
= lim

h →0

3

8. y = x – 1
a., b.

10. f (x) = x 3 – 3x
f (c + h ) – f (c )
mtan = lim
h
h→0
[(c + h)3 – 3(c + h)] – (c3 – 3c)
h
h→0

= lim

c3 + 3c 2 h + 3ch 2 + h3 – 3c – 3h – c3 + 3c
h
h→0

= lim

h(3c 2 + 3ch + h 2 − 3)
= 3c 2 – 3
h
h→0
At x = –2, m tan = 9
x = –1, m tan = 0
x = 0, m tan = –3
x = 1, m tan = 0
x = 2, m tan = 9
= lim

c.

m tan = 12

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Section 2.1

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11.

13. a.

16(12 ) –16(02 ) = 16 ft

b.

16(22 ) –16(12 ) = 48 ft

c.

Vave =

d.

f ( x) =
mtan

1
x +1

f (1 + h) – f (1)
= lim
h
h→0

−
= lim 2+ h 2
h
h →0
− 2(2h+ h)
= lim
h
h →0
1
= lim −
h→0 2(2 + h)
1

e.

b.

1
4
1
1
y – = – ( x –1)
2
4

=–

1
x –1

f (0 + h) − f (0)
= lim
h
h →0
1 +1
= lim h −1
h
h →0

12. f (x) =
mtan

= lim

16(3.01) 2 − 16(3)2
3.01 − 3
0.9616
=
0.01
= 96.16 ft/s
Vave =

f (t ) = 16t 2 ; v = 32c
v = 32(3) = 96 ft/s

1

14. a.

h

h −1

h →0

h
1
= lim
h →0 h − 1
= −1
y + 1 = –1(x – 0); y = –x – 1

Vave =

d.

(32 + 1) – (22 + 1)
= 5 m/sec
3– 2

[(2.003)2 + 1] − (22 + 1)
2.003 − 2
0.012009
=
0.003
= 4.003 m/sec
Vave =

Vave =

c.

144 – 64
= 80 ft/sec
3–2

[(2 + h) 2 + 1] – (22 + 1)
2+h–2

4h + h 2
h
= 4 +h
=

f (t ) = t2 + 1
f (2 + h) – f (2)
v = lim
h
h →0
[(2 + h)2 + 1] – (22 + 1)
h
h →0

= lim

4h + h 2
h
h →0
= lim (4 + h)
= lim

h →0

=4

96

Section 2.1

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15. a.

f (α + h) – f (α )
h

v = lim

h →0

2(α + h) + 1 – 2α + 1
h
h →0

= lim

2α + 2h + 1 – 2α + 1

h

= lim

h →0

= lim

( 2α + 2h + 1 – 2α + 1)( 2α + 2h + 1 + 2α + 1)
h( 2α + 2h + 1 + 2α + 1)

h →0

2h

= lim

2α + 2h + 1 + 2α + 1)

h →0 h(

2

=

2α + 1 + 2α + 1
1

b.

2α + 1

=

=

1
2α + 1

ft/s

1
2

2α + 1 = 2
3
2
The object reaches a velocity of 1 ft/s when t = 3 .
2
2

2 α + 1= 4; α =

16. f (t ) = – t2 + 4 t

18. a.

[–(c + h)2 + 4(c + h)] – (– c 2 + 4c)
h
h →0

v = lim

– c 2 – 2ch – h 2 + 4c + 4h + c 2 – 4c
h
h →0
h(–2c – h + 4)
= lim
= –2c + 4
h
h →0
–2c + 4 = 0 when c = 2
The particle comes to a momentary stop at
t = 2.

b.

1000(2.5)2 – 1000(2)2 2250
=
= 4500
2.5 – 2
0.5

c.

f (t ) = 1000t 2

= lim

17. a.

b.

c.

1000(2 + h)2 − 1000(2) 2
h
h→0

r = lim

4000 + 4000h + 1000h 2 – 4000
h
h→0
h(4000 + 1000h)
= lim
= 4000
h
h→0

= lim

⎡1
⎤ ⎡1 2 ⎤
2
⎢ 2 (2.01) + 1⎥ – ⎢ 2 (2) + 1⎥ = 0.02005 g
⎣
⎦ ⎣
⎦
rave

0.02005
=
= 2.005 g/hr
2.01 – 2

1 2
t +1
2
⎡ 1 (2 + h)2 + 1⎤ – ⎡ 1 22 + 1⎤
2
⎦ ⎣2
⎦
r = lim ⎣
h
h →0
f (t ) =

= lim

h→0

= lim

2 + 2h + 12 h 2 + 1 − 2 − 1

(

h 2 + 12 h

h
h→0
At t = 2, r = 2

h

)=2

Instructor’s Resource Manual

1000(3)2 – 1000(2)2 = 5000

19. a.
b.

dave =

53 – 33 98
=
= 49 g/cm
5–3
2

f (x) = x 3
(3 + h)3 – 33
h
h →0

d = lim

27 + 27h + 9h 2 + h3 – 27
h
h→0

= lim

h(27 + 9h + h 2 )
= 27 g/cm
h
h→0

= lim

Section 2.1

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20. MR = lim

h→0

R (c + h ) – R (c )
h

[0.4(c + h) – 0.001(c + h)2 ] – (0.4c – 0.001c 2 )
h

h→0

= lim

0.4c + 0.4h – 0.001c 2 – 0.002ch – 0.001h 2 – 0.4c + 0.001c 2
h
h→0
h(0.4 – 0.002c – 0.001h)
= lim
= 0.4 – 0.002c
h
h→0
When n = 10, MR = 0.38; when n = 100, MR = 0.2
= lim

2(1 + h)2 – 2(1) 2
h
h →0

21. a = lim

2 + 4h + 2h 2 – 2
h
h→0
h(4 + 2h)
= lim
=4
h
h→0
= lim

22. r = lim

h→0

p (c + h ) – p (c )
h

[120(c + h)2 – 2(c + h)3 ] – (120c 2 – 2c3 )
h
h →0

= lim

h(240c – 6c 2 + 120h – 6ch – 2h 2 )
h
h →0

= lim

= 240c – 6c 2

When

t = 10, r = 240(10) – 6(10) 2 = 1800
t = 20, r = 240(20) – 6(20)2 = 2400
t = 40, r = 240(40) – 6(40)2 = 0

100 – 800
175

=–
≈ –29.167
24 – 0
6
29,167 gal/hr
700 – 400
≈ −75
At 8 o’clock, r ≈
6 − 10
75,000 gal/hr

23. rave =

24. a. The elevator reached the seventh floor at time
t = 80 . The average velocity is
v avg = (84 − 0) / 80 = 1.05 feet per second
b. The slope of the line is approximately
60 − 12
= 1.2 . The velocity is
55 − 15
approximately 1.2 feet per second.

98

Section 2.1

c. The building averages 84/7=12 feet from
floor to floor. Since the velocity is zero for
two intervals between time 0 and time 85, the
elevator stopped twice. The heights are

approximately 12 and 60. Thus, the elevator
stopped at floors 1 and 5.
25. a.

A tangent line at t = 91 has slope
approximately (63 − 48) /(91 − 61) = 0.5 . The
normal high temperature increases at the rate
of 0.5 degree F per day.

b. A tangent line at t = 191 has approximate
slope (90 − 88) / 30 ≈ 0.067 . The normal
high temperature increases at the rate of
0.067 degree per day.
c. There is a time in January, about January 15,
when the rate of change is zero. There is also
a time in July, about July 15, when the rate of
change is zero.
d. The greatest rate of increase occurs around
day 61, that is, some time in March. The
greatest rate of decrease occurs between day
301 and 331, that is, sometime in November.
26. The slope of the tangent line at t = 1930 is
approximately (8 − 6) /(1945 − 1930) ≈ 0.13 . The
rate of growth in 1930 is approximately 0.13
million, or 130,000, persons per year. In 1990,
the tangent line has approximate slope
(24 − 16) /(20000 − 1980) ≈ 0.4 . Thus, the rate of
growth in 1990 is 0.4 million, or 400,000,
persons per year. The approximate percentage
growth in 1930 is 0.107 / 6 ≈ 0.018 and in 1990 it

is approximately 0.4 / 20 ≈ 0.02 .
27. In both (a) and (b), the tangent line is always
positive. In (a) the tangent line becomes steeper
and steeper as t increases; thus, the velocity is
increasing. In (b) the tangent line becomes flatter
and flatter as t increases; thus, the velocity is
decreasing.

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28.

2.2 Concepts Review

1
f (t ) = t 3 + t
3

f ( c + h ) – f (c )
h
h →0
3
1
⎡ ( c + h ) + (c + h ) ⎤ – 1 c 3 + c
3
⎦ 3

= lim ⎣
h
h→0

current = lim

(

= lim

(

)

) = c2 + 1

h c 2 + ch + 13 h 2 + 1

h
h→0
When t = 3, the current =10

c 2 + 1 = 20
2
c = 19
c = 19 ≈ 4.4
A 20-amp fuse will blow at t = 4.4 s.

1.

f (c + h) – f (c) f (t ) – f (c)
;
h
t –c

2.

f ′(c )

3. continuous; f ( x) = x
4.

=

f ′(1) = lim

h →0

f (1 + h) – f (1)
h

(1 + h)2 – 12
2h + h 2
= lim
= lim
h
h
h→0
h →0
= lim (2 + h ) = 2

h→0

2.

f ′(2) = lim

h →0

f (2 + h) – f (2)
h

[2(2 + h)]2 – [2(2)]2
h
h→0

4
1
30. V = π r 3 , r = t
3
4
1
V = π t3
48
rate =

dy
dx

Problem Set 2.2
1.

29. A = πr 2 , r = 2t
A = 4πt2
4π(3 + h)2 – 4π(3)2
rate = lim
h
h →0
h(24π + 4πh)
= lim
= 24π km2/day
h
h→0

f '( x);

= lim

16h + 4h 2
= lim (16 + 4h) = 16
h
h→0
h →0

= lim

1
(3 + h)3 − 33 27
π lim
= π
h

48 h→0
48

3.

9
π inch 3 / sec
16

m tan = 7

b. m tan = 0

c.

m tan = –1

d. m tan = 17. 92

32. y = f ( x) = sin x sin 2 x

h →0

5h + h 2
= lim (5 + h) = 5
h
h→0
h →0

= lim

4.

f ′(4) = lim

h →0

2

a.

m tan = –1.125 b. m tan ≈ –1.0315

c.

m tan = 0

d. m tan ≈ 1.1891

33. s = f (t ) = t + t cos 2 t
At t = 3, v ≈ 2.818
(t + 1)3
t+2
At t = 1.6, v ≈ 4.277

34. s = f (t ) =

Instructor’s Resource Manual

f (3 + h) – f (3)

h

[(3 + h)2 – (3 + h)] – (32 – 3)
= lim
h
h→0

31. y = f ( x) = x 3 – 2 x 2 + 1
a.

f ′(3) = lim

= lim

h →0

1
3+ h

f (4 + h) – f (4)
h

1
– 4–1

h

= lim

h →0

3–(3+ h )
3(3+ h )

h

–1
h →0 3(3 + h)

= lim

1
=–
9
s ( x + h) – s ( x )
h
h →0
[2( x + h) + 1] – (2 x + 1)
= lim
h
h →0
2h
= lim
=2
h →0 h

5. s ′( x) = lim

Section 2.2

99

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6.

f ( x + h) – f ( x )
h
[α ( x + h) + β ] – (α x + β )
= lim
h
h →0
αh
= lim
=α
h →0 h
f ′( x) = lim

12. g ′( x) = lim

h →0

h →0

[( x + h)4 + ( x + h) 2 ] – ( x 4 + x 2 )
h
h →0

= lim

4hx3 + 6h 2 x 2 + 4h3 x + h 4 + 2hx + h 2
h
h →0

= lim

= lim (4 x3 + 6hx 2 + 4h 2 x + h3 + 2 x + h)

r ( x + h) – r ( x )
7. r ′( x) = lim
h
h →0

h →0
3

= 4x + 2x

[3( x + h)2 + 4] – (3 x 2 + 4)
= lim
h
h →0
6 xh + 3h 2
= lim (6 x + 3h) = 6 x
h
h →0
h →0

f ′( x) = lim

h →0

f ( x + h) – f ( x )
h

[( x + h)2 + ( x + h) + 1] – ( x 2 + x + 1)
h
h →0

= lim
= lim

h →0

9.

2 xh + h + h
= lim (2 x + h + 1) = 2 x + 1
h
h →0
2

f ′( x) = lim

h →0

h( x + h) – h( x )
h

h →0
⎡⎛ 2
2 ⎞ 1⎤
= lim ⎢⎜
– ⎟⋅ ⎥
h → 0 ⎣⎝ x + h x ⎠ h ⎦

13. h′( x) = lim

= lim

8.

f ( x + h) – f ( x )
h

[a( x + h) 2 + b( x + h) + c] – (ax 2 + bx + c)
h
h →0

= lim

2axh + ah 2 + bh
= lim (2ax + ah + b)
h
h →0
h →0
= 2ax + b

⎡ –2h 1 ⎤

–2
2
= lim ⎢
⋅ ⎥ = lim
=–
h → 0 ⎣ x ( x + h ) h ⎦ h →0 x ( x + h )
x2
S ( x + h) – S ( x )
h
h →0
⎡⎛
1
1 ⎞ 1⎤
= lim ⎢⎜
–
⎟⋅ ⎥
h →0 ⎣⎝ x + h + 1 x + 1 ⎠ h ⎦

14. S ′( x) = lim

⎡
–h
1⎤
= lim ⎢
⋅ ⎥
h→0 ⎣ ( x + 1)( x + h + 1) h ⎦
–1
1
= lim
=−

h→0 ( x + 1)( x + h + 1)
( x + 1) 2

= lim

10.

f ′( x) = lim

h →0

f ( x + h) – f ( x )
h

( x + h) 4 – x 4
h
h →0

= lim

4hx3 + 6h 2 x 2 + 4h3 x + h 4
h
h →0

= lim

= lim (4 x3 + 6hx 2 + 4h 2 x + h3 ) = 4 x3
h →0

11.

F ( x + h) – F ( x )
h
h →0

15. F ′( x) = lim

⎡⎛
6
6 ⎞ 1⎤
–
= lim ⎢⎜
⎟⋅ ⎥
h→0 ⎢⎜⎝ ( x + h) 2 + 1 x 2 + 1 ⎟⎠ h ⎥
⎣
⎦
2
2
⎡ 6( x + 1) – 6( x + 2hx + h 2 + 1) 1 ⎤
= lim ⎢
⋅ ⎥
h ⎥⎦
h→0 ⎢⎣ ( x 2 + 1)( x 2 + 2hx + h 2 + 1)
⎡
–12hx – 6h 2
1⎤
= lim ⎢
⋅ ⎥
h→0 ⎢ ( x 2 + 1)( x 2 + 2hx + h 2 + 1) h ⎥
⎣

⎦
= lim

f ′( x) = lim

h →0

h →0 ( x 2

f ( x + h) – f ( x )
h

[( x + h)3 + 2( x + h)2 + 1] – ( x3 + 2 x 2 + 1)
= lim
h
h →0
3hx 2 + 3h 2 x + h3 + 4hx + 2h 2
h
h →0

= lim

= lim (3x + 3hx + h + 4 x + 2h) = 3 x + 4 x
2

h →0

2

g ( x + h) – g ( x )

h

2

–12 x – 6h
+ 1)( x + 2hx + h + 1)
2

2

=−

12 x
( x + 1)2
2

F ( x + h) – F ( x )
h
h →0
⎡⎛ x + h –1 x –1 ⎞ 1 ⎤
= lim ⎢⎜
–
⎟⋅ ⎥
h →0 ⎣ ⎝ x + h + 1 x + 1 ⎠ h ⎦

16. F ′( x) = lim

⎡ x 2 + hx + h –1 – ( x 2 + hx – h –1) 1 ⎤
= lim ⎢
⋅ ⎥

h ⎦⎥
( x + h + 1)( x + 1)
h →0 ⎢
⎣
⎡
2h
1⎤
2
= lim ⎢
⋅ ⎥=
h→0 ⎣ ( x + h + 1)( x + 1) h ⎦ ( x + 1) 2

100

Section 2.2

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G ( x + h) – G ( x )
h
⎡⎛ 2( x + h) –1 2 x –1 ⎞ 1 ⎤
= lim ⎢⎜
–
⎟⋅
x – 4 ⎠ h ⎥⎦
h→0 ⎣⎝ x + h – 4

17. G ′( x ) = lim

h →0

⎡ 2 x 2 + 2hx − 9 x − 8h + 4 − (2 x 2 + 2hx − 9 x − h + 4) 1 ⎤
⎡
–7h
1⎤
⋅ ⎥ = lim ⎢
⋅ ⎥
h
( x + h − 4)( x − 4)
⎥⎦ h→0 ⎣ ( x + h – 4)( x – 4) h ⎦
–7
7
= lim
=–
h→0 ( x + h – 4)( x – 4)
( x – 4)2
= lim ⎢
h →0 ⎢⎣

G ( x + h) – G ( x )
h
h →0

18. G ′( x ) = lim

⎡⎛

⎡ (2 x + 2h)( x 2 – x ) – 2 x( x 2 + 2 xh + h 2 – x – h) 1 ⎤
2( x + h)
2x ⎞ 1 ⎤
= lim ⎢⎜
⋅ ⎥ = lim ⎢
–
⋅ ⎥
⎟
h ⎦⎥
h→0 ⎢⎝⎜ ( x + h) 2 – ( x + h) x 2 – x ⎠⎟ h ⎥
( x 2 + 2hx + h 2 – x – h)( x 2 – x)
⎣
⎦ h→0 ⎣⎢
⎡
–2h 2 x – 2hx 2
1⎤
= lim ⎢
⋅ ⎥
2
2
2
h→0 ⎣⎢ ( x + 2hx + h – x – h)( x – x ) h ⎦⎥
–2hx – 2 x 2

= lim

h→0 ( x 2

=

+ 2hx + h 2 – x – h)( x 2 – x)

–2 x 2
2

( x – x)

2

19. g ′( x) = lim

h →0

=–

2
( x – 1) 2

g ( x + h) – g ( x )
h

3( x + h) – 3x
h
h→0

= lim

= lim

( 3x + 3h – 3x )( 3x + 3h + 3 x )

h( 3 x + 3h + 3x )

h→0

3h

= lim

h→0 h(

3 x + 3h + 3 x )

20. g ′( x) = lim

h →0

= lim

h →0

3
3x + 3h + 3x

=

3
2 3x

g ( x + h) – g ( x )
h

⎡⎛
1
1 ⎞ 1⎤
–
= lim ⎢⎜
⎟⋅ ⎥
h→0 ⎢⎜⎝ 3( x + h)
3 x ⎟⎠ h ⎦⎥
⎣
⎡ 3x – 3x + 3h 1 ⎤
= lim ⎢
⋅ ⎥
h ⎦⎥
h→0 ⎣⎢
9 x ( x + h)
⎡ ( 3 x – 3 x + 3h )( 3 x + 3 x + 3h ) 1 ⎤
= lim ⎢
⋅ ⎥
h ⎦⎥
h→0 ⎣⎢
9 x( x + h)( 3x + 3x + 3h )

= lim

h→0 h

–3h
9 x( x + h)( 3x + 3x + 3h )

Instructor’s Resource Manual

=

–3
3x ⋅ 2 3x

=–

1
2 x 3x

Section 2.2

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21. H ′( x ) = lim

h →0

H ( x + h) – H ( x )
h

⎡⎛
3
–

= lim ⎢⎜
h→0 ⎣⎝ x + h – 2

⎞ 1⎤
⎟⋅ ⎥
x – 2 ⎠ h⎦
3

⎡3 x – 2 – 3 x + h – 2 1 ⎤
= lim ⎢
⋅ ⎥
h→0 ⎣⎢
( x + h – 2)( x – 2) h ⎦⎥
= lim

h→0

3( x – 2 – x + h – 2)( x – 2 + x + h – 2)

h ( x + h – 2)( x – 2)( x – 2 + x + h – 2)
−3h

= lim

h→0 h[( x – 2)

–3

= lim

h→0 ( x – 2)

=–

x + h – 2 + ( x + h – 2) x – 2]

x + h – 2 + ( x + h – 2) x – 2

3
2( x – 2) x – 2

22. H ′( x) = lim

h →0

=−

3
2( x − 2)3 2

H ( x + h) – H ( x )
h

( x + h) 2 + 4 – x 2 + 4
h
h→0

= lim

⎛ x 2 + 2hx + h 2 + 4 – x 2 + 4 ⎞ ⎛ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞

⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
h →0
2
2
2
⎛
⎞
h ⎜ x + 2hx + h + 4 + x + 4 ⎟
⎝
⎠

= lim

h→0

= lim

2hx + h 2
h ⎛⎜ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞⎟
⎝
⎠
2x + h

h →0

x 2 + 2hx + h 2 + 4 + x 2 + 4
2x
x
=
=
2
2
2 x +4
x +4

23. f ′( x) = lim

t→x

f (t ) – f ( x)
t–x

(t − 3t ) – ( x – 3 x)
t–x
t→x

= lim

2

2

t 2 – x 2 – (3t – 3x)
t–x
t→x

(t – x)(t + x) – 3(t – x)
= lim
t–x
t→x
(t – x)(t + x – 3)
= lim
= lim (t + x – 3)
t–x
t→x
t→x
=2x–3
= lim

24.

f ′( x) = lim

t→x
3

f (t ) – f ( x)
t–x

(t + 5t ) – ( x3 + 5 x)
t–x
t→x

= lim

t 3 – x3 + 5t – 5 x

t–x
t→x

= lim

(t – x)(t 2 + tx + x 2 ) + 5(t – x)
t–x
t→x

= lim

(t – x)(t 2 + tx + x 2 + 5)
t–x
t→x

= lim

= lim (t 2 + tx + x 2 + 5) = 3x 2 + 5
t→x

102

Section 2.2

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25.

f (t ) – f ( x)
t–x
t→x
⎡⎛ t
x ⎞ ⎛ 1 ⎞⎤
= lim ⎢⎜
–
⎟⎜
⎟⎥
t → x ⎣⎝ t – 5 x – 5 ⎠ ⎝ t – x ⎠ ⎦
f ′( x) = lim

38. The slope of the tangent line is always −1 .

tx – 5t – tx + 5 x
t → x (t – 5)( x – 5)(t – x)

= lim

= lim

t → x (t

=−

26.

–5(t – x)

–5
= lim
– 5)( x – 5)(t – x) t → x (t – 5)( x – 5)

5

( x − 5)

2

39. The derivative is positive until x = 0 , then
becomes negative.

f (t ) – f ( x)
t–x
t→x
⎡⎛ t + 3 x + 3 ⎞ ⎛ 1 ⎞ ⎤
= lim ⎢⎜
–
⎟⎜
⎟
x ⎠ ⎝ t – x ⎠ ⎥⎦
t → x ⎣⎝ t
f ′( x) = lim

3x – 3t
–3
3
= lim
=–

t → x xt (t – x ) t → x xt
x2

= lim

27. f (x) = 2 x 3 at x = 5

40. The derivative is negative until x = 1 , then
becomes positive.

28. f (x) = x 2 + 2 x at x = 3
29. f (x) = x 2 at x = 2
30. f (x) = x 3 + x at x = 3
31. f (x) = x 2 at x
32. f (x) = x 3 at x
33. f (t ) =

2
at t
t

41. The derivative is −1 until x = 1 . To the right of
x = 1 , the derivative is 1. The derivative is
undefined at x = 1 .

34. f(y) = sin y at y
35. f(x) = cos x at x
36. f(t) = tan t at t
37. The slope of the tangent line is always 2.
42. The derivative is −2 to the left of x = −1 ; from

−1 to 1, the derivative is 2, etc. The derivative is
not defined at x = −1, 1, 3 .

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

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

53.

and 0 on ( 3, 4 ) . The derivative is undefined at
x = −2, − 1, 0, 1, 2, 3 .

1
1
Δy x +Δx +1 – x +1
=
Δx
Δx
⎛ x + 1 – ( x + Δx + 1) ⎞ ⎛ 1 ⎞
=⎜

⎟⎜ ⎟
⎝ ( x + Δx + 1)( x + 1) ⎠ ⎝ Δx ⎠

=

– Δx
( x + Δx + 1)( x + 1)Δx

=–

1
( x + Δx + 1)( x + 1)

⎡
⎤
dy
1
1
= lim −
=−
dx Δx →0 ⎢⎣ ( x + Δx + 1)( x + 1) ⎥⎦
( x + 1) 2

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

1
⎛ 1⎞
− ⎜1 + ⎟
x + Δx ⎝ x ⎠

Δx
−Δx
1
1
−
x ( x + Δx )
1
= x + Δx x =
=−
Δx
Δx
x ( x + Δx )

Δy
54.
=
Δx

1+

dy
1
1
= lim −
=− 2
dx Δx →0 x ( x + Δx )
x

55.
45. Δy = [3(1.5) + 2] – [3(1) + 2] = 1.5

46. Δy = [3(0.1) 2 + 2(0.1) + 1] – [3(0.0) 2 + 2(0.0) + 1]
= 0.23

x 2 + xΔx − x + x + Δx − 1 − ⎡ x 2 + xΔx − x + x − Δx − 1⎤ 1
⎣
⎦×
2
Δx
x + x Δx + x + x + Δ x + 1
2Δx
1
2
= 2
×
= 2
Δ
x
x + xΔx + x + x + Δx + 1
x + x Δx + x + x + Δ x + 1
2
2
2
dy
= lim
=
=
dx Δx →0 x 2 + xΔx + x + x + Δx + 1 x 2 + 2 x + 1 ( x + 1)2

47. Δy = 1/1.2 – 1/1 = – 0.1667

=

48. Δy = 2/(0.1+1) – 2/(0+1) = – 0.1818
49. Δy =

3
3
–
≈ 0.0081
2.31 + 1 2.34 + 1

50. Δy = cos[2(0.573)] – cos[2(0.571)] ≈ –0.0036
51.

52.

Δy ( x + Δx) 2 – x 2 2 xΔx + (Δx) 2
=
=
= 2 x + Δx
Δx
Δx
Δx
dy
= lim (2 x + Δx) = 2 x
dx Δx →0

56.

Δy

=
Δx

( x + Δx ) 2 − 1 − x 2 − 1
x + Δx
Δx

x

(

)

3 x 2 Δx + 3x(Δx)2 – 6 xΔx – 3(Δx) 2 + Δx3
=
Δx

⎡ x ( x + Δx )2 − x − ( x + Δx ) x 2 − 1 ⎤
⎥× 1
=⎢
⎢
⎥ Δx
x ( x + Δx )
⎣
⎦
2
2
3
⎡ x x + 2 xΔx + Δx − x − x + x 2 Δx − x − Δx
=⎢

⎢
x 2 + x Δx
⎣⎢

= 3x 2 + 3xΔx – 6 x – 3Δx + (Δx)2

=

Δy [( x + Δx)3 – 3( x + Δx) 2 ] – ( x3 – 3 x 2 )
=
Δx
Δx

dy
= lim (3 x 2 + 3 xΔx – 6 x – 3Δx + (Δx)2 )
dx Δx→0
= 3x2 – 6 x

104

x + Δx − 1 x − 1
−
Δy x + Δx + 1 x + 1
=
Δx
Δx
( x + 1)( x + Δx − 1) − ( x − 1)( x + Δx + 1) 1
=
×
Δx

( x + Δx + 1)( x + 1)

Section 2.2

(

( ))

x 2 Δx + x ( Δx ) + Δx
2

x + x Δx
2

×

(

) ⎤⎥ × 1

⎥ Δx
⎦⎥

1
x 2 + x Δx + 1
= 2
Δx
x + x Δx

dy

x 2 + xΔx + 1 x 2 + 1
= lim
= 2
Δ
x
→
0
dx
x 2 + x Δx
x

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57.

1
f ′(0) ≈ – ; f ′(2) ≈ 1
2
2
f ′(5) ≈ ; f ′(7) ≈ –3
3

58. g ′(–1) ≈ 2; g ′(1) ≈ 0

g ′(4) ≈ –2; g ′(6) ≈ –

1
3

63. The derivative is 0 at approximately t = 15 and
t = 201 . The greatest rate of increase occurs at
about t = 61 and it is about 0.5 degree F per day.
The greatest rate of decrease occurs at about
t = 320 and it is about 0.5 degree F per day. The
derivative is positive on (15,201) and negative on
(0,15) and (201,365).

59.

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.

60.

61. a.

5
3
f (2) ≈ ; f ′(2) ≈

2
2
f (0.5) ≈ 1.8; f ′(0.5) ≈ –0.6

b.

2.9 − 1.9
= 0.5
2.5 − 0.5

c.

x=5

d. x = 3, 5
e.

x = 1, 3, 5

f.

x=0

g.

x ≈ −0.7,

3
and 5 < x < 7
2

62. The derivative fails to exist at the corners of the
graph; that is, at t = 10, 15, 55, 60, 80 . The
derivative exists at all other points on the interval
(0,85) .

Instructor’s Resource Manual

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
positive (negative) when the slope of the tangent
line of the short-dash function is positive
(negative). Also, the long-dash function is
positive (negative) when the slope of the tangent
line of the solid function is positive (negative).
Thus, the short-dash function is f, the solid
function is f ' = g , and the dash function is g ' .
66. Note that since x = 0 + x, f(x) = f(0 + x) = f(0)f(x),
hence f(0) = 1.
f ( a + h) – f ( a )
f ′(a ) = lim
h
h →0
f ( a ) f ( h) – f ( a )
= lim
h

h→0
f ( h) – 1
f (h) – f (0)
= f (a ) lim
= f (a) lim
h
h
h →0
h →0
= f (a ) f ′(0)
f ′ ( a) exists since f ′ (0 ) exists.

Section 2.2

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67. If f is differentiable everywhere, then it is
continuous everywhere, so
lim − f ( x ) = lim− ( mx + b ) = 2 m + b = f (2) = 4
x→2

b. If f is an even function,
f (t ) − f ( x0 )
f ′(– x0 ) = lim
. Let u = –t, as
t + x0

t →− x0

x→ 2

f (−u ) − f ( x0 )
−u + x0
u → x0

and b = 4 – 2m.
For f to be differentiable everywhere,
f ( x) − f (2)
f ′(2) = lim
must exist.
x−2
x→2
f ( x) − f (2)
x2 − 4
= lim
= lim ( x + 2) = 4
x−2
x → 2+
x → 2+ x − 2
x → 2+
f ( x) − f (2)
mx + b − 4
lim
= lim
−
−
x

x−2
−
2
x→2
x→2

above, then f ′(− x0 ) = lim

f (u ) − f ( x0 )
f (u ) − f ( x0 )
= − lim
u − x0
u → x0 −(u − x0 )
u → x0
= − f ′ (x 0 ) = −m.

= lim

lim

mx + 4 − 2m − 4
m( x − 2)
= lim
=m
−
x−2
x−2
x →2
Thus m = 4 and b = 4 – 2(4) = –4

= lim

x → 2−

68.

69.

f ( x + h) – f ( x ) + f ( x ) – f ( x – h )
2h
h →0
⎡ f ( x + h ) – f ( x ) f ( x – h) – f ( x ) ⎤
= lim ⎢
+
⎥
2h
–2h
h→0 ⎣
⎦
1
f ( x + h) – f ( x ) 1
f [ x + (– h)] – f ( x)
= lim
+ lim
h
2 h →0
2 – h →0
–h
1
1

= f ′( x) + f ′( x ) = f ′( x).
2
2
For the converse, let f (x) = x . Then
h – –h
h–h
f s (0) = lim
= lim
=0
2h
h→0
h →0 2h
but f ′ (0) does not exist.
f s ( x) = lim

70. Say f(–x) = –f(x). Then
f (– x + h) – f (– x)
f ′(– x) = lim
h
h →0
– f ( x – h) + f ( x )
f ( x – h) – f ( x )
= lim
= – lim
h
h
h→0
h →0
f [ x + (– h)] − f ( x)
= lim

= f ′( x) so f ′ ( x ) is
–h
– h →0
an even function if f(x) is an odd function.
Say f(–x) = f(x). Then
f (– x + h) – f (– x)
f ′(– x) = lim
h
h →0
f ( x – h) – f ( x )
= lim
h
h→0
f [ x + (– h)] – f ( x)
= – lim
= – f ′( x) so f ′ (x)
–h
– h→0
is an odd function if f(x) is an even function.
71.

f (t ) − f ( x0 )
, so
t − x0
0
f (t ) − f (− x0 )
f ′(− x0 ) = lim
t − (− x0 )
t →− x
0

f ′( x0 ) = lim
t→x

f (t ) − f (− x0 )
= lim
t + x0
t →− x 0

a. If f is an odd function,
f (t ) − [− f (− x0 )]
f ′(− x0 ) = lim
t + x0
t →− x0
f (t ) + f (− x0 )
.
t + x0
Let u = –t. As t → − x0 , u → x 0 and so
f (−u ) + f ( x0 )
f ′(− x0 ) = lim
−u + x0
u → x0

a.

0< x<

8 ⎛ 8⎞
; ⎜ 0, ⎟
3 ⎝ 3⎠

b.

0≤ x≤

8 ⎡ 8⎤
; 0,
3 ⎢⎣ 3 ⎥⎦

c.

A function f(x) decreases as x increases when
f ′ ( x ) < 0.

a.

π < x < 6.8

c.

A function f(x) increases as x increases when
f ′ ( x ) > 0.

= lim

t →− x0

72.

− f (u ) + f ( x0 )
−[ f (u ) − f ( x0 )]

= lim
u
x
−
(
−
)
−(u − x0 )
u → x0
u → x0
0

= lim

f (u ) − f ( x0 )
= f ′( x0 ) = m.
u − x0
u → x0

= lim

106

Section 2.2

b. π < x < 6.8

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2.3 Concepts Review

13. Dx ( x 4 + x3 + x 2 + x + 1)

1. the derivative of the second; second;
f (x) g′ (x ) + g(x) f ′( x)
2. denominator; denominator; square of the
g ( x) f ′( x) – f ( x) g ′( x)
denominator;
g 2 ( x)
3. nx

n– 1

h ; nx

= 3Dx ( x 4 ) – 2 Dx ( x3 ) – 5Dx ( x 2 )

= 12 x3 – 6 x 2 –10 x + π

15. Dx (πx 7 – 2 x5 – 5 x –2 )
= πDx ( x7 ) – 2 Dx ( x5 ) – 5 Dx ( x –2 )

1. Dx (2 x ) = 2 Dx ( x ) = 2 ⋅ 2 x = 4 x
2

= π(7 x6 ) – 2(5 x 4 ) – 5(–2 x –3 )

= 7 πx 6 –10 x 4 + 10 x –3

2. Dx (3x3 ) = 3Dx ( x3 ) = 3 ⋅ 3x 2 = 9 x 2

16. Dx ( x12 + 5 x −2 − πx −10 )

3. Dx (πx ) = πDx ( x) = π ⋅1 = π
4. Dx (πx ) = πDx ( x ) = π ⋅ 3 x = 3πx
5.

14. Dx (3x 4 – 2 x3 – 5 x 2 + πx + π2 )

= 3(4 x3 ) – 2(3 x 2 ) – 5(2 x) + π(1) + 0

Problem Set 2.3

3

= 4 x3 + 3 x 2 + 2 x + 1

+ πDx ( x) + Dx (π2 )

n –1

4. kL(f); L(f) + L(g); Dx

2

= Dx ( x 4 ) + Dx ( x3 ) + Dx ( x 2 ) + Dx ( x) + Dx (1)

3

2

2

Dx (2 x –2 ) = 2 Dx ( x –2 ) = 2(–2 x –3 ) = –4 x –3

6. Dx (–3 x –4 ) = –3Dx ( x –4 ) = –3(–4 x –5 ) = 12 x –5
⎛π⎞
7. Dx ⎜ ⎟ = πDx ( x –1 ) = π(–1x –2 ) = – πx –2
⎝x⎠
π
=– 2
x

⎛α ⎞
8. Dx ⎜ ⎟ = α Dx ( x –3 ) = α (–3x –4 ) = –3α x –4
⎝ x3 ⎠
3α
=–
x4
⎛ 100 ⎞
9. Dx ⎜
= 100 Dx ( x –5 ) = 100(–5 x –6 )
5 ⎟
⎝ x ⎠
500
= –500 x –6 = –

x6
⎛ 3α ⎞ 3α
3α
10. Dx ⎜
Dx ( x –5 ) =
=
(–5 x –6 )
5⎟
4
⎝ 4x ⎠ 4
15α –6
15α
=–
x =–
4
4 x6

11. Dx ( x 2 + 2 x) = Dx ( x 2 ) + 2 Dx ( x ) = 2 x + 2

= Dx ( x12 ) + 5Dx ( x −2 ) − πDx ( x −10 )
= 12 x11 + 5(−2 x −3 ) − π(−10 x −11 )

= 12 x11 − 10 x −3 + 10πx −11
⎛ 3
⎞
17. Dx ⎜ + x –4 ⎟ = 3Dx ( x –3 ) + Dx ( x –4 )
3
⎝x
⎠
9

= 3(–3 x –4 ) + (–4 x –5 ) = –
– 4 x –5
x4

18. Dx (2 x –6 + x –1 ) = 2 Dx ( x –6 ) + Dx ( x –1 )
= 2(–6 x –7 ) + (–1x –2 ) = –12 x –7 – x –2
⎛2 1 ⎞
19. Dx ⎜ –
= 2 Dx ( x –1 ) – Dx ( x –2 )
2⎟
x
x ⎠
⎝
2
2
= 2(–1x –2 ) – (–2 x –3 ) = –
+
2
x
x3
⎛ 3
1 ⎞
–3
–4
20. Dx ⎜ –
⎟ = 3 Dx ( x ) – Dx ( x )
3
x4 ⎠
⎝x
9

4
= 3(–3 x –4 ) – (–4 x –5 ) = –
+
4
x
x5
⎛ 1
⎞ 1
21. Dx ⎜ + 2 x ⎟ = Dx ( x –1 ) + 2 Dx ( x)
⎝ 2x
⎠ 2
1
1
= (–1x –2 ) + 2(1) = –
+2
2
2 x2

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

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⎛ 2 2⎞ 2
⎛2⎞
22. Dx ⎜ – ⎟ = Dx ( x –1 ) – Dx ⎜ ⎟
⎝ 3x 3 ⎠ 3
⎝3⎠
2
2
= (–1x –2 ) – 0 = –
3
3x 2

23. Dx [ x( x 2 + 1)] = x Dx ( x 2 + 1) + ( x 2 + 1) Dx ( x)
= x(2 x) + ( x + 1)(1) = 3 x + 1
2

2

26. Dx [(–3 x + 2)2 ]
= (–3 x + 2) Dx (–3 x + 2) + (–3 x + 2) Dx (–3x + 2)
= (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 12
27. Dx [( x 2 + 2)( x3 + 1)]
= ( x 2 + 2) Dx ( x3 + 1) + ( x3 + 1) Dx ( x 2 + 2)
= ( x 2 + 2)(3x 2 ) + ( x3 + 1)(2 x)

24. Dx [3 x( x3 –1)] = 3 x Dx ( x3 –1) + ( x3 –1) Dx (3 x)
= 3x(3 x 2 ) + ( x3 –1)(3) = 12 x3 – 3

25. Dx [(2 x + 1) ]
= (2 x + 1) Dx (2 x + 1) + (2 x + 1) Dx (2 x + 1)

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

= 3x 4 + 6 x 2 + 2 x 4 + 2 x
= 5x4 + 6 x2 + 2 x

28. Dx [( x 4 –1)( x 2 + 1)]
= ( x 4 –1) Dx ( x 2 + 1) + ( x 2 + 1) Dx ( x 4 –1)
= ( x 4 –1)(2 x) + ( x 2 + 1)(4 x3 )

= 2 x5 – 2 x + 4 x5 + 4 x3 = 6 x 5 + 4 x3 – 2 x

29. Dx [( x 2 + 17)( x3 – 3 x + 1)]
= ( x 2 + 17) Dx ( x3 – 3 x + 1) + ( x3 – 3x + 1) Dx ( x 2 + 17)
= ( x 2 + 17)(3 x 2 – 3) + ( x3 – 3x + 1)(2 x)
= 3x 4 + 48 x 2 – 51 + 2 x 4 – 6 x 2 + 2 x

= 5 x 4 + 42 x 2 + 2 x – 51
30. Dx [( x 4 + 2 x)( x3 + 2 x 2 + 1)] = ( x 4 + 2 x) Dx ( x3 + 2 x 2 + 1) + ( x3 + 2 x 2 + 1) Dx ( x 4 + 2 x)
= ( x 4 + 2 x)(3 x 2 + 4 x) + ( x3 + 2 x 2 + 1)(4 x3 + 2)
= 7 x6 + 12 x5 + 12 x3 + 12 x 2 + 2

31. Dx [(5 x 2 – 7)(3x 2 – 2 x + 1)] = (5 x 2 – 7) Dx (3x 2 – 2 x + 1) + (3x 2 – 2 x + 1) Dx (5 x 2 – 7)
= (5 x 2 – 7)(6 x – 2) + (3 x 2 – 2 x + 1)(10 x)

= 60 x3 – 30 x 2 – 32 x + 14
32. Dx [(3 x 2 + 2 x)( x 4 – 3 x + 1)] = (3 x 2 + 2 x) Dx ( x 4 – 3 x + 1) + ( x 4 – 3 x + 1) Dx (3x 2 + 2 x)
= (3 x 2 + 2 x)(4 x3 – 3) + ( x 4 – 3 x + 1)(6 x + 2)

= 18 x5 + 10 x 4 – 27 x 2 – 6 x + 2

⎛ 1 ⎞ (3x 2 + 1) Dx (1) – (1) Dx (3 x 2 + 1)
33. Dx ⎜
⎟=
(3 x 2 + 1)2
⎝ 3x2 + 1 ⎠
=

(3 x 2 + 1)(0) – (6 x)
(3 x + 1)
2

2

=–

6x
(3x + 1) 2
2

⎛ 2 ⎞ (5 x 2 –1) Dx (2) – (2) Dx (5 x 2 –1)
34. Dx ⎜
⎟=
(5 x 2 –1) 2
⎝ 5 x 2 –1 ⎠
=

(5 x 2 –1)(0) – 2(10 x)
2

(5 x –1)

108

Section 2.3

2

=–

20 x
(5 x 2 –1)2

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⎛
⎞ (4 x 2 – 3x + 9) Dx (1) – (1) Dx (4 x 2 – 3 x + 9)
1
35. Dx ⎜
⎟=
(4 x 2 – 3x + 9)2
⎝ 4 x 2 – 3x + 9 ⎠
=

=

(4 x 2 – 3x + 9)(0) – (8 x – 3)

(4 x – 3 x + 9)
−8 x + 3
2

=–

2

8x − 3
(4 x – 3x + 9)2
2

(4 x 2 – 3x + 9)2

⎛
⎞ (2 x3 – 3 x) Dx (4) – (4) Dx (2 x3 – 3 x)
4
36. Dx ⎜
⎟ =
(2 x3 – 3 x)2
⎝ 2 x3 – 3x ⎠
=

(2 x3 – 3 x)(0) – 4(6 x 2 – 3)
(2 x3 – 3 x)2

=

–24 x 2 + 12
(2 x3 – 3x) 2

⎛ x –1 ⎞ ( x + 1) Dx ( x –1) – ( x –1) Dx ( x + 1)
37. Dx ⎜
⎟=
⎝ x +1⎠
( x + 1)2
=

( x + 1)(1) – ( x –1)(1)
( x + 1)

2

=

2
( x + 1)2

⎛ 2 x –1 ⎞ ( x –1) Dx (2 x –1) – (2 x –1) Dx ( x –1)
38. Dx ⎜
⎟=
⎝ x –1 ⎠
( x –1) 2
=

( x –1)(2) – (2 x –1)(1)
( x –1)

2

=–

1
( x –1) 2

⎛ 2 x 2 – 1 ⎞ (3 x + 5) Dx (2 x 2 –1) – (2 x 2 –1) Dx (3 x + 5)
39. Dx ⎜
⎟ =
⎜ 3x + 5 ⎟
(3 x + 5)2
⎝
⎠
=
=

(3 x + 5)(4 x) – (2 x 2 – 1)(3)
(3x + 5) 2
6 x 2 + 20 x + 3
(3x + 5)2

⎛ 5 x – 4 ⎞ (3 x 2 + 1) Dx (5 x – 4) – (5 x – 4) Dx (3x 2 + 1)
40. Dx ⎜
⎟=
(3 x 2 + 1) 2
⎝ 3x2 + 1 ⎠
=
=

(3 x 2 + 1)(5) – (5 x – 4)(6 x)
(3x 2 + 1)2

−15 x 2 + 24 x + 5
(3x 2 + 1)2

⎛ 2 x 2 – 3 x + 1 ⎞ (2 x + 1) Dx (2 x 2 – 3x + 1) – (2 x 2 – 3x + 1) Dx (2 x + 1)
41. Dx ⎜
⎟ =
⎜ 2x +1 ⎟
(2 x + 1)2
⎝
⎠
=
=

(2 x + 1)(4 x – 3) – (2 x 2 – 3 x + 1)(2)
(2 x + 1)2
4 x2 + 4 x – 5
(2 x + 1) 2

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⎛ 5 x 2 + 2 x – 6 ⎞ (3 x – 1) Dx (5 x 2 + 2 x – 6) – (5 x 2 + 2 x – 6) Dx (3 x – 1)
42. Dx ⎜

⎟=
⎜
3 x – 1 ⎟⎠
(3 x – 1)2
⎝
=
=

(3 x – 1)(10 x + 2) – (5 x 2 + 2 x – 6)(3)
(3x – 1)2
15 x 2 – 10 x + 16
(3x – 1)2

⎛ x 2 – x + 1 ⎞ ( x 2 + 1) Dx ( x 2 – x + 1) – ( x 2 – x + 1) Dx ( x 2 + 1)
43. Dx ⎜
⎟ =
⎜ x2 + 1 ⎟
( x 2 + 1)2
⎝
⎠
=
=

( x 2 + 1)(2 x – 1) – ( x 2 – x + 1)(2 x)
( x 2 + 1)2
x2 – 1
( x 2 + 1)2

⎛ x 2 – 2 x + 5 ⎞ ( x 2 + 2 x – 3) Dx ( x 2 – 2 x + 5) – ( x 2 – 2 x + 5) Dx ( x 2 + 2 x – 3)
44. Dx ⎜

⎟=
⎜ x2 + 2 x – 3 ⎟
( x 2 + 2 x – 3) 2
⎝
⎠
=
=

45. a.

( x 2 + 2 x – 3)(2 x – 2) – ( x 2 – 2 x + 5)(2 x + 2)
( x 2 + 2 x – 3) 2
4 x 2 – 16 x – 4
( x 2 + 2 x – 3) 2
( f ⋅ g )′(0) = f (0) g ′(0) + g (0) f ′(0)
= 4(5) + (–3)(–1) = 23

b.

( f + g )′(0) = f ′(0) + g ′(0) = –1 + 5 = 4

c.

( f g )′(0) =

=

g 2 (0)

–3(–1) – 4(5)

(–3)

46. a.

g (0) f ′(0) – f (0) g ′(0)

2

=–

17
9

( f – g )′(3) = f ′(3) – g ′(3) = 2 – (–10) = 12

b.

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

c.

( g f )′(3) =

f (3) g ′(3) – g (3) f ′(3)
2

f (3)

=

7(–10) – 6(2)
(7)

2

=–

82
49

47. Dx [ f ( x )]2 = Dx [ f ( x ) f ( x)]
= f ( x) Dx [ f ( x)] + f ( x) Dx [ f ( x)]
= 2 ⋅ f ( x ) ⋅ Dx f ( x )
48. Dx [ f ( x) g ( x)h( x)] = f ( x) Dx [ g ( x)h( x)] + g ( x)h( x) Dx f ( x)
= f ( x)[ g ( x) Dx h( x) + h( x) Dx g ( x)] + g ( x)h( x) Dx f ( x)
= f ( x) g ( x) Dx h( x ) + f ( x)h( x) Dx g ( x) + g ( x)h( x) Dx f ( x)

110

Section 2.3

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54. Proof #1:

49. Dx ( x 2 – 2 x + 2) = 2 x – 2

At x = 1: m tan = 2 (1) – 2 = 0
Tangent line: y = 1

Dx [ f ( x) − g ( x) ] = Dx [ f ( x) + (−1) g ( x) ]
= Dx [ f ( x) ] + Dx [ (−1) g ( x) ]

⎛ 1 ⎞ ( x + 4) Dx (1) – (1) Dx ( x + 4)
50. Dx ⎜
⎟=
( x 2 + 4)2
⎝ x2 + 4 ⎠
2

=

( x 2 + 4)(0) – (2 x)
( x 2 + 4)2

At x = 1: mtan = −

=–

2x
( x 2 + 4) 2

2(1)

=–

2

25

(1 + 4)
1
2
Tangent line: y – = – ( x –1)
5
25
2
7
y = – x+
25
25
2

2

51. Dx ( x3 – x 2 ) = 3 x 2 – 2 x
The tangent line is horizontal when m tan = 0:
mtan = 3x 2 – 2 x = 0
x(3 x − 2) = 0
2
x = 0 and x =
3
4 ⎞
⎛2
(0, 0) and ⎜ , – ⎟
⎝ 3 27 ⎠
⎛1
⎞

52. Dx ⎜ x3 + x 2 – x ⎟ = x 2 + 2 x –1
3
⎝
⎠
mtan = x + 2 x –1 = 1
2

x2 + 2 x – 2 = 0
–2 ± 4 – 4(1)(–2) –2 ± 12
x=
=
2
2
= –1 – 3, –1 + 3
x = –1 ± 3
5
5
⎛
⎞ ⎛
⎞
⎜ −1 + 3, − 3 ⎟ , ⎜ −1 − 3, + 3 ⎟
3
3
⎝
⎠ ⎝
⎠

53.

= Dx f ( x) − Dx g ( x)

2

y = 100 / x5 = 100 x −5
y ' = −500 x −6

Set y ' equal to −1 , the negative reciprocal of
the slope of the line y = x . Solving for x gives
x = ±5001/ 6 ≈ ±2.817
y = ±100(500)−5 / 6 ≈ ±0.563

Proof #2:
Let F ( x) = f ( x) − g ( x) . Then
F '( x) = lim

[ f ( x + h) − g ( x + h) ] − [ f ( x ) − g ( x ) ]

h →0

h

⎡ f ( x + h) − f ( x ) g ( x + h ) − g ( x ) ⎤
= lim ⎢
−
⎥
h →0 ⎣
h
h
⎦
= f '( x) − g '( x)

55. a.

Dt (–16t 2 + 40t + 100) = –32t + 40
v = –32(2) + 40 = –24 ft/s

b. v = –32t + 40 = 0
t=5 s
4
56. Dt (4.5t 2 + 2t ) = 9t + 2
9t + 2 = 30
28
t=
s
9
57. mtan = Dx (4 x – x 2 ) = 4 – 2 x
The line through (2,5) and (x 0 , y0 ) has slope
y0 − 5
.
x0 − 2
4 – 2 x0 =

4 x0 – x0 2 – 5
x0 – 2

–2 x02 + 8 x0 – 8 = – x0 2 + 4 x0 – 5
x0 2 – 4 x0 + 3 = 0
( x0 – 3)( x0 –1) = 0
x 0 = 1, x0 = 3

At x 0 = 1: y0 = 4(1) – (1)2 = 3
mtan = 4 – 2(1) = 2
Tangent line: y – 3 = 2(x – 1); y = 2x + 1
At x0 = 3 : y0 = 4(3) – (3)2 = 3
mtan = 4 – 2(3) = –2
Tangent line: y – 3 = –2(x – 3); y = –2x + 9

The points are (2.817,0.563) and
(−2.817,−0.563) .

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58. Dx ( x 2 ) = 2 x
The line through (4, 15) and ( x0 , y0 ) has slope

61. The watermelon has volume

of the rind is

y0 − 15
. If (x 0 , y0 ) is on the curve y = x 2 , then
x0 − 4

mtan = 2 x0 =

4 3
πr ; the volume
3

3

V=

x02 –15
.
x0 – 4

r ⎞
4 3 4 ⎛
271 3
πr – π ⎜ r – ⎟ =
πr .
3
3 ⎝ 10 ⎠
750

At the end of the fifth week r = 10, so
271 2 271
542π
DrV =
πr =
π(10)2 =
≈ 340 cm3

250
250
5
per cm of radius growth. Since the radius is
growing 2 cm per week, the volume of the rind is
542π
(2) ≈ 681 cm3 per
growing at the rate of
5
week.

2 x0 2 – 8 x0 = x02 –15
x0 2 – 8 x0 + 15 = 0
( x0 – 3)( x0 – 5) = 0

At x0 = 3 : y0 = (3)2 = 9
She should shut off the engines at (3, 9). (At
x 0 = 5 she would not go to (4, 15) since she is
moving left to right.)

2.4 Concepts Review
59. Dx (7 – x ) = –2 x
The line through (4, 0) and ( x0 , y0 ) has
2

slope

1.

y0 − 0

. If the fly is at ( x0 , y0 ) when the
x0 − 4

spider sees it, then mtan = –2 x0 =

2

7 – x0 – 0
.
x0 – 4

–2 x02 + 8 x0 = 7 – x02
x 02 – 8x 0 + 7 = 0
( x0 – 7)( x0 –1) = 0
At x0 = 1: y0 = 6
d = (4 – 1)2 + (0 – 6) 2 = 9 + 36 = 45 = 3 5
≈ 6. 7
They are 6.7 units apart when they see each
other.
1
⎛ 1⎞
60. P(a, b) is ⎜ a, ⎟ . Dx y = –
so the slope of
a
⎝
⎠
x2
1
the tangent line at P is – . The tangent line is
a2

1
1
1
y– =–
( x – a ) or y = –
( x – 2a ) which
2
a
a
a2
has x-intercept (2a, 0).
1
1
d (O, P ) = a 2 + , d ( P, A) = (a – 2a )2 +
2
a
a2

= a2 +

1

= d (O, P ) so AOP is an isosceles
a2
triangle. The height of AOP is a while the base,
1
OA has length 2a, so the area is (2 a)(a) = a2 .
2

112

Section 2.4

sin( x + h) – sin( x)
h

2. 0; 1
3. cos x; –sin x
4. cos

π 1
3 1⎛
π⎞
= ;y–
= ⎜x– ⎟
3 2
2
2⎝
3⎠

Problem Set 2.4
1. Dx(2 sin x + 3 cos x) = 2 Dx(sin x) + 3 Dx(cos x)
= 2 cos x – 3 sin x
2. Dx (sin 2 x) = sin x Dx (sin x ) + sin x Dx (sin x)
= sin x cos x + sin x cos x = 2 sin x cos x = sin 2x
3. Dx (sin 2 x + cos 2 x) = Dx (1) = 0
4. Dx (1 – cos 2 x) = Dx (sin 2 x)
= sin x Dx (sin x) + sin x Dx (sin x)
= sin x cos x + sin x cos x
= 2 sin x cos x = sin 2x

⎛ 1 ⎞
5. Dx (sec x) = Dx ⎜
⎟
⎝ cos x ⎠
cos x Dx (1) – (1) Dx (cos x )
=
cos 2 x
sin x
1 sin x
=
=
⋅
= sec x tan x
2
x cos x
cos
cos x

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⎛ 1 ⎞
6. Dx (csc x) = Dx ⎜
⎟
⎝ sin x ⎠
sin x Dx (1) − (1) Dx (sin x )
=

sin 2 x
– cos x
–1 cos x
=
=
⋅
= – csc x cot x
2
sin
x sin x
sin x
⎛ sin x ⎞
7. Dx (tan x) = Dx ⎜
⎟
⎝ cos x ⎠
cos x Dx (sin x) − sin x Dx (cos x)
=
cos 2 x

=

cos 2 x + sin 2 x
cos2 x

1

=

cos 2 x

=

− sin x – cos x
2

2

=

sin x

=

cos x(cos x – sin x) – (– sin 2 x – sin x cos x)
cos 2 x + sin 2 x
cos2 x

=

cos 2 x

cos 2 x
1
cos x
− cos x −
−
sin x
sin x sin 2 x

12. Dx ( sin x tan x ) = sin xDx [ tan x ] + tan xDx [sin x ]

2

2

cos 2 x
1

tan 2 x
⎛
sin 2 x sin x
1 ⎞ ⎛ sin 2 x ⎞
–
–
= ⎜ sin x –
⎟ ÷⎜
⎟
2
cos x cos x cos x ⎠ ⎝ cos 2 x ⎠
⎝

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

⎛ sin x + cos x ⎞
9. Dx ⎜
⎟
cos x
⎝
⎠
cos x Dx (sin x + cos x) − (sin x + cos x) Dx (cos x)
=

cos 2 x

=

tan x(cos x – sin x) – sec2 x(sin x + cos x)

11. Dx ( sin x cos x ) = sin xDx [ cos x ] + cos xDx [sin x ]

–(sin x + cos x)

sin x
1
=–
= – csc2 x
2
sin x

=

=

= sec2 x

2

⎛ sin x + cos x ⎞
Dx ⎜
⎟
tan x
⎝

⎠
tan x Dx (sin x + cos x) − (sin x + cos x) Dx (tan x )
=
tan 2 x

⎛
sin 2 x sin x
1 ⎞⎛ cos 2 x ⎞
= ⎜ sin x −
−
−
⎟⎜
⎟
2
cos x cos x cos x ⎠⎝ sin 2 x ⎠
⎝

⎛ cos x ⎞
8. Dx (cot x) = Dx ⎜
⎟
⎝ sin x ⎠
sin x Dx (cos x) − cos x Dx (sin x)
=
sin 2 x
2

10.

= sec2 x

(

)

= sin x sec2 x + tan x ( cos x )
⎛ 1 ⎞ sin x
= sin x ⎜
( cos x )
⎟+
⎝ cos 2 x ⎠ cos x
= tan x sec x + sin x
⎛ sin x ⎞ xDx ( sin x ) − sin xDx ( x )
13. Dx ⎜
⎟=
x2
⎝ x ⎠
x cos x − sin x
=
x2

⎛ 1 − cos x ⎞ xDx (1 − cos x ) − (1 − cos x ) Dx ( x )
14. Dx ⎜
⎟=
x
x2
⎝
⎠
x sin x + cos x − 1
=
x2

15. Dx ( x 2 cos x) = x 2 Dx (cos x) + cos x Dx ( x 2 )

= − x 2 sin x + 2 x cos x

⎛ x cos x + sin x ⎞
16. Dx ⎜
⎟
x2 + 1
⎝
⎠
=
=
=

( x 2 + 1) Dx ( x cos x + sin x) − ( x cos x + sin x) Dx ( x 2 + 1)
( x 2 + 1) 2
( x 2 + 1)(– x sin x + cos x + cos x) – 2 x( x cos x + sin x)
( x 2 + 1)2
– x3 sin x – 3 x sin x + 2 cos x
( x 2 + 1) 2

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y = tan 2 x = (tan x)(tan x)

17.

b. Dt(20 sin t) = 20 cos t
π
π
At t = : rate = 20 cos = 10 3 ≈ 17. 32 ft/s
6
6

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

18.

2

Dx y = (sec 2 x) sec x tan x + (sec x) Dx (sec2 x)

25.

When y = 0 , y = tan 0 = 0 and y ' = sec 2 0 = 1 .
The tangent line at x = 0 is y = x .

= sec3 x tan x + 2sec3 x tan x
= 3sec2 x tan x

26.

= 2 tan x sec 2 x
Now, sec2 x is never 0, but tan x = 0 at
x = kπ where k is an integer.

27.

π
At x = : mtan = –2;
4
y=1

= 9 ⎡sin 2 x − cos 2 x ⎤
⎣
⎦
= 9 [ − cos 2 x ]

The tangent line is horizontal when y ' = 0 or, in
this case, where cos 2 x = 0 . This occurs when

21. Dx sin 2 x = Dx (2sin x cos x)
= 2 ⎣⎡sin x Dx cos x + cos x Dx sin x ⎦⎤

x=

= −2sin x + 2 cos x
2

2

28.
22. Dx cos 2 x = Dx (2 cos x − 1) = 2 Dx cos x − Dx 1
2

2

= −2sin x cos x

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

(

= 30 −2sin 2 t + 2 cos 2 t

)

= 60 cos 2t
30sin 2t = 15
1
sin 2t =
2

π

→ t=

π

6

12
π
⎛ π ⎞
At t = ; 60 cos ⎜ 2 ⋅ ⎟ = 30 3 ft/sec
12
⎝ 12 ⎠

The seat is moving to the left at the rate of 30 3
ft/s.
24. The coordinates of the seat at time t are
(20 cos t, 20 sin t).
a.

π
π⎞
⎛
⎜ 20 cos , 20sin ⎟ = (10 3, 10)
6
6⎠
⎝
≈ (17.32, 10)

Section 2.4

y = 9sin x cos x
y ' = 9 [sin x(− sin x) + cos x(cos x) ]

π⎞
⎛
Tangent line: y –1 = –2 ⎜ x – ⎟

4⎠
⎝

2t =

y = tan 2 x = (tan x)(tan x)
y ' = (tan x)(sec 2 x) + (tan x)(sec2 x)

19. Dx(cos x) = –sin x
At x = 1: mtan = – sin1 ≈ –0.8415
y = cos 1 ≈ 0.5403
Tangent line: y – 0.5403 = –0.8415(x – 1)
20. Dx (cot x) = – csc2 x

y = tan x
y ' = sec2 x

= sec3 x tan x + sec x(sec x ⋅ sec x tan x
+ sec x ⋅ sec x tan x)

114

The fastest rate 20 cos t can obtain is
20 ft/s.

c.

y = sec x = (sec x)(sec x)
3

π
4

+k

π
2

where k is an integer.

f ( x) = x − sin x
f '( x) = 1 − cos x
f '( x) = 0 when cos x = 1 ; i.e. when x = 2kπ
where k is an integer.
f '( x) = 2 when x = (2k + 1)π where k is an
integer.

29. The curves intersect when 2 sin x = 2 cos x,
sin x = cos x at x = π for 0 < x < π .
4
2
π
Dx ( 2 sin x) = 2 cos x ; 2 cos = 1
4
π
Dx ( 2 cos x) = – 2 sin x ; − 2 sin = −1
4
1(–1) = –1 so the curves intersect at right angles.
30. v = Dt (3sin 2t ) = 6 cos 2t
At t = 0: v = 6 cm/s

π
t = : v = −6 cm/s
2
t = π : v = 6 cm/s

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sin( x + h) 2 – sin x 2
h
h→0

31. Dx (sin x 2 ) = lim

sin( x 2 + 2 xh + h 2 ) – sin x 2
h
h→0
sin x 2 cos(2 xh + h 2 ) + cos x 2 sin(2 xh + h 2 ) – sin x 2
sin x 2 [cos(2 xh + h 2 ) – 1] + cos x 2 sin(2 xh + h 2 )
= lim
= lim
h →0
h
h
h→0

= lim

2
⎡
cos(2 xh + h 2 ) – 1
2 sin(2 xh + h ) ⎤
cos
x
= lim(2 x + h) ⎢sin x 2
+
⎥ = 2 x(sin x 2 ⋅ 0 + cos x 2 ⋅1) = 2 x cos x 2
h →0
2 xh + h 2
2 xh + h 2 ⎥⎦
⎢⎣

sin(5( x + h)) – sin 5 x
h
h →0
sin(5 x + 5h) – sin 5 x
= lim
h
h →0
sin 5 x cos 5h + cos 5 x sin 5h – sin 5 x
= lim
h
h →0
cos 5h – 1
sin 5h ⎤
⎡
= lim ⎢sin 5 x

+ cos 5 x
h
h ⎥⎦
h→0 ⎣
cos 5h – 1
sin 5h ⎤
⎡
= lim ⎢5sin 5 x
+ 5cos 5 x
5h
5h ⎥⎦
h→0 ⎣
= 0 + 5cos 5 x ⋅1 = 5cos 5 x

32. Dx (sin 5 x) = lim

33. f(x) = x sin x
a.

34.

f ( x ) = cos3 x − 1.25cos 2 x + 0.225

x0 ≈ 1.95
f ′ (x 0 ) ≈ –1. 24

2.5 Concepts Review
1. Dt u; f ′( g (t )) g ′(t )
2. Dv w; G ′( H ( s )) H ′( s )

b. f(x) = 0 has 6 solutions on [π , 6π ]
f ′ (x) = 0 has 5 solutions on [π , 6π ]
c.

f(x) = x sin x is a counterexample.
Consider the interval [ 0, π ] .
f ( −π ) = f (π ) = 0 and f ( x ) = 0 has

exactly two solutions in the interval (at 0 and
π ). However, f ' ( x ) = 0 has two solutions
in the interval, not 1 as the conjecture
indicates it should have.
d. The maximum value of f ( x) – f ′( x) on
[π , 6π ] is about 24.93.

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

Problem Set 2.5
1. y = u15 and u = 1 + x
Dx y = Du y ⋅ Dx u
= (15u14 )(1)
= 15(1 + x )14

2. y = u5 and u = 7 + x
Dx y = Du y ⋅ Dx u
= (5u 4 )(1)
= 5(7 + x)4

3. y = u5 and u = 3 – 2x
Dx y = Du y ⋅ Dx u
= (5u 4 )(–2) = –10(3 – 2 x) 4

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4. y = u7 and u = 4 + 2 x 2
Dx y = Du y ⋅ Dx u

10. y = cos u and u = 3 x 2 – 2 x
Dx y = Du y ⋅ Dx u
= (–sin u)(6x – 2)

= (7u 6 )(4 x) = 28 x(4 + 2 x 2 )6

= –(6 x – 2) sin(3x 2 – 2 x)

5. y = u11 and u = x3 – 2 x 2 + 3 x + 1
Dx y = Du y ⋅ Dx u

11. y = u 3 and u = cos x

Dx y = Du y ⋅ Dx u

= (11u10 )(3x 2 – 4 x + 3)

= (3u 2 )(– sin x)

= 11(3 x 2 – 4 x + 3)( x3 – 2 x 2 + 3x + 1)10

= –3sin x cos 2 x

6. y = u –7 and u = x 2 – x + 1
Dx y = Du y ⋅ Dx u

12. y = u 4 , u = sin v, and v = 3 x 2
Dx y = Du y ⋅ Dv u ⋅ Dx v

= (–7u –8 )(2 x – 1)

= (4u 3 )(cos v )(6 x)

= –7(2 x – 1)( x 2 – x + 1) –8

= 24 x sin 3 (3 x 2 ) cos(3 x 2 )

7. y = u –5 and u = x + 3
Dx y = Du y ⋅ Dx u

x +1
x –1
Dx y = Du y ⋅ Dx u

13. y = u 3 and u =

= (–5u –6 )(1) = –5( x + 3) –6 = –

5
( x + 3)6

= (3u 2 )

8. y = u and u = 3x + x – 3
Dx y = Du y ⋅ Dx u
–9

( x –1) 2

2

⎛ x +1⎞
= 3⎜
⎟
⎝ x –1⎠

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

( x –1) Dx ( x + 1) – ( x + 1) Dx ( x –1)
–2
⎜⎜

2
⎝ ( x – 1)

14. y = u −3 and u =

9(6 x + 1)

⎞
6( x + 1) 2
⎟⎟ = −
( x – 1)4
⎠

x−2
x−π

Dx y = Du y ⋅ Dx u

(3 x 2 + x – 3)10

= (−3u −4 ) ⋅

9. y = sin u and u = x + x
Dx y = Du y ⋅ Dx u
= (cos u)(2x + 1)
2

( x − π) Dx ( x − 2) − ( x − 2) Dx ( x − π)

⎛ x−2⎞

= −3 ⎜
⎟
⎝ x−π⎠

= (2 x + 1) cos( x + x)
2

15. y = cos u and u =

2⎛

( x − π) 2
−4

(2 − π)
( x − π) 2

= −3

( x − π)2
( x − 2)4

(2 − π)

3x 2
x+2

Dx y = Du y ⋅ Dx u = (– sin u )

( x + 2) Dx (3 x 2 ) – (3x 2 ) Dx ( x + 2)

( x + 2) 2

⎛ 3x 2 ⎞ ( x + 2)(6 x) – (3x 2 )(1)
⎛ 3x 2 ⎞
3 x 2 + 12 x
= – sin ⎜
=–
sin ⎜
⎟
⎟
⎜ x+2⎟
⎜
⎟
( x + 2)2
( x + 2)2
⎝
⎠
⎝ x+2⎠

16. y = u 3 , u = cos v, and v =

x2
1– x

Dx y = Du y ⋅ Dv u ⋅ Dx v = (3u 2 )(− sin v)

(1 – x) Dx ( x 2 ) – ( x 2 ) Dx (1 − x)
(1 – x) 2

⎛ x2 ⎞ ⎛ x2 ⎞

⎛ x 2 ⎞ ⎛ x 2 ⎞ (1 – x)(2 x) – ( x 2 )(–1)
–3(2 x – x 2 )
=
= –3cos 2 ⎜
cos 2 ⎜
⎟ sin ⎜
⎟
⎟ sin ⎜
⎟
⎜1– x ⎟ ⎜1– x ⎟
⎜1– x ⎟ ⎜1– x ⎟
(1 – x)2
(1 – x)2
⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
⎠

116

Section 2.5

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

18. Dx [(2 – 3x 2 )4 ( x 7 + 3)3 ] = (2 – 3 x 2 )4 Dx ( x 7 + 3)3 + ( x 7 + 3)3 Dx (2 – 3x 2 ) 4
= (2 – 3 x 2 )4 (3)( x 7 + 3) 2 (7 x 6 ) + ( x 7 + 3)3 (4)(2 – 3 x 2 )3 (–6 x) = 3x (3 x 2 – 2)3 ( x 7 + 3) 2 (29 x 7 – 14 x5 + 24)
⎡ ( x + 1)2 ⎤ (3x – 4) Dx ( x + 1)2 – ( x + 1)2 Dx (3x – 4)
(3 x – 4)(2)( x + 1)(1) – ( x + 1)2 (3) 3 x 2 – 8 x – 11
19. Dx ⎢
=
=
⎥=
(3x – 4) 2
(3x – 4) 2
(3 x – 4)2
⎣⎢ 3x – 4 ⎦⎥

=

( x + 1)(3 x − 11)
(3x − 4)2

⎡ 2 x – 3 ⎤ ( x 2 + 4) 2 Dx (2 x – 3) – (2 x – 3) Dx ( x 2 + 4)2
20. Dx ⎢
⎥=
2
2
( x 2 + 4) 4
⎣⎢ ( x + 4) ⎦⎥
=

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

(

)(

) (

=

)

−6 x 2 + 12 x + 8
( x 2 + 4)3

(

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

)

22. y ′ = 2(x + sin x )(x + sin x )′ = 2(x + sin x )(1 + cos x )
3

2

⎛ 3t – 2 ⎞

⎛ 3t – 2 ⎞ (t + 5) Dt (3t – 2) – (3t – 2) Dt (t + 5)
23. Dt ⎜
`⎟ = 3 ⎜
⎟
t
5
+
⎝
⎠
⎝ t +5 ⎠
(t + 5)2
2

51(3t – 2)2
⎛ 3t – 2 ⎞ (t + 5)(3) – (3t – 2)(1)
=
= 3⎜
⎟
⎝ t +5 ⎠
(t + 5) 4
(t + 5)2
⎛ s 2 – 9 ⎞ ( s + 4) Ds ( s 2 – 9) – ( s 2 – 9) Ds ( s + 4)
( s + 4)(2s ) – ( s 2 – 9)(1) s 2 + 8s + 9
24. Ds ⎜
=
=
⎟=
⎜ s+4 ⎟
( s + 4)2
( s + 4) 2

( s + 4)2
⎝
⎠

d ⎛ (3t − 2)3 ⎞
25.
⎜
⎟=
dt ⎜⎝ t + 5 ⎟⎠
=

26.

(t + 5)

d
d
(3t − 2)3 − (3t − 2)3 (t + 5)
(t + 5)(3)(3t – 2)2 (3) – (3t – 2)3 (1)
dt
dt
=
(t + 5)2
(t + 5) 2

(6t + 47)(3t – 2)2
(t + 5) 2

d
(sin 3 θ ) = 3sin 2 θ cos θ

dθ
3

2

2

dy d ⎛ sin x ⎞
⎛ sin x ⎞ d sin x
⎛ sin x ⎞
= ⎜
= 3⎜
27.
⎟ = 3⎜
⎟ ⋅
⎟ ⋅
dx dx ⎝ cos 2 x ⎠
⎝ cos 2 x ⎠ dx cos 2 x
⎝ cos 2 x ⎠

(cos 2 x)

d
d
(sin x) − (sin x) (cos 2 x)
dx
dx
cos 2 2 x

2

2
3
⎛ sin x ⎞ cos x cos 2 x + 2sin x sin 2 x 3sin x cos x cos 2 x + 6sin x sin 2 x
= 3⎜
=
⎟
⎝ cos 2 x ⎠
cos 2 2 x
cos 4 2 x

=

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

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28.

dy d
d

d
= [sin t tan(t 2 + 1)] = sin t ⋅ [tan(t 2 + 1)] + tan(t 2 + 1) ⋅ (sin t )
dt dt
dt
dt
= (sin t )[sec 2 (t 2 + 1)](2t ) + tan(t 2 + 1) cos t = 2t sin t sec2 (t 2 + 1) + cos t tan(t 2 + 1)
2

29.

2

⎛ x 2 + 1 ⎞ ( x + 2) Dx ( x 2 + 1) – ( x 2 + 1) Dx ( x + 2)
⎛ x 2 + 1 ⎞ 2 x 2 + 4 x – x 2 – 1 3( x 2 + 1)2 ( x 2 + 4 x – 1)
f ′( x) = 3 ⎜
=
= 3⎜
⎟
⎟
⎜ x+2 ⎟
⎜ x+2 ⎟
( x + 2) 2
( x + 2) 2
( x + 2)4
⎝
⎠
⎝
⎠
f ′(3) = 9.6

30. G ′(t ) = (t 2 + 9)3 Dt (t 2 – 2)4 + (t 2 – 2) 4 Dt (t 2 + 9)3 = (t 2 + 9)3 (4)(t 2 – 2)3 (2t ) + (t 2 – 2) 4 (3)(t 2 + 9)2 (2t )
= 2t (7t 2 + 30)(t 2 + 9)2 (t 2 – 2)3
G ′(1) = –7400

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

F ′(1) = 5cos 5 ≈ 1.4183

32. g ′( s ) = (cos πs ) Ds (sin 2 πs ) + (sin 2 πs ) Ds (cos πs ) = (cos πs )(2sin πs )(cos πs )(π) + (sin 2 πs )(– sin πs )(π)
= π sin πs[2 cos 2 πs – sin 2 πs ]
⎛1⎞
g ′ ⎜ ⎟ = –π
⎝2⎠

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

34. Dt [cos5 (4t – 19)] = 5cos 4 (4t – 19) Dt cos(4t – 19) = 5cos 4 (4t – 19)[– sin(4t – 19)]Dt (4t – 19)
4
= –5cos 4 (4t – 19) sin(4t – 19)(4) = –20 cos (4t – 19) sin(4t – 19)

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

⎡
⎛ u + 1 ⎞⎤
⎛ u +1⎞
⎛ u + 1 ⎞⎤ ⎛ u + 1 ⎞
3 ⎛ u +1⎞

3 ⎛ u +1⎞ ⎡
36 . Du ⎢cos4 ⎜
⎟ ⎥ = 4 cos ⎜
⎟ Du cos ⎜
⎟ = 4 cos ⎜
⎟ ⎢ – sin ⎜
⎟ ⎥ Du ⎜
⎟
⎝ u –1 ⎠ ⎦
⎝ u –1 ⎠
⎝ u –1 ⎠
⎝ u –1 ⎠ ⎣
⎝ u –1 ⎠ ⎦ ⎝ u –1 ⎠
⎣
8
⎛ u +1⎞ ⎛ u +1⎞
⎛ u + 1 ⎞ ⎛ u + 1 ⎞ (u –1) Du (u + 1) – (u + 1) Du (u –1)
cos3 ⎜
=
= –4 cos3 ⎜
⎟ sin ⎜
⎟
⎟ sin ⎜
⎟
2
2
u
–1
u
–1

⎝ u –1 ⎠ ⎝ u –1 ⎠
⎝
⎠ ⎝
⎠
(u –1)
(u –1)
37. Dθ [cos 4 (sin θ 2 )] = 4 cos3 (sin θ 2 ) Dθ cos(sin θ 2 ) = 4 cos3 (sin θ 2 )[– sin(sin θ 2 )]Dθ (sin θ 2 )
= –4 cos3 (sin θ 2 ) sin(sin θ 2 )(cosθ 2 ) Dθ (θ 2 ) = –8θ cos3 (sin θ 2 ) sin(sin θ 2 )(cos θ 2 )

38. Dx [ x sin 2 (2 x)] = x Dx sin 2 (2 x) + sin 2 (2 x) Dx x = x[2sin(2 x) Dx sin(2 x)] + sin 2 (2 x)(1)
= x[2sin(2 x ) cos(2 x) Dx (2 x)] + sin 2 (2 x) = x[4sin(2 x) cos(2 x)] + sin 2 (2 x) = 2 x sin(4 x) + sin 2 (2 x)

39. Dx {sin[cos(sin 2 x)]} = cos[cos(sin 2 x)]Dx cos(sin 2 x) = cos[cos(sin 2 x)][– sin(sin 2 x)]Dx (sin 2 x)
= – cos[cos(sin 2 x)]sin(sin 2 x)(cos 2 x) Dx (2 x) = –2 cos[cos(sin 2 x)]sin(sin 2 x)(cos 2 x)
40. Dt {cos 2 [cos(cos t )]} = 2 cos[cos(cos t )]Dt cos[cos(cos t )] = 2 cos[cos(cos t )]{– sin[cos(cos t )]}Dt cos(cos t )
= –2 cos[cos(cos t )]sin[cos(cos t )][– sin(cos t )]Dt (cos t ) = 2 cos[cos(cos t )]sin[cos(cos t )]sin(cos t )(– sin t )
= –2sin t cos[cos(cos t )]sin[cos(cos t )]sin(cos t )

118

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