51543 0131469657 ism-2

94 Section 2.1 Instructor’s Resource Manual
CHAPTER 2 The Derivative
2.1 Concepts Review
1. tangent line
2. secant line
3.
( ) ( )f c h f c
h
+ −
4. average velocity
Problem Set 2.1
1. Slope
3
2
5 – 3
4
2 –
= =
2.
6 – 4
Slope –2
4 – 6
= =
3.
Slope 2≈ −
4.
Slope 1.5≈
5.
5
Slope
2
≈
6.
3
Slope –
2
≈
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Instructor’s Resource Manual Section 2.1 95
7. y = x 2
+ 1
a., b.
c. mtan = 2
d.
2
sec
(1.01) 1.0 2
1.01 1
0.0201
.01
2.01
m
+ −
=
−
=
=
e. tan
0
(1 ) – (1)
lim
h
f h f
m
h→
+
=
2 2
0
[(1 ) 1] – (1 1)
lim
h
h
h→
+ + +
=
2
0
0
2 2 2
lim
(2 )
lim
h
h
h h
h
h h
h
→
→
+ + −
=
+
=
0
lim (2 ) 2
h
h
→
= + =
8. y = x 3
–1
a., b.
c. mtan = 12
d.
3
sec
[(2.01) 1.0] 7
2.01 2
0.120601
0.01
m
− −
=
−
=
= 12.0601
e. tan
0
(2 ) – (2)
lim
h
f h f
m
h→
+
=
3 3
0
[(2 ) –1] – (2 1)
lim
h
h
h→
+ −
=
2 3
0
2
0
12 6
lim
(12 6 )
lim
h
h
h h h
h
h h h
h
→
→
+ +
=
+ +
=
= 12
9. f (x) = x2
– 1
tan
0
( ) – ( )
lim
h
f c h f c
m
h→
+
=
2 2
0
[( ) –1] – ( –1)
lim
h
c h c
h→
+
=
2 2 2
0
2 –1– 1
lim
h
c ch h c
h→
+ + +
=
0
(2 )
lim 2
h
h c h
c
h→
+
= =
At x = –2, mtan = –4
x = –1, mtan = –2
x = 1, mtan = 2
x = 2, mtan = 4
10. f (x) = x3
– 3x
tan
0
( ) – ( )
lim
h
f c h f c
m
h→
+
=
3 3
0
[( ) – 3( )] – ( – 3 )
lim
h
c h c h c c
h→
+ +
=
3 2 2 3 3
0
3 3 – 3 – 3 – 3
lim
h
c c h ch h c h c c
h→
+ + + +
=
2 2
2
0
(3 3 3)
lim 3 – 3
h
h c ch h
c
h→
+ + −
= =
At x = –2, mtan = 9
x = –1, mtan = 0
x = 0, mtan = –3
x = 1, mtan = 0
x = 2, mtan = 9

11.
1
( )
1
f x
x
=
+
tan
0
(1 ) – (1)
lim
h
f h f
m
h→
+
=
1 1
2 2
0
2(2 )
0
0
lim
lim
1
lim
2(2 )
h
h
h
h
h
h
h
h
h
+
→
+
→
→
−
=
−
=
= −
+
1
–
4
=
1 1
– – ( –1)
2 4
y x=
12. f (x) =
1
x –1
tan
0
1
1
0
1
0
0
(0 ) (0)
lim
1
lim
lim
1
lim
1
1
h
h
h
h
h
h
h
f h f
m
h
h
h
h
→
−
→
−
→
→
+ −
=
+
=
=
=
−
= −
y + 1 = –1(x – 0); y = –x – 1
13. a. 2 2
16(1 ) –16(0 ) 16 ft=
b. 2 2
16(2 ) –16(1 ) 48 ft=
c. ave
144 – 64
V 80
3 – 2
= = ft/sec
d.
2 2
ave
16(3.01) 16(3)
V
3.01 3
0.9616
0.01
−
=
−
=
= 96.16 ft/s
e. 2
( ) 16 ; 32f t t v c= =
v = 32(3) = 96 ft/s
14. a.
2 2
ave
(3 1) – (2 1)
V 5
3 – 2
+ +
= = m/sec
b.
2 2
ave
[(2.003) 1] (2 1)
V
2.003 2
0.012009
0.003
+ − +
=
−
=
= 4.003 m/sec
c.
2 2
ave
2
[(2 ) 1] – (2 1)
V
2 – 2
4
h
h
h h
h
+ + +
=
+
+
=
= 4 +h
d. f (t) = t2
+ 1
0
(2 ) – (2)
lim
h
f h f
v
h→
+
=
2 2
0
[(2 ) 1] – (2 1)
lim
h
h
h→
+ + +
=
2
0
0
4
lim
lim (4 )
4
h
h
h h
h
h
→
→
+
=
= +
=

15. a.
0
( ) – ( )
lim
h
f h f
v
h
α α
→
+
=
0
2( ) 1 – 2 1
lim
h
h
h
α α
→
+ + +
=
0
2 2 1 – 2 1
lim
h
h
h
α α
→
+ + +
=
0
( 2 2 1 – 2 1)( 2 2 1 2 1)
lim
( 2 2 1 2 1)h
h h
h h
α α α α
α α→
+ + + + + + +
=
+ + + +
0
2
lim
( 2 2 1 2 1)h
h
h hα α→
=
+ + + +
2 1
2 1 2 1 2 1α α α
= =
+ + + +
ft/s
b.
1 1
22 1α
=
+
2 1 2α + =
2α +1= 4; α =
3
2
The object reaches a velocity of 1
2
ft/s when t =
3
2
.
16. f (t) = –t2
+ 4t
2 2
0
[–( ) 4( )] – (– 4 )
lim
h
c h c h c c
v
h→
+ + + +
=
2 2 2
0
– – 2 – 4 4 – 4
lim
h
c ch h c h c c
h→
+ + +
=
0
(–2 – 4)
lim –2 4
h
h c h
c
h→
+
= = +
–2c + 4 = 0 when c = 2
The particle comes to a momentary stop at
t = 2.
17. a. 2 21 1
(2.01) 1 – (2) 1 0.02005
2 2
⎡ ⎤ ⎡ ⎤
+ + =⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
g
b. ave
0.02005
2.005
2.01– 2
r = = g/hr
c. 21
( ) 1
2
f t t= +
2 21 1
2 2
0
(2 ) 1 – 2 1
lim
h
h
r
h→
⎡ ⎤ ⎡ ⎤+ + +
⎣ ⎦ ⎣ ⎦=
21
2
0
2 2 1 2 1
lim
h
h h
h→
+ + + − −
=
( )1
2
0
2
lim 2
h
h h
h→
+
= =
At t = 2, r = 2
18. a. 2 2
1000(3) –1000(2) 5000=
b.
2 2
1000(2.5) –1000(2) 2250
4500
2.5 – 2 0.5
= =
c. f (t) = 1000t2
2 2
0
1000(2 ) 1000(2)
lim
h
h
r
h→
+ −
=
2
0
4000 4000 1000 – 4000
lim
h
h h
h→
+ +
=
0
(4000 1000 )
lim 4000
h
h h
h→
+
= =
19. a.
3 3
ave
5 – 3 98
49
5 – 3 2
d = = = g/cm
b. f (x) = x3
3 3
0
(3 ) – 3
lim
h
h
d
h→
+
=
2 3
0
27 27 9 – 27
lim
h
h h h
h→
+ + +
=
2
0
(27 9 )
lim 27
h
h h h
h→
+ +
= = g/cm

20.
0
( ) – ( )
lim
h
R c h R c
MR
h→
+
=
2 2
0
[0.4( ) – 0.001( ) ] – (0.4 – 0.001 )
lim
h
c h c h c c
h→
+ +
=
2 2 2
0
0.4 0.4 – 0.001 – 0.002 – 0.001 – 0.4 0.001
lim
h
c h c ch h c c
h→
+ +
=
0
(0.4 – 0.002 – 0.001 )
lim 0.4 – 0.002
h
h c h
c
h→
= =
When n = 10, MR = 0.38; when n = 100, MR = 0.2
21.
2 2
0
2(1 ) – 2(1)
lim
h
h
a
h→
+
=
2
0
0
2 4 2 – 2
lim
(4 2 )
lim 4
h
h
h h
h
h h
h
→
→
+ +
=
+
= =
22.
0
( ) – ( )
lim
h
p c h p c
r
h→
+
=
2 3 2 3
0
[120( ) – 2( ) ] – (120 – 2 )
lim
h
c h c h c c
h→
+ +
=
2 2
0
(240 – 6 120 – 6 – 2 )
lim
h
h c c h ch h
h→
+
=
2
240 – 6c c=
When t = 10, 2
240(10) – 6(10) 1800r = =
t = 20, 2
240(20) – 6(20) 2400r = =
t = 40, 2
240(40) – 6(40) 0r = =
23. ave
100 – 800 175
– –29.167
24 – 0 6
r = = ≈
29,167 gal/hr
At 8 o’clock,
700 – 400
75
6 10
r ≈ ≈ −
−
75,000 gal/hr
24. a. The elevator reached the seventh floor at time
80=t . The average velocity is
05.180/)084( =−=avgv feet per second
b. The slope of the line is approximately
2.1
1555
1260
=
−
−
. The velocity is
approximately 1.2 feet per second.
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 91=t has slope
approximately 5.0)6191/()4863( =−− . The
normal high temperature increases at the rate
of 0.5 degree F per day.
b. A tangent line at 191=t has approximate
slope 067.030/)8890( ≈− . 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 1930=t 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 02.020/4.0 ≈ .
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.

28. 31
( )
3
f t t t= +
0
( ) – ( )
current lim
h
f c h f c
h→
+
=
( )3 31 1
3 3
0
( ) ( ) –
lim
h
c h c h c c
h→
⎡ ⎤+ + + +
⎣ ⎦=
( )2 21
3 2
0
1
lim 1
h
h c ch h
c
h→
+ + +
= = +
When t = 3, the current =10
2
1 20c + =
c
2
= 19
19 4.4c = ≈
A 20-amp fuse will blow at t = 4.4 s.
29. 2
,A r= π r = 2t
A = 4πt2
2 2
0
4 (3 ) – 4 (3)
rate lim
h
h
h→
π + π
=
0
(24 4 )
lim 24
h
h h
h→
π + π
= = π km2/day
30. 3
3
3 3
0
3
4 1
,
3 4
1
48
1 (3 ) 3 27
rate lim
48 48
9
inch /sec
16
h
V r r t
V t
h
h
π
π
π π
π
→
= =
=
+ −
= =
=
31. 3 2
( ) – 2 1y f x x x= = +
a. mtan = 7 b. mtan = 0
c. mtan = –1 d. mtan = 17. 92
32. 2
( ) sin sin 2y f x x x= =
a. mtan = –1.125 b. mtan ≈ –1.0315
c. mtan = 0 d. mtan ≈1.1891
33. 2
( ) coss f t t t t= = +
At t = 3, v ≈ 2.818
34.
3
( 1)
( )
2
t
s f t
t
+
= =
+
At t = 1.6, v ≈ 4.277
2.2 Concepts Review
1.
( ) – ( ) ( ) – ( )
;
–
f c h f c f t f c
h t c
+
2. ( )cf ′
3. continuous; ( )f x x=
4. ( )' ;
dy
f x
dx
Problem Set 2.2
1.
0
(1 ) – (1)
(1) lim
h
f h f
f
h→
+
′ =
2 2 2
0 0
(1 ) –1 2
lim lim
h h
h h h
h h→ →
+ +
= =
= lim
h→0
(2 + h) = 2
2.
0
(2 ) – (2)
(2) lim
h
f h f
f
h→
+
′ =
2 2
0
[2(2 )] –[2(2)]
lim
h
h
h→
+
=
2
0 0
16 4
lim lim (16 4 ) 16
h h
h h
h
h→ →
+
= = + =
3.
0
(3 ) – (3)
(3) lim
h
f h f
f
h→
+
′ =
2 2
0
[(3 ) – (3 )] – (3 – 3)
lim
h
h h
h→
+ +
=
2
0 0
5
lim lim (5 ) 5
h h
h h
h
h→ →
+
= = + =
4.
0
(4 ) – (4)
(4) lim
h
f h f
f
h→
+
′ =
3–(3 )1 1
3(3 )3 4–1
0 0 0
– –1
lim lim lim
3(3 )
h
hh
h h hh h h
+
++
→ → →
= = =
+
= –
1
9
5.
0
( ) – ( )
( ) lim
h
s x h s x
s x
h→
+
′ =
0
[2( ) 1] – (2 1)
lim
h
x h x
h→
+ + +
=
0
2
lim 2
h
h
h→
= =

6.
0
( ) – ( )
( ) lim
h
f x h f x
f x
h→
+
′ =
0
[ ( ) ] – ( )
lim
h
x h x
h
α β α β
→
+ + +
=
0
lim
h
h
h
α
α
→
= =
7.
0
( ) – ( )
( ) lim
h
r x h r x
r x
h→
+
′ =
2 2
0
[3( ) 4] – (3 4)
lim
h
x h x
h→
+ + +
=
2
0 0
6 3
lim lim (6 3 ) 6
h h
xh h
x h x
h→ →
+
= = + =
8.
0
( ) – ( )
( ) lim
h
f x h f x
f x
h→
+
′ =
2 2
0
[( ) ( ) 1] – ( 1)
lim
h
x h x h x x
h→
+ + + + + +
=
2
0 0
2
lim lim (2 1) 2 1
h h
xh h h
x h x
h→ →
+ +
= = + + = +
9.
0
( ) – ( )
( ) lim
h
f x h f x
f x
h→
+
′ =
2 2
0
[ ( ) ( ) ] – ( )
lim
h
a x h b x h c ax bx c
h→
+ + + + + +
=
2
0 0
2
lim lim (2 )
h h
axh ah bh
ax ah b
h→ →
+ +
= = + +
= 2ax + b
10.
0
( ) – ( )
( ) lim
h
f x h f x
f x
h→
+
′ =
4 4
0
( ) –
lim
h
x h x
h→
+
=
3 2 2 3 4
0
4 6 4
lim
h
hx h x h x h
h→
+ + +
=
3 2 2 3 3
0
lim (4 6 4 ) 4
h
x hx h x h x
→
= + + + =
11.
0
( ) – ( )
( ) lim
h
f x h f x
f x
h→
+
′ =
3 2 3 2
0
[( ) 2( ) 1] – ( 2 1)
lim
h
x h x h x x
h→
+ + + + + +
=
2 2 3 2
0
3 3 4 2
lim
h
hx h x h hx h
h→
+ + + +
=
2 2 2
0
lim (3 3 4 2 ) 3 4
h
x hx h x h x x
→
= + + + + = +
12.
0
( ) – ( )
( ) lim
h
g x h g x
g x
h→
+
′ =
4 2 4 2
0
[( ) ( ) ] – ( )
lim
h
x h x h x x
h→
+ + + +
=
3 2 2 3 4 2
0
4 6 4 2
lim
h
hx h x h x h hx h
h→
+ + + + +
=
3 2 2 3
0
lim (4 6 4 2 )
h
x hx h x h x h
→
= + + + + +
= 4x3
+2x
13.
0
( ) – ( )
( ) lim
h
h x h h x
h x
h→
+
′ =
0
2 2 1
lim –
h x h x h→
⎡ ⎤⎛ ⎞
= ⋅⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦
0 0
–2 1 –2
lim lim
( ) ( )h h
h
x x h h x x h→ →
⎡ ⎤
= ⋅ =⎢ ⎥+ +⎣ ⎦ 2
2
–
x
=
14.
0
( ) – ( )
( ) lim
h
S x h S x
S x
h→
+
′ =
0
1 1 1
lim –
1 1h x h x h→
⎡ ⎤⎛ ⎞
= ⋅⎜ ⎟⎢ ⎥+ + +⎝ ⎠⎣ ⎦
0
– 1
lim
( 1)( 1)h
h
x x h h→
⎡ ⎤
= ⋅⎢ ⎥
+ + +⎣ ⎦
20
–1 1
lim
( 1)( 1) ( 1)h x x h x→
= = −
+ + + +
15.
0
( ) – ( )
( ) lim
h
F x h F x
F x
h→
+
′ =
2 20
6 6 1
lim –
( ) 1 1h hx h x→
⎡ ⎤⎛ ⎞
= ⋅⎢ ⎥⎜ ⎟⎜ ⎟+ + +⎢ ⎥⎝ ⎠⎣ ⎦
2 2 2
2 2 20
6( 1) – 6( 2 1) 1
lim
( 1)( 2 1)h
x x hx h
hx x hx h→
⎡ ⎤+ + + +
= ⋅⎢ ⎥
+ + + +⎢ ⎥⎣ ⎦
2
2 2 20
–12 – 6 1
lim
( 1)( 2 1)h
hx h
hx x hx h→
⎡ ⎤
= ⋅⎢ ⎥
+ + + +⎢ ⎥⎣ ⎦
2 2 2 2 20
–12 – 6 12
lim
( 1)( 2 1) ( 1)h
x h x
x x hx h x→
= = −
+ + + + +
16.
0
( ) – ( )
( ) lim
h
F x h F x
F x
h→
+
′ =
0
–1 –1 1
lim –
1 1h
x h x
x h x h→
⎡ + ⎤⎛ ⎞
= ⋅⎜ ⎟⎢ ⎥+ + +⎝ ⎠⎣ ⎦
2 2
0
–1– ( – –1) 1
lim
( 1)( 1)h
x hx h x hx h
x h x h→
⎡ ⎤+ + +
= ⋅⎢ ⎥
+ + +⎢ ⎥⎣ ⎦
20
2 1 2
lim
( 1)( 1) ( 1)h
h
x h x h x→
⎡ ⎤
= ⋅ =⎢ ⎥+ + + +⎣ ⎦

17.
0
( ) – ( )
( ) lim
h
G x h G x
G x
h→
+
′ =
0
2( ) –1 2 –1 1
lim –
– 4 – 4h
x h x
x h x h→
⎡ + ⎤⎛ ⎞
= ⋅⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦
2 2
2 2 9 8 4 (2 2 9 4) 1
lim
( 4)( 4)0
x hx x h x hx x h
hx h xh
+ − − + − + − − +
= ⋅
+ − −→
⎡ ⎤
⎢ ⎥
⎢ ⎥⎣ ⎦ 0
–7 1
lim
( – 4)( – 4)h
h
x h x h→
⎡ ⎤
= ⋅⎢ ⎥+⎣ ⎦
20
–7 7
lim –
( – 4)( – 4) ( – 4)h x h x x→
= =
+
18.
0
( ) – ( )
( ) lim
h
G x h G x
G x
h→
+
′ =
2 20
2( ) 2 1
lim –
( ) – ( ) –h
x h x
hx h x h x x→
⎡ ⎤⎛ ⎞+
= ⋅⎢ ⎥⎜ ⎟⎜ ⎟+ +⎢ ⎥⎝ ⎠⎣ ⎦
2 2 2
2 2 20
(2 2 )( – ) – 2 ( 2 – – ) 1
lim
( 2 – – )( – )h
x h x x x x xh h x h
hx hx h x h x x→
⎡ ⎤+ + +
= ⋅⎢ ⎥
+ +⎢ ⎥⎣ ⎦
2 2
2 2 20
–2 – 2 1
lim
( 2 – – )( – )h
h x hx
hx hx h x h x x→
⎡ ⎤
= ⋅⎢ ⎥
+ +⎢ ⎥⎣ ⎦
2
2 2 20
–2 – 2
lim
( 2 – – )( – )h
hx x
x hx h x h x x→
=
+ +
2
2 2 2
–2 2
–
( – ) ( –1)
x
x x x
= =
19.
0
( ) – ( )
( ) lim
h
g x h g x
g x
h→
+
′ =
0
3( ) – 3
lim
h
x h x
h→
+
=
0
( 3 3 – 3 )( 3 3 3 )
lim
( 3 3 3 )h
x h x x h x
h x h x→
+ + +
=
+ +
0 0
3 3
lim lim
( 3 3 3 ) 3 3 3h h
h
h x h x x h x→ →
= =
+ + + +
3
2 3x
=
20.
0
( ) – ( )
( ) lim
h
g x h g x
g x
h→
+
′ =
0
1 1 1
lim –
3( ) 3h hx h x→
⎡ ⎤⎛ ⎞
= ⋅⎢ ⎥⎜ ⎟⎜ ⎟+⎢ ⎥⎝ ⎠⎣ ⎦
0
3 – 3 3 1
lim
9 ( )h
x x h
hx x h→
⎡ ⎤+
= ⋅⎢ ⎥
+⎢ ⎥⎣ ⎦
0
( 3 – 3 3 )( 3 3 3 ) 1
lim
9 ( )( 3 3 3 )h
x x h x x h
hx x h x x h→
⎡ ⎤+ + +
= ⋅⎢ ⎥
+ + +⎢ ⎥⎣ ⎦
0
–3 –3
lim
9 ( )( 3 3 3 ) 3 2 3h
h
h x x h x x h x x→
= =
+ + + ⋅
1
–
2 3x x
=

21.
0
( ) – ( )
( ) lim
h
H x h H x
H x
h→
+
′ =
0
3 3 1
lim –
– 2 – 2h hx h x→
⎡ ⎤⎛ ⎞
= ⋅⎢ ⎥⎜ ⎟
+⎝ ⎠⎣ ⎦
0
3 – 2 – 3 – 2 1
lim
( – 2)( – 2)h
x x h
hx h x→
⎡ ⎤+
= ⋅⎢ ⎥
+⎢ ⎥⎣ ⎦
0
3( – 2 – – 2)( – 2 – 2)
lim
( – 2)( – 2)( – 2 – 2)h
x x h x x h
h x h x x x h→
+ + +
=
+ + +
0
3
lim
[( – 2) – 2 ( – 2) – 2]h
h
h x x h x h x→
−
=
+ + +
0
–3
lim
( – 2) – 2 ( – 2) – 2h x x h x h x→
=
+ + +
3 2
3 3
–
2( – 2) – 2 2( 2)x x x
= = −
−
22.
0
( ) – ( )
( ) lim
h
H x h H x
H x
h→
+
′ =
2 2
0
( ) 4 – 4
lim
h
x h x
h→
+ + +
=
2 2 2 2 2 2
0 2 2 2
2 4 – 4 2 4 4
lim
2 4 4h
x hx h x x hx h x
h x hx h x→
⎛ ⎞⎛ ⎞+ + + + + + + + +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠=
⎛ ⎞+ + + + +⎜ ⎟
⎝ ⎠
2
0 2 2 2
2
lim
2 4 4h
hx h
h x hx h x→
+
=
⎛ ⎞+ + + + +⎜ ⎟
⎝ ⎠
2 2 20
2
lim
2 4 4h
x h
x hx h x→
+
=
+ + + + +
2 2
2
2 4 4
x x
x x
= =
+ +
23.
( ) – ( )
( ) lim
–t x
f t f x
f x
t x→
′ =
2 2
( 3 ) – ( – 3 )
lim
–t x
t t x x
t x→
−
=
2 2
– – (3 – 3 )
lim
–t x
t x t x
t x→
=
( – )( ) – 3( – )
lim
–t x
t x t x t x
t x→
+
=
( – )( – 3)
lim lim( – 3)
–t x t x
t x t x
t x
t x→ →
+
= = +
= 2 x – 3
24.
( ) – ( )
( ) lim
–t x
f t f x
f x
t x→
′ =
3 3
( 5 ) – ( 5 )
lim
–t x
t t x x
t x→
+ +
=
3 3
– 5 – 5
lim
–t x
t x t x
t x→
+
=
2 2
( – )( ) 5( – )
lim
–t x
t x t tx x t x
t x→
+ + +
=
2 2
( – )( 5)
lim
–t x
t x t tx x
t x→
+ + +
=
2 2 2
lim( 5) 3 5
t x
t tx x x
→
= + + + = +

25.
( ) – ( )
( ) lim
–t x
f t f x
f x
t x→
′ =
1
lim –
– 5 – 5 –t x
t x
t x t x→
⎡ ⎤⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟⎢ ⎥
⎝ ⎠⎝ ⎠⎣ ⎦
– 5 – 5
lim
( – 5)( – 5)( – )t x
tx t tx x
t x t x→
+
=
–5( – ) –5
lim lim
( – 5)( – 5)( – ) ( – 5)( – 5)t x t x
t x
t x t x t x→ →
= =
( )
2
5
5x
= −
−
26.
( ) – ( )
( ) lim
–t x
f t f x
f x
t x→
′ =
3 3 1
lim –
–t x
t x
t x t x→
⎡ + + ⎤⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟⎢ ⎥
⎝ ⎠⎝ ⎠⎣ ⎦
2
3 – 3 –3 3
lim lim –
( – )t x t x
x t
xt t x xt x→ →
= = =
27. f (x) = 2x3
at x = 5
28. f (x) = x2
+ 2x at x = 3
29. f (x) = x2
at x = 2
30. f (x) = x3
+ x at x = 3
31. f (x) = x2
at x
32. f (x) = x3
at x
33. f (t) =
2
t
at t
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.
38. The slope of the tangent line is always 1− .
39. The derivative is positive until 0x = , then
becomes negative.
40. The derivative is negative until 1x = , then
becomes positive.
41. The derivative is 1− until 1x = . To the right of
1x = , the derivative is 1. The derivative is
undefined at 1x = .
42. The derivative is 2− to the left of 1x = − ; from
1− to 1, the derivative is 2, etc. The derivative is
not defined at 1, 1, 3x = − .

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
and 0 on ( )3,4 . The derivative is undefined at
2, 1, 0, 1, 2, 3x = − − .
44. The derivative is 1 except at 2, 0, 2x = − where
it is undefined.
45. Δy = [3(1.5) + 2] – [3(1) + 2] = 1.5
46. 2 2
[3(0.1) 2(0.1) 1] –[3(0.0) 2(0.0) 1]yΔ = + + + +
= 0.23
47. Δy = 1/1.2 – 1/1 = – 0.1667
48. Δy = 2/(0.1+1) – 2/(0+1) = – 0.1818
49.
3 3
– 0.0081
2.31 1 2.34 1
yΔ = ≈
+ +
50. cos[2(0.573)] – cos[2(0.571)] –0.0036yΔ = ≈
51.
2 2 2
( ) – 2 ( )
2
y x x x x x x
x x
x x x
Δ + Δ Δ + Δ
= = = + Δ
Δ Δ Δ
0
lim (2 ) 2
x
dy
x x x
dx Δ →
= + Δ =
52.
3 2 3 2
[( ) – 3( ) ] – ( – 3 )y x x x x x x
x x
Δ + Δ + Δ
=
Δ Δ
2 2 2 3
3 3 ( ) – 6 – 3( )x x x x x x x x
x
Δ + Δ Δ Δ + Δ
=
Δ
2 2
3 3 – 6 – 3 ( )x x x x x x= + Δ Δ + Δ
2 2
0
lim (3 3 – 6 – 3 ( ) )
x
dy
x x x x x x
dx Δ →
= + Δ Δ + Δ
2
3 – 6x x=
53.
1 1
1 1
–x x xy
x x
+Δ + +Δ
=
Δ Δ
1– ( 1) 1
( 1)( 1)
x x x
x x x x
⎛ ⎞+ + Δ + ⎛ ⎞
= ⎜ ⎟⎜ ⎟
+ Δ + + Δ⎝ ⎠⎝ ⎠
–
( 1)( 1)
x
x x x x
Δ
=
+ Δ + + Δ
1
–
( 1)( 1)x x x
=
+ Δ + +
20
1 1
lim
( 1)( 1) ( 1)x
dy
dx x x x xΔ →
⎡ ⎤
= − = −⎢ ⎥+ Δ + + +⎣ ⎦
54.
( )
( )
( ) 20
1 1
1 1
1 1
1
1 1
lim
x
y x x x
x x
x
x x xx x x
x x x x x
dy
dx x x x xΔ →
⎛ ⎞
+ − +⎜ ⎟Δ + Δ ⎝ ⎠=
Δ Δ
−Δ
− + Δ+ Δ= = = −
Δ Δ + Δ
= − = −
+ Δ
55.
( )( ) ( )( )
( )( )
( )
2 2
2
2 2
2 2 20
1 1
1 1
1 1 1 1 1
1 1
1 1 1
1
2 1 2
1 1
2 2 2
lim
1 2 1 1x
x x x
y x x x
x x
x x x x x x
x x x x
x x x x x x x x x x x x
xx x x x x x
x
xx x x x x x x x x x x x
dy
dx x x x x x x x x xΔ →
+ Δ − −
−
Δ + Δ + +=
Δ Δ
+ + Δ − − − + Δ +
= ×
+ Δ + + Δ
⎡ ⎤+ Δ − + + Δ − − + Δ − + − Δ −
⎣ ⎦= ×
Δ+ Δ + + + Δ +
Δ
= × =
Δ+ Δ + + + Δ + + Δ + + + Δ +
= = =
+ Δ + + + Δ + + + +
56.
( )2 21 1x x x
y x x x
x x
+ Δ − −
−
Δ + Δ=
Δ Δ
( ) ( )( )
( )
( )( ) ( )
( )
2 2
2 2 3 2
2
22 2
2 2
2 2
2 20
1 1
2 1
1 1
1 1
lim
x
x x x x x x x
x x x x
x x x x x x x x x x x
xx x x
x x x x x x x x
xx x x x x x
dy x x x x
dx x x x xΔ →
⎡ ⎤+ Δ − − + Δ −
⎢ ⎥= ×
⎢ ⎥+ Δ Δ
⎣ ⎦
⎡ ⎤+ Δ + Δ − − + Δ − − Δ
⎢ ⎥= ×
⎢ ⎥ Δ+ Δ
⎢ ⎥⎣ ⎦
Δ + Δ + Δ + Δ +
= × =
Δ+ Δ + Δ
+ Δ + +
= =
+ Δ

57.
1
(0) – ; (2) 1
2
f f′ ′≈ ≈
2
(5) ; (7) –3
3
f f′ ′≈ ≈
58. (–1) 2; (1) 0g g′ ′≈ ≈
1
(4) –2; (6) –
3
g g′ ′≈ ≈
59.
60.
61. a.
5 3
(2) ; (2)
2 2
f f ′≈ ≈
(0.5) 1.8; (0.5) –0.6f f ′≈ ≈
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.
3
0.7,
2
x ≈ − and 5 < x < 7
62. The derivative fails to exist at the corners of the
graph; that is, at 80,60,55,15,10=t . The
derivative exists at all other points on the interval
)85,0( .
63. The derivative is 0 at approximately 15=t and
201=t . The greatest rate of increase occurs at
about 61=t and it is about 0.5 degree F per day.
The greatest rate of decrease occurs at about
320=t 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).
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 1.2=x , where the solid
function is zero. The solid function has a tangent
line with zero slope at about 2.1,4.0=x 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 gf =' , 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.
0
( ) – ( )
( ) lim
h
f a h f a
f a
h→
+
′ =
0
( ) ( ) – ( )
lim
h
f a f h f a
h→
=
0 0
( ) –1 ( ) – (0)
( ) lim ( ) lim
h h
f h f h f
f a f a
h h→ →
= =
( ) (0)f a f ′=
′f (a) exists since ′f (0) exists.

67. If f is differentiable everywhere, then it is
continuous everywhere, so
lim
x→2
−
f( x) = lim
x→2
−
(mx + b) = 2m + b = f (2) = 4
and b = 4 – 2m.
For f to be differentiable everywhere,
2
( ) (2)
(2) lim
2x
f x f
f
x→
−
′ =
−
must exist.
2
2 2 2
( ) (2) 4
lim lim lim ( 2) 4
2 2x x x
f x f x
x
x x+ + +→ → →
− −
= = + =
− −
2 2
( ) (2) 4
lim lim
2 2x x
f x f mx b
x x− −→ →
− + −
=
− −
2 2
4 2 4 ( 2)
lim lim
2 2x x
mx m m x
m
x x− −→ →
+ − − −
= = =
− −
Thus m = 4 and b = 4 – 2(4) = –4
68.
0
( ) – ( ) ( ) – ( – )
( ) lim
2
s
h
f x h f x f x f x h
f x
h→
+ +
=
0
( ) – ( ) ( – ) – ( )
lim
2 –2h
f x h f x f x h f x
h h→
+⎡ ⎤
= +⎢ ⎥
⎣ ⎦
0 – 0
1 ( ) – ( ) 1 [ (– )] – ( )
lim lim
2 2 –h h
f x h f x f x h f x
h h→ →
+ +
= +
1 1
( ) ( ) ( ).
2 2
f x f x f x′ ′ ′= + =
For the converse, let f (x) = x . Then
0 0
– – –
(0) lim lim 0
2 2
s
h h
h h h h
f
h h→ →
= = =
but ′f (0) does not exist.
69. 0
0
00
( ) ( )
( ) lim ,
t x
f t f x
f x
t x→
−
′ =
−
so
0
0
00
( ) ( )
( ) lim
( )t x
f t f x
f x
t x→−
− −
′ − =
− −
0
00
( ) ( )
lim
t x
f t f x
t x→−
− −
=
+
a. If f is an odd function,
0
0
0 0
( ) [ ( )]
( ) lim
t x
f t f x
f x
t x→−
− − −
′ − =
+
0
0 0
( ) ( )
lim
t x
f t f x
t x→−
+ −
=
+
.
Let u = –t. As t → −x0 , u → x0 and so
0
0
0 0
( ) ( )
( ) lim
u x
f u f x
f x
u x→
− +
′ − =
− +
0 0
0 00 0
( ) ( ) [ ( ) ( )]
lim lim
( ) ( )u x u x
f u f x f u f x
u x u x→ →
− + − −
= =
− − − −
0
0
0 0
( ) ( )
lim ( ) .
u x
f u f x
f x m
u x→
−
′= = =
−
b. If f is an even function,
0
0
0 0
( ) ( )
(– ) lim
t x
f t f x
f x
t x→−
−
′ =
+
. Let u = –t, as
above, then 0
0
0 0
( ) ( )
( ) lim
u x
f u f x
f x
u x→
− −
′ − =
− +
0 0
0 00 0
( ) ( ) ( ) ( )
lim lim
( )u x u x
f u f x f u f x
u x u x→ →
− −
= = −
− − −
= − ′f (x0 ) = −m.
70. Say f(–x) = –f(x). Then
0
(– ) – (– )
(– ) lim
h
f x h f x
f x
h→
+
′ =
0 0
– ( – ) ( ) ( – ) – ( )
lim – lim
h h
f x h f x f x h f x
h h→ →
+
= =
– 0
[ (– )] ( )
lim ( )
–h
f x h f x
f x
h→
+ −
′= = so ′f (x) is
an even function if f(x) is an odd function.
Say f(–x) = f(x). Then
0
(– ) – (– )
(– ) lim
h
f x h f x
f x
h→
+
′ =
0
( – ) – ( )
lim
h
f x h f x
h→
=
– 0
[ (– )] – ( )
– lim – ( )
–h
f x h f x
f x
h→
+
′= = so ′f (x)
is an odd function if f(x) is an even function.
71.
a.
8
0
3
x< < ;
8
0,
3
⎛ ⎞
⎜ ⎟
⎝ ⎠
b.
8
0
3
x≤ ≤ ;
8
0,
3
⎡ ⎤
⎢ ⎥
⎣ ⎦
c. A function f(x) decreases as x increases when
′f (x) < 0.
72.
a. 6.8xπ < < b. 6.8xπ < <
c. A function f(x) increases as x increases when
′f (x) > 0.

2.3 Concepts Review
1. the derivative of the second; second;
f (x) ′g (x ) + g(x) ′f ( x)
2. denominator; denominator; square of the
denominator;
2
( ) ( ) – ( ) ( )
( )
g x f x f x g x
g x
′ ′
3. nx n–1
h; nxn–1
4. kL(f); L(f) + L(g); Dx
Problem Set 2.3
1. 2 2
(2 ) 2 ( ) 2 2 4x xD x D x x x= = ⋅ =
2. 3 3 2 2
(3 ) 3 ( ) 3 3 9x xD x D x x x= = ⋅ =
3. ( ) ( ) 1x xD x D xπ = π = π⋅ = π
4. 3 3 2 2
( ) ( ) 3 3x xD x D x x xπ = π = π⋅ = π
5. –2 –2 –3 –3
(2 ) 2 ( ) 2(–2 ) –4x xD x D x x x= = =
6. –4 –4 –5 –5
(–3 ) –3 ( ) –3(–4 ) 12x xD x D x x x= = =
7. –1 –2 –2
( ) (–1 ) –x xD D x x x
x
π⎛ ⎞
= π = π = π⎜ ⎟
⎝ ⎠
= –
π
x2
8. –3 –4 –4
3
( ) (–3 ) –3x xD D x x x
x
α
α α α
⎛ ⎞
= = =⎜ ⎟
⎝ ⎠
4
3
–
x
α
=
9. –5 –6
5
100
100 ( ) 100(–5 )x xD D x x
x
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
–6
6
500
–500 –x
x
= =
10. –5 –6
5
3 3 3
( ) (–5 )
4 44
x xD D x x
x
α α α⎛ ⎞
= =⎜ ⎟
⎝ ⎠
–6
6
15 15
– –
4 4
x
x
α α
= =
11. 2 2
( 2 ) ( ) 2 ( ) 2 2x x xD x x D x D x x+ = + = +
12. 4 3 4 3
(3 ) 3 ( ) ( )x x xD x x D x D x+ = +
3 2 3 2
3(4 ) 3 12 3x x x x= + = +
13. 4 3 2
( 1)xD x x x x+ + + +
4 3 2
( ) ( ) ( ) ( ) (1)x x x x xD x D x D x D x D= + + + +
3 2
4 3 2 1x x x= + + +
14. 4 3 2 2
(3 – 2 – 5 )xD x x x x+ π + π
4 3 2
2
3 ( ) – 2 ( ) – 5 ( )
( ) ( )
x x x
x x
D x D x D x
D x D
=
+ π + π
3 2
3(4 ) – 2(3 ) – 5(2 ) (1) 0x x x= + π +
3 2
12 – 6 –10x x x= + π
15. 7 5 –2
( – 2 – 5 )xD x x xπ
7 5 –2
( ) – 2 ( ) – 5 ( )x x xD x D x D x= π
6 4 –3
(7 ) – 2(5 ) – 5(–2 )x x x= π
6 4 –3
7 –10 10x x x= π +
16. 12 2 10
( 5 )xD x x x− −
+ − π
12 2 10
( ) 5 ( ) ( )x x xD x D x D x− −
= + − π
11 3 11
12 5( 2 ) ( 10 )x x x− −
= + − − π −
11 3 11
12 10 10x x x− −
= − + π
17. –4 –3 –4
3
3
3 ( ) ( )x x xD x D x D x
x
⎛ ⎞
+ = +⎜ ⎟
⎝ ⎠
–4 –5 –5
4
9
3(–3 ) (–4 ) – – 4x x x
x
= + =
18. –6 –1 –6 –1
(2 ) 2 ( ) ( )x x xD x x D x D x+ = +
–7 –2 –7 –2
2(–6 ) (–1 ) –12 –x x x x= + =
19. –1 –2
2
2 1
– 2 ( ) – ( )x x xD D x D x
x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
–2 –3
2 3
2 2
2(–1 ) – (–2 ) –x x
x x
= = +
20. –3 –4
3 4
3 1
– 3 ( ) – ( )x x xD D x D x
x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
–4 –5
4 5
9 4
3(–3 ) – (–4 ) –x x
x x
= = +
21. –11 1
2 ( ) 2 ( )
2 2
x x xD x D x D x
x
⎛ ⎞
+ = +⎜ ⎟
⎝ ⎠
–2
2
1 1
(–1 ) 2(1) – 2
2 2
x
x
= + = +

22. –12 2 2 2
– ( ) –
3 3 3 3
x x xD D x D
x
⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
–2
2
2 2
(–1 ) – 0 –
3 3
x
x
= =
23. 2 2 2
[ ( 1)] ( 1) ( 1) ( )x x xD x x x D x x D x+ = + + +
2 2
(2 ) ( 1)(1) 3 1x x x x= + + = +
24. 3 3 3
[3 ( –1)] 3 ( –1) ( –1) (3 )x x xD x x x D x x D x= +
2 3 3
3 (3 ) ( –1)(3) 12 – 3x x x x= + =
25. 2
[(2 1) ]xD x +
(2 1) (2 1) (2 1) (2 1)x xx D x x D x= + + + + +
(2 1)(2) (2 1)(2) 8 4x x x= + + + = +
26. 2
[(–3 2) ]xD x +
(–3 2) (–3 2) (–3 2) (–3 2)x xx D x x D x= + + + + +
= (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 12
27. 2 3
[( 2)( 1)]xD x x+ +
2 3 3 2
( 2) ( 1) ( 1) ( 2)x xx D x x D x= + + + + +
2 2 3
( 2)(3 ) ( 1)(2 )x x x x= + + +
4 2 4
3 6 2 2x x x x= + + +
4 2
5 6 2x x x= + +
28. 4 2
[( –1)( 1)]xD x x +
4 2 2 4
( –1) ( 1) ( 1) ( –1)x xx D x x D x= + + +
4 2 3
( –1)(2 ) ( 1)(4 )x x x x= + +
5 5 3 5 3
2 – 2 4 4 6 4 – 2x x x x x x x= + + = +
29. 2 3
[( 17)( – 3 1)]xD x x x+ +
2 3 3 2
( 17) ( – 3 1) ( – 3 1) ( 17)x xx D x x x x D x= + + + + +
2 2 3
( 17)(3 – 3) ( – 3 1)(2 )x x x x x= + + +
4 2 4 2
3 48 – 51 2 – 6 2x x x x x= + + +
4 2
5 42 2 – 51x x x= + +
30. 4 3 2
[( 2 )( 2 1)]xD x x x x+ + + 4 3 2 3 2 4
( 2 ) ( 2 1) ( 2 1) ( 2 )x xx x D x x x x D x x= + + + + + + +
4 2 3 2 3
( 2 )(3 4 ) ( 2 1)(4 2)x x x x x x x= + + + + + +
6 5 3 2
7 12 12 12 2x x x x= + + + +
31. 2 2
[(5 – 7)(3 – 2 1)]xD x x x + 2 2 2 2
(5 – 7) (3 – 2 1) (3 – 2 1) (5 – 7)x xx D x x x x D x= + + +
2 2
(5 – 7)(6 – 2) (3 – 2 1)(10 )x x x x x= + +
3 2
60 – 30 – 32 14x x x= +
32. 2 4
[(3 2 )( – 3 1)]xD x x x x+ + 2 4 4 2
(3 2 ) ( – 3 1) ( – 3 1) (3 2 )x xx x D x x x x D x x= + + + + +
2 3 4
(3 2 )(4 – 3) ( – 3 1)(6 2)x x x x x x= + + + +
5 4 2
18 10 – 27 – 6 2x x x x= + +
33.
2 2
2 2 2
(3 1) (1) – (1) (3 1)1
3 1 (3 1)
x x
x
x D D x
D
x x
+ +⎛ ⎞
=⎜ ⎟
+ +⎝ ⎠
2
2 2 2 2
(3 1)(0) – (6 ) 6
–
(3 1) (3 1)
x x x
x x
+
= =
+ +
34.
2 2
2 2 2
(5 –1) (2) – (2) (5 –1)2
5 –1 (5 –1)
x x
x
x D D x
D
x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
2
2 2 2 2
(5 –1)(0) – 2(10 ) 20
–
(5 –1) (5 –1)
x x x
x x
= =

35.
2
1
4 – 3 9
xD
x x
⎛ ⎞
⎜ ⎟
+⎝ ⎠
2 2
2 2
(4 – 3 9) (1) – (1) (4 – 3 9)
(4 – 3 9)
x xx x D D x x
x x
+ +
=
+
2
2 2 2 2
(4 – 3 9)(0) – (8 – 3) 8 3
–
(4 – 3 9) (4 – 3 9)
x x x x
x x x x
+ −
= =
+ +
2 2
8 3
(4 – 3 9)
x
x x
− +
=
+
36.
3
4
2 – 3
xD
x x
⎛ ⎞
⎜ ⎟
⎝ ⎠
3 3
3 2
(2 – 3 ) (4) – (4) (2 – 3 )
(2 – 3 )
x xx x D D x x
x x
=
3 2 2
3 2 3 2
(2 – 3 )(0) – 4(6 – 3) –24 12
(2 – 3 ) (2 – 3 )
x x x x
x x x x
+
= =
37.
2
( 1) ( –1) – ( –1) ( 1)–1
1 ( 1)
x x
x
x D x x D xx
D
x x
+ +⎛ ⎞
=⎜ ⎟
+⎝ ⎠ +
2 2
( 1)(1) – ( –1)(1) 2
( 1) ( 1)
x x
x x
+
= =
+ +
38.
2 –1
–1
x
x
D
x
⎛ ⎞
⎜ ⎟
⎝ ⎠ 2
( –1) (2 –1) – (2 –1) ( –1)
( –1)
x xx D x x D x
x
=
2 2
( –1)(2) – (2 –1)(1) 1
–
( –1) ( –1)
x x
x x
= =
39.
2
2 –1
3 5
x
x
D
x
⎛ ⎞
⎜ ⎟
⎜ ⎟+⎝ ⎠
2 2
2
(3 5) (2 –1) – (2 –1) (3 5)
(3 5)
x xx D x x D x
x
+ +
=
+
2
2
(3 5)(4 ) – (2 –1)(3)
(3 5)
x x x
x
+
=
+
2
2
6 20 3
(3 5)
x x
x
+ +
=
+
40.
2
5 – 4
3 1
x
x
D
x
⎛ ⎞
⎜ ⎟
+⎝ ⎠
2 2
2 2
(3 1) (5 – 4) – (5 – 4) (3 1)
(3 1)
x xx D x x D x
x
+ +
=
+
2
2 2
(3 1)(5) – (5 – 4)(6 )
(3 1)
x x x
x
+
=
+
2
2 2
15 24 5
(3 1)
x x
x
− + +
=
+
41.
2
2 – 3 1
2 1
x
x x
D
x
⎛ ⎞+
⎜ ⎟
⎜ ⎟+⎝ ⎠
2 2
2
(2 1) (2 – 3 1) – (2 – 3 1) (2 1)
(2 1)
x xx D x x x x D x
x
+ + + +
=
+
2
2
(2 1)(4 – 3) – (2 – 3 1)(2)
(2 1)
x x x x
x
+ +
=
+
2
2
4 4 – 5
(2 1)
x x
x
+
=
+

42.
2
5 2 – 6
3 –1
x
x x
D
x
⎛ ⎞+
⎜ ⎟
⎜ ⎟
⎝ ⎠
2 2
2
(3 –1) (5 2 – 6) – (5 2 – 6) (3 –1)
(3 –1)
x xx D x x x x D x
x
+ +
=
2
2
(3 –1)(10 2) – (5 2 – 6)(3)
(3 –1)
x x x x
x
+ +
=
2
2
15 –10 16
(3 –1)
x x
x
+
=
43.
2
2
– 1
1
x
x x
D
x
⎛ ⎞+
⎜ ⎟
⎜ ⎟+⎝ ⎠
2 2 2 2
2 2
( 1) ( – 1) – ( – 1) ( 1)
( 1)
x xx D x x x x D x
x
+ + + +
=
+
2 2
2 2
( 1)(2 –1) – ( – 1)(2 )
( 1)
x x x x x
x
+ +
=
+
2
2 2
–1
( 1)
x
x
=
+
44.
2
2
– 2 5
2 – 3
x
x x
D
x x
⎛ ⎞+
⎜ ⎟
⎜ ⎟+⎝ ⎠
2 2 2 2
2 2
( 2 – 3) ( – 2 5) – ( – 2 5) ( 2 – 3)
( 2 – 3)
x xx x D x x x x D x x
x x
+ + + +
=
+
2 2
2 2
( 2 – 3)(2 – 2) – ( – 2 5)(2 2)
( 2 – 3)
x x x x x x
x x
+ + +
=
+
2
2 2
4 –16 – 4
( 2 – 3)
x x
x x
=
+
45. a. ( ) (0) (0) (0) (0) (0)f g f g g f′ ′ ′⋅ = +
= 4(5) + (–3)(–1) = 23
b. ( ) (0) (0) (0) –1 5 4f g f g′ ′ ′+ = + = + =
c.
2
(0) (0) – (0) (0)
( ) (0)
(0)
g f f g
f g
g
′ ′
′ =
2
–3(–1) – 4(5) 17
–
9(–3)
= =
46. a. ( – ) (3) (3) – (3) 2 – (–10) 12f g f g′ ′ ′= = =
b. ( ) (3) (3) (3) (3) (3)f g f g g f′ ′ ′⋅ = + = 7(–10) + 6(2) = –58
c.
2
(3) (3) – (3) (3)
( ) (3)
(3)
f g g f
g f
f
′ ′
′ =
2
7(–10) – 6(2) 82
–
49(7)
= =
47. 2
[ ( )] [ ( ) ( )]x xD f x D f x f x=
( ) [ ( )] ( ) [ ( )]x xf x D f x f x D f x= +
2 ( ) ( )xf x D f x= ⋅ ⋅
48. [ ( ) ( ) ( )] ( ) [ ( ) ( )] ( ) ( ) ( )x x xD f x g x h x f x D g x h x g x h x D f x= +
( )[ ( ) ( ) ( ) ( )] ( ) ( ) ( )x x xf x g x D h x h x D g x g x h x D f x= + +
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x x xf x g x D h x f x h x D g x g x h x D f x= + +

49. 2
( – 2 2) 2 – 2xD x x x+ =
At x = 1: mtan = 2(1) – 2 = 0
Tangent line: y = 1
50.
2 2
2 2 2
( 4) (1) – (1) ( 4)1
4 ( 4)
x x
x
x D D x
D
x x
+ +⎛ ⎞
=⎜ ⎟
+ +⎝ ⎠
2
2 2 2 2
( 4)(0) – (2 ) 2
–
( 4) ( 4)
x x x
x x
+
= =
+ +
At x = 1: tan 2 2
2(1) 2
–
25(1 4)
m = − =
+
Tangent line:
1 2
– – ( –1)
5 25
y x=
2 7
–
25 25
y x= +
51. 3 2 2
( – ) 3 – 2xD x x x x=
The tangent line is horizontal when mtan = 0:
2
tan 3 – 2 0m x x= =
(3 2) 0x x − =
x = 0 and x =
2
3
(0, 0) and
2 4
, –
3 27
⎛ ⎞
⎜ ⎟
⎝ ⎠
52. 3 2 21
– 2 –1
3
xD x x x x x
⎛ ⎞
+ = +⎜ ⎟
⎝ ⎠
2
tan 2 –1 1m x x= + =
2
2 – 2 0x x+ =
–2 4 – 4(1)(–2) –2 12
2 2
x
± ±
= =
–1– 3, –1 3= +
–1 3x = ±
5 5
1 3, 3 , 1 3, 3
3 3
⎛ ⎞ ⎛ ⎞
− + − − − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
53. 5 5
6
100/ 100
' 500
y x x
y x
−
−
= =
= −
Set 'y equal to 1− , the negative reciprocal of
the slope of the line xy = . Solving for x gives
1/ 6
5/ 6
500 2.817
100(500) 0.563
x
y −
= ± ≈ ±
= ± ≈ ±
The points are (2.817,0.563) and
)563.0,817.2( −− .
54. Proof #1:
[ ] [ ]
[ ] [ ]
( ) ( ) ( ) ( 1) ( )
( ) ( 1) ( )
( ) ( )
x x
x x
x x
D f x g x D f x g x
D f x D g x
D f x D g x
− = + −
= + −
= −
Proof #2:
Let ( ) ( ) ( )F x f x g x= − . Then
[ ] [ ]
0
0
( ) ( ) ( ) ( )
'( ) lim
( ) ( ) ( ) ( )
lim
'( ) '( )
h
h
f x h g x h f x g x
F x
h
f x h f x g x h g x
h h
f x g x
→
→
+ − + − −
=
+ − + −⎡ ⎤
= −⎢ ⎥
⎣ ⎦
= −
55. a. 2
(–16 40 100) –32 40tD t t t+ + = +
v = –32(2) + 40 = –24 ft/s
b. v = –32t + 40 = 0
t = 5
4
s
56. 2
(4.5 2 ) 9 2tD t t t+ = +
9t + 2 = 30
t =
28
9
s
57. 2
tan (4 – ) 4 – 2xm D x x x= =
The line through (2,5) and (x0, y0) has slope
0
0
5
.
2
y
x
−
−
2
0 0
0
0
4 – – 5
4 – 2
– 2
x x
x
x
=
2 2
0 0 0 0–2 8 – 8 – 4 – 5x x x x+ = +
2
0 0– 4 3 0x x + =
0 0( – 3)( –1) 0x x =
x0 = 1, x0 = 3
At x0 = 1: 2
0 4(1) – (1) 3y = =
tan 4 – 2(1) 2m = =
Tangent line: y – 3 = 2(x – 1); y = 2x + 1
At 2
0 03: 4(3) – (3) 3x y= = =
tan 4 – 2(3) –2m = =
Tangent line: y – 3 = –2(x – 3); y = –2x + 9

58. 2
( ) 2xD x x=
The line through (4, 15) and 0 0( , )x y has slope
0
0
15
.
4
y
x
−
−
If (x0, y0) is on the curve y = x 2
, then
2
0
tan 0
0
–15
2
– 4
x
m x
x
= = .
2 2
0 0 02 – 8 –15x x x=
2
0 0– 8 15 0x x + =
0 0( – 3)( – 5) 0x x =
At 2
0 03: (3) 9x y= = =
She should shut off the engines at (3, 9). (At
x0 = 5 she would not go to (4, 15) since she is
moving left to right.)
59. 2
(7 – ) –2xD x x=
The line through (4, 0) and 0 0( , )x y has
slope 0
0
0
.
4
y
x
−
−
If the fly is at 0 0( , )x y when the
spider sees it, then
2
0
tan 0
0
7 – – 0
–2
– 4
x
m x
x
= = .
2 2
0 0 0–2 8 7 –x x x+ =
x0
2 – 8x0 + 7 = 0
0 0( – 7)( –1) 0x x =
At 0 01: 6x y= =
2 2
(4 –1) (0 – 6) 9 36 45 3 5d = + = + = =
≈ 6. 7
They are 6.7 units apart when they see each
other.
60. P(a, b) is
1
, .a
a
⎛ ⎞
⎜ ⎟
⎝ ⎠ 2
1
–xD y
x
= so the slope of
the tangent line at P is
2
1
– .
a
The tangent line is
2
1 1
– – ( – )y x a
a a
= or
2
1
– ( – 2 )y x a
a
= which
has x-intercept (2a, 0).
2 2
2 2
1 1
( , ) , ( , ) ( – 2 )d O P a d P A a a
a a
= + = +
2
2
1
( , )a d O P
a
= + = so AOP is an isosceles
triangle. The height of AOP is a while the base,
OA has length 2a, so the area is
1
2
(2a)(a) = a2
.
61. The watermelon has volume
4
3
πr3
; the volume
of the rind is
3
3 34 4 271
– – .
3 3 10 750
r
V r r r
⎛ ⎞
= π π = π⎜ ⎟
⎝ ⎠
At the end of the fifth week r = 10, so
2 2271 271 542
(10) 340
250 250 5
rD V r
π
= π = π = ≈ cm3
per cm of radius growth. Since the radius is
growing 2 cm per week, the volume of the rind is
growing at the rate of 3542
(2) 681 cm
5
π
≈ per
week.
2.4 Concepts Review
1.
sin( ) – sin( )x h x
h
+
2. 0; 1
3. cos x; –sin x
4.
1 3 1
cos ; – –
3 2 2 2 3
y x
π π⎛ ⎞
= = ⎜ ⎟
⎝ ⎠
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. 2
(sin ) sin (sin ) sin (sin )x x xD x x D x x D x= +
= sin x cos x + sin x cos x = 2 sin x cos x = sin 2x
3. 2 2
(sin cos ) (1) 0x xD x x D+ = =
4. 2 2
(1– cos ) (sin )x xD x D x=
sin (sin ) sin (sin )x xx D x x D x= +
= sin x cos x + sin x cos x
= 2 sin x cos x = sin 2x
5.
2
1
(sec )
cos
cos (1) – (1) (cos )
cos
x x
x x
D x D
x
x D D x
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
2
sin 1 sin
sec tan
cos coscos
x x
x x
x xx
= = ⋅ =

6.
2
1
(csc )
sin
sin (1) (1) (sin )
sin
x x
x x
D x D
x
x D D x
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
−
=
2
– cos –1 cos
– csc cot
sin sinsin
x x
x x
x xx
= = ⋅ =
7.
2
sin
(tan )
cos
cos (sin ) sin (cos )
cos
x x
x x
x
D x D
x
x D x x D x
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
−
=
2 2
2
2 2
cos sin 1
sec
cos cos
x x
x
x x
+
= = =
8.
2
cos
(cot )
sin
sin (cos ) cos (sin )
sin
x x
x x
x
D x D
x
x D x x D x
x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
−
=
2 2 2 2
2 2
sin – cos –(sin cos )
sin sin
x x x x
x x
− +
= =
2
2
1
– – csc
sin
x
x
= =
9.
2
sin cos
cos
cos (sin cos ) (sin cos ) (cos )
cos
x
x x
x x
D
x
x D x x x x D x
x
+⎛ ⎞
⎜ ⎟
⎝ ⎠
+ − +
=
2
2
cos (cos – sin ) – (–sin – sin cos )
cos
x x x x x x
x
=
2 2
2
2 2
cos sin 1
sec
cos cos
x x
x
x x
+
= = =
10.
2
sin cos
tan
tan (sin cos ) (sin cos ) (tan )
tan
x
x x
x x
D
x
x D x x x x D x
x
+⎛ ⎞
⎜ ⎟
⎝ ⎠
+ − +
=
2
2
tan (cos – sin ) – sec (sin cos )
tan
x x x x x x
x
+
=
2 2
2 2
2 2
2 2
sin sin 1 sin
sin – – –
cos coscos cos
sin sin 1 cos
sin
cos coscos sin
x x x
x
x xx x
x x x
x
x xx x
⎛ ⎞ ⎛ ⎞
= ÷⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞⎛ ⎞
= − − −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
2
2
cos 1 cos
cos
sin sin sin
x x
x
x x x
= − − −
11. ( ) [ ] [ ]
( ) ( ) 2 2
sin cos sin cos cos sin
sin sin cos cos cos sin
x x xD x x xD x xD x
x x x x x x
= +
= − + = −
12. ( ) [ ] [ ]
( ) ( )
( )
2
2
sin tan sin tan tan sin
sin sec tan cos
1 sin
sin cos
coscos
tan sec sin
x x xD x x xD x xD x
x x x x
x
x x
xx
x x x
= +
= +
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
= +
13.
( ) ( )
2
2
sin sinsin
cos sin
x x
x
xD x xD xx
D
x x
x x x
x
−⎛ ⎞
=⎜ ⎟
⎝ ⎠
−
=
14.
( ) ( ) ( )
2
2
1 cos 1 cos1 cos
sin cos 1
x x
x
xD x x D xx
D
x x
x x x
x
− − −−⎛ ⎞
=⎜ ⎟
⎝ ⎠
+ −
=
15. 2 2 2
2
( cos ) (cos ) cos ( )
sin 2 cos
x x xD x x x D x x D x
x x x x
= +
= − +
16. 2
2 2
2 2
cos sin
1
( 1) ( cos sin ) ( cos sin ) ( 1)
( 1)
x
x x
x x x
D
x
x D x x x x x x D x
x
+⎛ ⎞
⎜ ⎟
+⎝ ⎠
+ + − + +
=
+
2
2 2
( 1)(– sin cos cos ) – 2 ( cos sin )
( 1)
x x x x x x x x x
x
+ + + +
=
+
3
2 2
– sin – 3 sin 2cos
( 1)
x x x x x
x
+
=
+

17. 2
2 2
2
tan (tan )(tan )
(tan )(sec ) (tan )(sec )
2tan sec
x
y x x x
D y x x x x
x x
= =
= +
=
18. 3 2
2 2
3
3 3
2
sec (sec )(sec )
(sec )sec tan (sec ) (sec )
sec tan sec (sec sec tan
sec sec tan )
sec tan 2sec tan
3sec tan
x x
y x x x
D y x x x x D x
x x x x x x
x x x
x x x x
x x
= =
= +
= + ⋅
+ ⋅
= +
=
19. Dx(cos x) = –sin x
At x = 1: tan –sin1 –0.8415m = ≈
y = cos 1 ≈ 0.5403
Tangent line: y – 0.5403 = –0.8415(x – 1)
20. 2
(cot ) – cscxD x x=
tanAt : –2;
4
x m
π
= =
y = 1
Tangent line: –1 –2 –
4
y x
π⎛ ⎞
= ⎜ ⎟
⎝ ⎠
21.
2 2
sin 2 (2sin cos )
2 sin cos cos sin
2sin 2cos
x x
x x
D x D x x
x D x x D x
x x
=
= +⎡ ⎤⎣ ⎦
= − +
22. 2 2
cos2 (2cos 1) 2 cos 1
2sin cos
x x x xD x D x D x D
x x
= − = −
= −
23.
( )2 2
(30sin 2 ) 30 (2sin cos )
30 2sin 2cos
60cos2
t tD t D t t
t t
t
=
= − +
=
30sin 2 15
1
sin 2
2
2
6 12
t
t
t t
π π
=
=
= → =
At ; 60cos 2 30 3 ft/sec
12 12
t
ππ ⎛ ⎞
= ⋅ =⎜ ⎟
⎝ ⎠
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. 20cos , 20sin (10 3,10)
6 6
π π⎛ ⎞
=⎜ ⎟
⎝ ⎠
≈ (17.32, 10)
b. Dt(20 sin t) = 20 cos t
At :rate 20cos 10 3
6 6
t
π π
= = = ≈ 17.32 ft/s
c. The fastest rate 20 cos t can obtain is
20 ft/s.
25.
2
tan
' sec
y x
y x
=
=
When 0y = , tan 0 0y = = and 2
' sec 0 1y = = .
The tangent line at 0x = is y x= .
26. 2
2 2
2
tan (tan )(tan )
' (tan )(sec ) (tan )(sec )
2tan sec
y x x x
y x x x x
x x
= =
= +
=
Now, 2
sec x is never 0, but tan 0x = at
x kπ= where k is an integer.
27.
[ ]
[ ]
2 2
9sin cos
' 9 sin ( sin ) cos (cos )
9 sin cos
9 cos2
y x x
y x x x x
x x
x
=
= − +
⎡ ⎤= −
⎣ ⎦
= −
The tangent line is horizontal when ' 0y = or, in
this case, where cos2 0x = . This occurs when
4 2
x k
π π
= + where k is an integer.
28. ( ) sin
'( ) 1 cos
f x x x
f x x
= −
= −
'( ) 0f x = when cos 1x = ; i.e. when 2x kπ=
where k is an integer.
'( ) 2f x = when (2 1)x k π= + where k is an
integer.
29. The curves intersect when 2 sin 2 cos ,x x=
sin x = cos x at x = π
4
for 0 < x < π
2
.
( 2 sin ) 2 cosxD x x= ; 2 cos 1
4
π
=
( 2 cos ) – 2 sinxD x x= ; 2 sin 1
4
π
− = −
1(–1) = –1 so the curves intersect at right angles.
30. v = (3sin 2 ) 6cos2tD t t=
At t = 0: v = 6 cm/s
t =
π
2
: v = 6− cm/s
t π= : v = 6 cm/s

31.
2 2
2
0
2 2 2
0
sin( ) – sin
(sin ) lim
sin( 2 ) – sin
lim
x
h
h
x h x
D x
h
x xh h x
h
→
→
+
=
+ +
=
2 2 2 2 2
0
sin cos(2 ) cos sin(2 ) – sin
lim
h
x xh h x xh h x
h→
+ + +
=
2 2 2 2
0
sin [cos(2 ) –1] cos sin(2 )
lim
h
x xh h x xh h
h→
+ + +
=
2 2
2 2
2 20
cos(2 ) –1 sin(2 )
lim(2 ) sin cos
2 2h
xh h xh h
x h x x
xh h xh h→
⎡ ⎤+ +
= + +⎢ ⎥
+ +⎢ ⎥⎣ ⎦
2 2 2
2 (sin 0 cos 1) 2 cosx x x x x= ⋅ + ⋅ =
32.
0
sin(5( )) – sin5
(sin5 ) limx
h
x h x
D x
h→
+
=
0
sin(5 5 ) – sin5
lim
h
x h x
h→
+
=
0
sin5 cos5 cos5 sin5 – sin5
lim
h
x h x h x
h→
+
=
0
cos5 –1 sin5
lim sin5 cos5
h
h h
x x
h h→
⎡ ⎤
= +⎢ ⎥
⎣ ⎦
0
cos5 –1 sin5
lim 5sin5 5cos5
5 5h
h h
x x
h h→
⎡ ⎤
= +⎢ ⎥
⎣ ⎦
0 5cos5 1 5cos5x x= + ⋅ =
33. f(x) = x sin x
a.
b. f(x) = 0 has 6 solutions on [ ,6 ]π π
′f (x) = 0 has 5 solutions on [ ,6 ]π π
c. f(x) = x sin x is a counterexample.
Consider the interval [ ]0,π .
( ) ( ) 0f fπ π− = = and ( ) 0f x = has
exactly two solutions in the interval (at 0 and
π ). However, ( )' 0f x = 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
[ ,6 ]π π is about 24.93.
34. ( ) 3 2
cos 1.25cos 0.225f x x x= − +
0 1.95x ≈
′f (x0) ≈ –1.24
2.5 Concepts Review
1. ; ( ( )) ( )tD u f g t g t′ ′
2. ; ( ( )) ( )vD w G H s H s′ ′
3. 2 2
( ( )) ;( ( ))f x f x
4.
2 2
2 cos( );6(2 1)x x x +
Problem Set 2.5
1. y = u15
and u = 1 + x
x u xD y D y D u= ⋅
14
(15 )(1)u=
14
15(1 )x= +
2. y = u5
and u = 7 + x
= (5u4
)(1)
4
5(7 )x= +
3. y = u5
and u = 3 – 2x
4 4
(5 )(–2) –10(3 – 2 )u x= =

4. y = u7
and 2
4 2u x= +
6 2 6
(7 )(4 ) 28 (4 2 )u x x x= = +
5. y = u11
and 3 2
– 2 3 1u x x x= + +
10 2
(11 )(3 – 4 3)u x x= +
2 3 2 10
11(3 – 4 3)( – 2 3 1)x x x x x= + + +
6. –7 2
and – 1y u u x x= = +
–8
(–7 )(2 –1)u x=
2 –8
–7(2 –1)( – 1)x x x= +
7. –5
and 3y u u x= = +
–6 –6
6
5
(–5 )(1) –5( 3) –
( 3)
u x
x
= = + =
+
8. –9 2
and 3 – 3y u u x x= = +
–10
(–9 )(6 1)u x= +
2 –10
–9(6 1)(3 – 3)x x x= + +
2 10
9(6 1)
–
(3 – 3)
x
x x
+
=
+
9. y = sin u and 2
u x x= +
= (cos u)(2x + 1)
2
(2 1)cos( )x x x= + +
10. y = cos u and 2
3 – 2u x x=
= (–sin u)(6x – 2)
2
–(6 – 2)sin(3 – 2 )x x x=
11. 3
y u= and u = cos x
2
(3 )(–sin )u x=
2
–3sin cosx x=
12. 4
y u= , u = sin v, and 2
3v x=
x u v xD y D y D u D v= ⋅ ⋅
3
(4 )(cos )(6 )u v x=
3 2 2
24 sin (3 )cos(3 )x x x=
13. 3 1
and
–1
x
y u u
x
+
= =
2
2
( –1) ( 1) – ( 1) ( –1)
(3 )
( –1)
x xx D x x D x
u
x
+ +
=
2 2
2 4
1 –2 6( 1)
3
–1 ( –1) ( –1)
x x
x x x
⎛ ⎞+ +⎛ ⎞
= = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
14. 3
y u−
= and
2x
u
x
−
=
− π
4
2
( ) ( 2) ( 2) ( )
3 )
( )
x xx D x x D x
u
x
− − π − − − − π
= (− ⋅
− π
4 2
2 4
2 (2 ) ( )
3 3 (2 )
( ) ( 2)
x x
x x x
−
− − π − π⎛ ⎞
= − = − − π⎜ ⎟
− π⎝ ⎠ − π −
15. y = cos u and
2
3
2
x
u
x
=
+
2 2
2
( 2) (3 ) – (3 ) ( 2)
(–sin )
( 2)
x xx D x x D x
u
x
+ +
=
+
2 2
2
3 ( 2)(6 ) – (3 )(1)
–sin
2 ( 2)
x x x x
x x
⎛ ⎞ +
= ⎜ ⎟
⎜ ⎟+ +⎝ ⎠
2 2
2
3 12 3
– sin
2( 2)
x x x
xx
⎛ ⎞+
= ⎜ ⎟
⎜ ⎟++ ⎝ ⎠
16.
2
3
, cos , and
1–
x
y u u v v
x
= = =
x u v xD y D y D u D v= ⋅ ⋅
2 2
2
2
(1– ) ( ) – ( ) (1 )
(3 )( sin )
(1– )
x xx D x x D x
u v
x
−
= −
2 2 2
2
2
(1– )(2 ) – ( )(–1)
–3cos sin
1– 1– (1– )
x x x x x
x x x
⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2 2 2
2
2
–3(2 – )
cos sin
1– 1–(1– )
x x x x
x xx
⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

17. 2 2 2 2 2 2 2 2 2
[(3 – 2) (3 – ) ] (3 – 2) (3 – ) (3 – ) (3 – 2)x x xD x x x D x x D x= +
2 2 2 2
(3 – 2) (2)(3 – )(–2 ) (3 – ) (2)(3 – 2)(3)x x x x x= +
2 2
2(3 2)(3 )[(3 2)( 2 ) (3 )(3)]x x x x x= − − − − + − 2 2
2(3 2)(3 )(9 4 9 )x x x x= − − + −
18. 2 4 7 3 2 4 7 3 7 3 2 4
[(2 – 3 ) ( 3) ] (2 – 3 ) ( 3) ( 3) (2 – 3 )x x xD x x x D x x D x+ = + + +
2 4 7 2 6 7 3 2 3
(2 – 3 ) (3)( 3) (7 ) ( 3) (4)(2 – 3 ) (–6 )x x x x x x= + + + 2 3 7 2 7 5
3 (3 – 2) ( 3) (29 –14 24)x x x x x= + +
19.
2 22
2
(3 – 4) ( 1) – ( 1) (3 – 4)( 1)
3 – 4 (3 – 4)
x x
x
x D x x D xx
D
x x
⎡ ⎤ + ++
=⎢ ⎥
⎢ ⎥⎣ ⎦
2 2
2 2
(3 – 4)(2)( 1)(1) – ( 1) (3) 3 – 8 –11
(3 – 4) (3 – 4)
x x x x x
x x
+ +
= =
2
( 1)(3 11)
(3 4)
x x
x
+ −
=
−
20.
2 2 2 2
2 2 2 4
( 4) (2 – 3) – (2 – 3) ( 4)2 – 3
( 4) ( 4)
x x
x
x D x x D xx
D
x x
⎡ ⎤ + +
=⎢ ⎥
+ +⎢ ⎥⎣ ⎦
2 2 2
2 4
( 4) (2) – (2 – 3)(2)( 4)(2 )
( 4)
x x x x
x
+ +
=
+
2
2 3
6 12 8
( 4)
x x
x
− + +
=
+
21. ( )( ) ( )( ) ( )44242442 2222
+=+=
′
++=′ xxxxxxy
22. ( )( ) ( )( )xxxxxxxy cos1sin2sinsin2 ++=′++=′
23.
3 2
2
( 5) (3 – 2) – (3 – 2) ( 5)3 – 2 3 – 2
` 3
5 5 ( 5)
t t
t
t D t t D tt t
D
t t t
+ +⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟
+ +⎝ ⎠ ⎝ ⎠ +
2
2
3 – 2 ( 5)(3) – (3 – 2)(1)
3
5 ( 5)
t t t
t t
+⎛ ⎞
= ⎜ ⎟
+⎝ ⎠ +
2
4
51(3 – 2)
( 5)
t
t
=
+
24.
2 22
2
( 4) ( – 9) – ( – 9) ( 4)– 9
4 ( 4)
s s
s
s D s s D ss
D
s s
⎛ ⎞ + +
=⎜ ⎟
⎜ ⎟+ +⎝ ⎠
2 2
2 2
( 4)(2 ) – ( – 9)(1) 8 9
( 4) ( 4)
s s s s s
s s
+ + +
= =
+ +
25.
3 3
3
2
( 5) (3 2) (3 2) ( 5)
(3 2)
5 ( 5)
d d
t t t t
d t dt dt
dt t t
+ − − − +⎛ ⎞−
=⎜ ⎟
⎜ ⎟+ +⎝ ⎠
2 3
2
( 5)(3)(3 – 2) (3) – (3 – 2) (1)
( 5)
t t t
t
+
=
+
2
2
(6 47)(3 – 2)
( 5)
t t
t
+
=
+
26. 3 2
(sin ) 3sin cos
d
d
θ θ θ
θ
=
27.
3 2 2
2
2 2 3
2 4
2
(cos2 ) (sin ) (sin ) (cos2 )
sin sin sin sin
3 3
cos2 cos2 cos2 cos2 cos 2
sin cos cos2 2sin sin 2 3sin cos cos2 6sin sin 2
3
cos2 cos 2 cos 2
3(sin )
d d
x x x x
dy d x x d x x dx dx
dx dx x x dx x x x
x x x x x x x x x x
x x x
x
−
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= = ⋅ = ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
+ +⎛ ⎞
= =⎜ ⎟
⎝ ⎠
=
4
(cos cos2 2sin sin 2 )
cos 2
x x x x
x
+

28. 2 2 2
2 2 2 2 2 2
[sin tan( 1)] sin [tan( 1)] tan( 1) (sin )
(sin )[sec ( 1)](2 ) tan( 1)cos 2 sin sec ( 1) cos tan( 1)
dy d d d
t t t t t t
dt dt dt dt
t t t t t t t t t t
= + = ⋅ + + + ⋅
= + + + = + + +
29.
2 2 22
2
( 2) ( 1) – ( 1) ( 2)1
( ) 3
2 ( 2)
x xx D x x D xx
f x
x x
⎛ ⎞ + + + ++
′ = ⎜ ⎟
⎜ ⎟+ +⎝ ⎠
22 2 2
2
1 2 4 – –1
3
2 ( 2)
x x x x
x x
⎛ ⎞+ +
= ⎜ ⎟
⎜ ⎟+ +⎝ ⎠
2 2 2
4
3( 1) ( 4 –1)
( 2)
x x x
x
+ +
=
+
(3) 9.6f ′ =
30. 2 3 2 4 2 4 2 3
( ) ( 9) ( – 2) ( – 2) ( 9)t tG t t D t t D t′ = + + + 2 3 2 3 2 4 2 2
( 9) (4)( – 2) (2 ) ( – 2) (3)( 9) (2 )t t t t t t= + + +
2 2 2 2 3
2 (7 30)( 9) ( – 2)t t t t= + +
(1) –7400G′ =
31. 2
( ) [cos( 3 1)](2 3)F t t t t′ = + + + 2
(2 3)cos( 3 1)t t t= + + + ; (1) 5cos5 1.4183F′ = ≈
32. 2 2
( ) (cos ) (sin ) (sin ) (cos )s sg s s D s s D s′ = π π + π π 2
(cos )(2sin )(cos )( ) (sin )(–sin )( )s s s s s= π π π π + π π π
2 2
sin [2cos – sin ]s s s= π π π π
1
–
2
g
⎛ ⎞′ = π⎜ ⎟
⎝ ⎠
33. 4 2 3 2 2
[sin ( 3 )] 4sin ( 3 ) sin( 3 )x xD x x x x D x x+ = + + 3 2 2 2
4sin ( 3 )cos( 3 ) ( 3 )xx x x x D x x= + + +
3 2 2
4sin ( 3 )cos( 3 )(2 3)x x x x x= + + + 3 2 2
4(2 3)sin ( 3 )cos( 3 )x x x x x= + + +
34. 5 4
[cos (4 –19)] 5cos (4 –19) cos(4 –19)t tD t t D t= 4
5cos (4 –19)[–sin(4 –19)] (4 –19)tt t D t=
4
–5cos (4 –19)sin(4 –19)(4)t t=
4
–20cos (4 –19)sin(4 –19)t t=
35. 3 2
[sin (cos )] 3sin (cos ) sin(cos )t tD t t D t= 2
3sin (cos )cos(cos ) (cos )tt t D t=
2
3sin (cos )cos(cos )(–sin )t t t=
2
–3sin sin (cos )cos(cos )t t t=
36 . 4 31 1 1
cos 4cos cos
–1 –1 –1
u u
u u u
D D
u u u
⎡ + ⎤ + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
3 1 1 1
4cos –sin
–1 –1 –1
u
u u u
D
u u u
+ ⎡ + ⎤ +⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
3
2
( –1) ( 1) – ( 1) ( –1)1 1
–4cos sin
–1 –1 ( –1)
u uu D u u D uu u
u u u
+ ++ +⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3
2
8 1 1
cos sin
–1 –1( –1)
u u
u uu
+ +⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
37. 4 2 3 2 2
[cos (sin )] 4cos (sin ) cos(sin )D Dθ θθ θ θ= 3 2 2 2
4cos (sin )[–sin(sin )] (sin )Dθθ θ θ=
3 2 2 2 2
–4cos (sin )sin(sin )(cos ) ( )Dθθ θ θ θ= 3 2 2 2
–8 cos (sin )sin(sin )(cos )θ θ θ θ=
38. 2 2 2
[ sin (2 )] sin (2 ) sin (2 )x x xD x x x D x x D x= + 2
[2sin(2 ) sin(2 )] sin (2 )(1)xx x D x x= +
2
[2sin(2 )cos(2 ) (2 )] sin (2 )xx x x D x x= + 2
[4sin(2 )cos(2 )] sin (2 )x x x x= + 2
2 sin(4 ) sin (2 )x x x= +
39. {sin[cos(sin 2 )]} cos[cos(sin 2 )] cos(sin 2 )x xD x x D x= cos[cos(sin 2 )][–sin(sin 2 )] (sin 2 )xx x D x=
– cos[cos(sin 2 )]sin(sin 2 )(cos2 ) (2 )xx x x D x= –2cos[cos(sin 2 )]sin(sin 2 )(cos2 )x x x=
40. 2
{cos [cos(cos )]} 2cos[cos(cos )] cos[cos(cos )]t tD t t D t= 2cos[cos(cos )]{–sin[cos(cos )]} cos(cos )tt t D t=
–2cos[cos(cos )]sin[cos(cos )][–sin(cos )] (cos )tt t t D t= 2cos[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=

41. ( ) (4) (4) (4)f g f g′ ′ ′+ = +
1 3
2
2 2
≈ + ≈
42. ( ) ( ) ( ) ( ) ( )
( ) ( )
( )
2 2 2 2 2
2 2 2
1 2 0 1
f g f g
f g
′ ′
′− = −
′′= −
= − =
43. ( ) ( ) ( )( ) ( ) ( ) 1110222 =+=′+′=′ fggffg
44.
2
(2) (2) – (2) (2)
( ) (2)
(2)
g f f g
f g
g
′ ′
′ =
2
(1)(1) – (3)(0)
1
(1)
≈ =
45. ( ) (6) ( (6)) (6)f g f g g′ ′ ′=
(2) (6) (1)( 1) –1f g′ ′= ≈ − =
46. ( ) (3) ( (3)) (3)g f g f f′ ′ ′=
3 3
(4) (3) (1)
2 2
g f
⎛ ⎞′ ′= ≈ =⎜ ⎟
⎝ ⎠
47. ( ) ( ) ( ) ( )xFxDxFxFD xx 22222 ′=′=
48.
( ) ( ) ( )
( )12
111
2
222
+′=
++′=+
xFx
xDxFxFD xx
49. ( )( )[ ] ( )( ) ( )tFtFtFDt ′−= −− 32
2
50.
( )( )
( )( ) ( )zFzF
zFdz
d
′−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −3
2
2
1
51. ( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( ) ( )
2
1 2 2 1 2 1 2
2 1 2 2 2 4 1 2 2
d d
F z F z F z
dz dz
F z F z F z F z
⎡ ⎤+ = + +
⎢ ⎥⎣ ⎦
′ ′= + = +
52.
( )
( )( )
( )
( )
( )( )
( )
( )( )
1
2 2
2
2
2 2
2
2
2
2
2
1
2
2
2 2
2 1
d d
y y F y
dy dyF y
yF yd
y F y y y
dy
F y
F y
y
F y
−⎡ ⎤
⎡ ⎤⎢ ⎥+ = + ⎢ ⎥⎢ ⎥ ⎣ ⎦
⎣ ⎦
′
′= − = −
⎛ ⎞
′⎜ ⎟
= −⎜ ⎟
⎜ ⎟
⎝ ⎠
53. ( ) ( ) ( )
( )
cos cos cos
sin cos
d d
F x F x x
dx dx
xF x
′=
′= −
54. ( )( ) ( )( ) ( )
( ) ( )( )
cos sin
sin
d d
F x F x F x
dx dx
F x F x
= −
′= −
55. ( )( ) ( )( ) ( )
( )( ) ( ) [ ]
( ) ( )( )
2
2
2
tan 2 sec 2 2
sec 2 2 2
2 2 sec 2
x x
x
D F x F x D F x
F x F x D x
F x F x
⎡ ⎤ = ⎡ ⎤⎣ ⎦⎣ ⎦
′= × ×
′=
56. ( ) ( )
( )( )
( )
2
2
tan 2 ' tan 2 tan 2
' tan 2 sec 2 2
2 ' tan 2 sec 2
d d
g x g x x
dx dx
g x x
g x x
= ⋅⎡ ⎤⎣ ⎦
= ⋅
=
57. ( ) ( )
( ) ( ) ( ) ( )
2
2 2
sin
sin sin
x
x x
D F x F x
F x D F x F x D F x
⎡ ⎤
⎣ ⎦
⎡ ⎤= × + ×⎣ ⎦
( ) ( ) ( )
( ) ( )
( ) ( ) ( )( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
2
2
2
2sin sin
sin
2sin cos
sin
2 sin cos
sin
x
x
F x F x D F x
F x F x
F x F x F x D F x
F x F x
F x F x F x F x
F x F x
= × × ⎡ ⎤⎣ ⎦
′+
= × × × ⎡ ⎤⎣ ⎦
′+
′=
′+
58. ( ) ( ) ( )
( ) ( ) ( ) [ ]
( ) ( ) ( )
3 2
2
3
sec 3sec sec
3sec sec tan
3 sec tan
x x
x
D F x F x D F x
F x F x F x D x
F x F x F x
⎡ ⎤ = ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦
= ⎡ ⎤⎣ ⎦
′=
59. ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
' sin sin
0 0 sin 0 2sin1 1.683
xg x f x D f x f x f x
g f f
′= − = −
′ ′= − = − ≈ −
60. ( )
( )( ) ( )( )
( )( )
( )( ) ( ) ( ) ( )
( )( )
2
2
1 sec 2 1 sec 2
1 sec 2
1 sec 2 2 2 sec 2 tan 2
1 sec 2
d d
F x x x F x
dx dxG x
F x
F x xF x F x F x
F x
+ − +
′ =
+
′+ −
=
+
( )
( )
( )( )
( )
( )( )
( )
2 2
1 sec 0 0 1 sec 0
0
1 sec 0 1 sec 0
1 1
0.713
1 sec 0 1 sec2
F F
G
F F
F
+ − +
′ = =
+ +
= = ≈ −
+ +

61. ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( )
sin cos
1 1 1 sin 1 1 cos 1
2 1 sin 0 1cos0 1
F x f x g x g x f x g x
F f g g f g
′ ′ ′= − +
′ ′ ′= − +
= − + − = −
62.
( )
1 sin3 ; 3 cos3 sin3
/3 3 cos3 sin 0
3 3 3
1 /3
/3 1
y x x y x x x
y
y x
y x
π π π
π π π
π π
π π
′= + = +
′ = + = − + = −
− = − −
= − − +
The line crosses the x-axis at
3
3
x
π−
= .
63. 2
sin ; 2sin cos sin 2 1
/ 4 , 0, 1, 2,...
y x y x x x
x k kπ π
′= = = =
= + = ± ±
64. ( ) ( ) ( ) ( )
( )( ) ( ) ( )
( ) ( )( ) ( )( ) ( )
3 2 22 4 3 2 4
3 2 23 4 2 4 2
3 2 2
1 2 1 3 1 1
2 1 1 3 1 1
1 2 2 2 3 1 2 2 32 48 80
32 80 1, 80 31
y x x x x x x
x x x x x x
y
y x y x
′ = + + + + +
= + + + + +
′ = + = + =
− = − = +
65. ( ) ( ) ( )
( ) ( )( )
3 32 2
3
2 1 2 4 1
1 4 1 1 1 1/ 2
1 1 1 1 3
,
4 2 2 2 4
y x x x x
y
y x y x
− −
−
′ = − + = − +
′ = − + = −
− = − + = − +
66. ( ) ( ) ( )
( ) ( )
2 2
2
3 2 1 2 6 2 1
0 6 1 6
1 6 0, 6 1
y x x
y
y x y x
′ = + = +
′ = =
− = − = +
The line crosses the x-axis at 1/ 6x = − .
67. ( ) ( ) ( )
( ) ( )
3 32 2
3
2 1 2 4 1
1 4 2 1/ 2
1 1 1 1 3
,
4 2 2 2 4
y x x x x
y
y x y x
− −
−
′ = − + = − +
′ = − = −
− = − + = − +
Set 0y = and solve for x. The line crosses the
x-axis at 3/ 2x = .
68. a.
2 2 2 2
2 2
4cos2 7sin 2
4 7 4 7
cos 2 sin 2 1
x y t t
t t
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + =
b. 2 2 2 2
( – 0) ( – 0)L x y x y= + = +
2 2
(4cos2 ) (7sin 2 )t t= +
2 2
16cos 2 49sin 2t t= +
c. 2 2
2 2
1
(16cos 2 49sin 2 )
2 16cos 2 49sin 2
tt D t t
t t
D L = +
+
2 2
32cos2 (cos2 ) 98sin 2 (sin 2 )
2 16cos 2 49sin 2
t tt D t t D t
t t
+
=
+
2 2
–64cos2 sin 2 196sin 2 cos2
2 16cos 2 49sin 2
t t t t
t t
+
=
+
2 2
16sin 4 49sin 4
16cos 2 49sin 2
t t
t t
− +
=
+
2 2
33sin 4
16cos 2 49sin 2
t
t t
=
+
At t =
π
8
: rate =
1 1
2 2
33
16 49⋅ + ⋅
≈ 5.8 ft/sec.
69. a. (10cos8 ,10sin8 )t tπ π
b. (10sin8 ) 10cos(8 ) (8 )t tD t t D tπ = π π
80 cos(8 )t= π π
At t = 1: rate = 80π ≈ 251 cm/s
P is rising at the rate of 251 cm/s.
70. a. (cos 2t, sin 2t)
b. 2 2 2
(0 – cos2 ) ( – sin 2 ) 5 ,t y t+ = so
2
sin 2 25 – cos 2y t t= +
c. 2
2
sin 2 25 cos 2
1
2cos2 4cos2 sin 2
2 25 cos 2
tD t t
t t t
t
⎛ ⎞+ −⎜ ⎟
⎝ ⎠
= + ⋅
−
2
sin 2
2cos2 1
25 – cos 2
t
t
t
⎛ ⎞
⎜ ⎟= +
⎜ ⎟
⎝ ⎠
71. 60 revolutions per minute is 120π radians per
minute or 2π radians per second.
a. (cos2 ,sin 2 )t tπ π
b. 2 2 2
(0 – cos2 ) ( – sin 2 ) 5 ,t y tπ + π = so
2
sin 2 25 – cos 2y t t= π + π
c. 2
2
sin 2 25 cos 2
2 cos2
1
4 cos2 sin 2
2 25 cos 2
tD t t
t
t t
t
⎛ ⎞π + − π⎜ ⎟
⎝ ⎠
= π π
+ ⋅ π π π
− π
2
sin 2
2 cos2 1
25 – cos 2
t
t
t
⎛ ⎞π
⎜ ⎟= π π +
⎜ ⎟
π⎝ ⎠

72. The minute hand makes 1 revolution every hour,
so at t minutes after the hour, it makes an angle
of
30
tπ
radians with the vertical. By the Law of
Cosines, the length of the elastic string is
2 2
10 10 – 2(10)(10)cos
30
t
s
π
= +
10 2 – 2cos
30
tπ
=
1
10 sin
15 30
2 2 – 2cos
30
ds t
dt t
π π
= ⋅ ⋅
π
30
30
sin
3 2 – 2cos
t
t
π
π
π
=
At 12:15, the string is stretching at the rate of
2
2
sin
0.74
3 23 2 – 2cos
π
π
π π
= ≈ cm/min
73. The minute hand makes 1 revolution every hour,
so at t minutes after noon it makes an angle of
30
tπ
radians with the vertical. Similarly, at t
minutes after noon the hour hand makes an angle
of
360
tπ
with the vertical. Thus, by the Law of
Cosines, the distance between the tips of the
hands is
2 2
6 8 – 2 6 8cos –
30 360
t t
s
π π⎛ ⎞
= + ⋅ ⋅ ⎜ ⎟
⎝ ⎠
11
100 – 96cos
360
tπ
=
11
360
1 44 11
sin
15 3602 100 – 96cos t
ds t
dt π
π π
= ⋅
11
360
11
360
22 sin
15 100 – 96cos
t
t
π
π
π
=
At 12:20,
11
18
11
18
22 sin
0.38
15 100 – 96cos
ds
dt
π
π
π
= ≈ in./min
74. From Problem 73,
11
360
11
360
22 sin
.
15 100 – 96cos
t
t
ds
dt
π
π
π
=
Using a computer algebra system or graphing
utility to view
ds
dt
for 0 60t≤ ≤ ,
ds
dt
is largest
when t ≈ 7.5. Thus, the distance between the tips
of the hands is increasing most rapidly at about
12:08.
75. 0 0sin sin 2x x=
0 0 0sin 2sin cosx x x=
0 0
1
cos [if sin 0]
2
x x= ≠
x0 =
π
3
Dx(sin x) = cos x, Dx(sin 2x) = 2cos 2x, so at x0,
the tangent lines to y = sin x and y = sin 2x have
slopes of m1 =
1
2
and 2
1
2 – –1,
2
m
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
respectively. From Problem 40 of Section 0.7,
2 1
1 2
–
tan
1
m m
m m
θ =
+
where θ is the angle between
the tangent lines.
( )
31
2 2
11
22
–1– –
tan –3,
1 (–1)
θ = = =
+
so θ ≈ –1.25. The curves intersect at an angle of
1.25 radians.
76.
1
sin
2 2
t
AB OA=
21
cos cos sin
2 2 2 2
t t t
D OA AB OA= ⋅ =
E = D + area (semi-circle)
2
2 1 1
cos sin
2 2 2 2
t t
OA AB
⎛ ⎞
= + π⎜ ⎟
⎝ ⎠
2 2 21
cos sin sin
2 2 2 2
t t t
OA OA= + π
2 1
sin cos sin
2 2 2 2
t t t
OA
⎛ ⎞
= + π⎜ ⎟
⎝ ⎠
2
1
2 2
cos
cos sin
2
t
t
D
tE
=
+ π
0
1
lim 1
1 0t
D
E+→
= =
+
cos( / 2)
lim lim
cos( / 2) sin( / 2)
2
0
0
0
2
t t
D t
E t tπ π π
π
− −→ →
=
+
= =
+

77. 2
andy u u x= =
Dx y = Duy ⋅ Dxu
2
1 2
2
2 2
xx x
x
x xu x
= ⋅ = = =
78.
2
2 2
2
–1
–1 ( –1)
–1
x x
x
D x D x
x
=
2 2
2 2
–1 2 –1
(2 )
–1 –1
x x x
x
x x
= =
79.
sin
sin (sin )
sin
x x
x
D x D x
x
=
sin
cos cot sin
sin
x
x x x
x
= =
80. a. ( ) ( ) ( )2 2 2
2
1 2
' 2x xD L x L x D x x
xx
= = ⋅ =
b. 4 4 4
(cos ) sec (cos )x xD L x x D x=
4 3
sec (4cos ) (cos )xx x D x=
( )
4 3
3
4
4sec cos ( sin )
1
4 cos sin
cos
x x x
x x
x
= −
= ⋅ ⋅ ⋅ −
–4sec sin 4tanx x x= = −
81. [ ( ( ( (0))))]
( ( ( (0)))) ( ( (0))) ( (0)) (0)
f f f f
f f f f f f f f f f
′
′ ′ ′ ′= ⋅ ⋅ ⋅
= 2 ⋅2 ⋅2 ⋅ 2 = 16
82. a. [2]
[1] [1]
'( ( )) '( )
'( ) ( )
d
f f f x f x
dx
d
f f f x
dx
= ⋅
= ⋅
b. [3]
[2] [1] [1]
[2] [2]
'( ( ( ))) '( ( )) '( )
'( ( )) '( ( )) ( )
'( ( )) ( )
d
f f f f x f f x f x
dx
d
f f x f f x f x
dx
d
f f x f x
dx
= ⋅ ⋅
= ⋅ ⋅
= ⋅
c. Conjecture:
[ ] [ 1] [ 1]
( ) '( ( )) ( )n n nd d
f x f f x f x
dx dx
− −
= ⋅
83. ( ) ( )1 1 1
2 1 2 1
2 2
( ) 1
( ) ( ) ( ( )) ( ) ( ( )) ( ( )) ( )
( ) ( )
( ) ( 1)( ( )) ( ) ( ( )) ( ) ( )( ( )) ( ) ( ( )) ( )
( ) ( ) ( ) ( ) ( )
( )( ) ( )
x x x x x
x x x x
x x x
f x
D D f x D f x g x f x D g x g x D f x
g x g x
f x g x D g x g x D f x f x g x D g x g x D f x
f x D g x D f x f x D g x
g xg x g x
− − −
− − − −
⎛ ⎞ ⎛ ⎞
= ⋅ = ⋅ = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= ⋅ − + = − +
− −
= + = 2 2
2
( ) ( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
x x x
x x
D f x f x D g x g x D f xg x
g x g x g x g x
g x D f x f x D g x
g x
−
+ ⋅ = +
−
=
84. ( ) ( )( )( )( ) ( )( )( ) ( )( ) ( )g x f f f f x f f f x f f x f x′ ′ ′ ′ ′=
( ) ( )( )( )( ) ( )( )( ) ( )( ) ( )
( )( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( )
( ) ( )( )( )( ) ( )( )( ) ( )( ) ( )
( )( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( )
1 1 1 1 1
2 2 2 1 1 1 2 1
2 2
1 2
2 2 2 2 2
1 1 1 2 2 2 1 2
2 2
1 2 1
g x f f f f x f f f x f f x f x
f f f x f f x f x f x f f x f x f x f x
f x f x
g x f f f f x f f f x f f x f x
f f f x f f x f x f x f f x f x f x f x
f x f x g x
′ ′ ′ ′ ′=
′ ′ ′ ′ ′ ′ ′ ′= =
′ ′= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
′ ′ ′ ′ ′=
′ ′ ′ ′ ′ ′ ′ ′= =
′ ′ ′= =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

2.6 Concepts Review
1.
3
3
3
( ), ,x
d y
f x D y
dx
′′′ , '''y
2.
2
2
; ;
ds ds d s
dt dt dt
3. ( ) 0f t′ >
4. 0; < 0
Problem Set 2.6
1. 2
3 6 6
dy
x x
dx
= + +
2
2
6 6
d y
x
dx
= +
3
3
6
d y
dx
=
2. 4 3
5 4
dy
x x
dx
= +
d2y
dx 2
= 20x3
+12x 2
3
2
3
60 24
d y
x x
dx
= +
3. 2 2
3(3 5) (3) 9(3 5)
dy
x x
dx
= + = +
2
2
18(3 5)(3) 162 270
d y
x x
dx
= + = +
3
3
162
d y
dx
=
4. 4 4
5(3 – 5 ) (–5) –25(3 – 5 )
dy
x x
dx
= =
2
3 3
2
–100(3 – 5 ) (–5) 500(3 – 5 )
d y
x x
dx
= =
3
2 2
3
1500(3 – 5 ) (–5) –7500(3 – 5 )
d y
x x
dx
= =
5. 7cos(7 )
dy
x
dx
=
2
2
2
–7 sin(7 )
d y
x
dx
=
3
3
3
–7 cos(7 ) –343cos(7 )
d y
x x
dx
= =
6. 2 3
3 cos( )
dy
x x
dx
=
2
2 2 3 3
2
3 [–3 sin( )] 6 cos( )
d y
x x x x x
dx
= + 4 3 3
–9 sin( ) 6 cos( )x x x x= +
3
4 3 2 3 3 3 2 3
3
–9 cos( )(3 ) sin( )(–36 ) 6 [–sin( )(3 )] 6cos( )
d y
x x x x x x x x x
dx
= + + +
6 3 3 3 3 3 3
–27 cos( ) – 36 sin( ) –18 sin( ) 6cos( )x x x x x x x= + 6 3 3 3
(6 – 27 )cos( ) – 54 sin( )x x x x=
7.
2 2
( –1)(0) – (1)(1) 1
–
( –1) ( –1)
dy x
dx x x
= =
2 2
2 4 3
( –1) (0) – 2( –1) 2
( –1) ( –1)
d y x x
dx x x
= − =
3 3 2
3 6
4
( 1) (0) 2[3( 1) ]
( 1)
6
( 1)
d y x x
dx x
x
− − −
=
−
= −
−
8.
2 2
(1– )(3) – (3 )(–1) 3
(1– ) ( –1)
dy x x
dx x x
= =
2 2
2 4 3
( –1) (0) – 3[2( –1)] 6
–
( –1) ( –1)
d y x x
dx x x
= =
3 3 2
3 6
4
( 1) (0) 6(3)( 1)
( 1)
18
( 1)
d y x x
dx x
x
− − −
= −
−
=
−

9. ( ) 2 ; ( ) 2; (2) 2f x x f x f′ ′′ ′′= = =
10. 2
( ) 15 4 1f x x x′ = + +
( ) 30 4f x x′′ = +
(2) 64f ′′ =
11.
2
2
( ) –f t
t
′ =
3
4
( )f t
t
′′ =
4 1
(2)
8 2
f ′′ = =
12.
2 2
2 2
(5 – )(4 ) – (2 )(–1) 20 – 2
( )
(5 – ) (5 – )
u u u u u
f u
u u
′ = =
2 2
4
(5 – ) (20 – 4 ) – (20 – 2 )2(5 – )(–1)
( )
(5 – )
u u u u u
f u
u
′′ =
3
100
(5 – )u
=
3
100 100
(2)
273
f ′′ = =
13. –3
( ) –2(cos ) (–sin )f θ θ θ′ = π π π = 2π(cosθπ)–3
(sinθπ)
–3 –4
( ) 2 [(cos ) ( )(cos ) (sin )(–3)(cos ) (–sin )( )]f θ θ θ θ θ θ′′ = π π π π + π π π π 2 2 2 4
2 [(cos ) 3sin (cos ) ]θ θ θ− −
= π π + π π
2 2
(2) 2 [1 3(0)(1)] 2f ′′ = π + = π
14.
2
( ) cos – sinf t t
t tt
π π π⎛ ⎞⎛ ⎞ ⎛ ⎞′ = +⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
– cos sin
t t t
π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 2 2
( ) – –sin – cos – cosf t
t t t tt t t
⎡ ⎤π π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
′′ = + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
2
3
– sin
tt
π π⎛ ⎞
= ⎜ ⎟
⎝ ⎠
2 2
(2) – sin – –1.23
8 2 8
f
π π π⎛ ⎞′′ = = ≈⎜ ⎟
⎝ ⎠
15. 2 2 2 3
( ) (3)(1– ) (–2 ) (1– )f s s s s s′ = + 2 2 2 2 3
–6 (1– ) (1– )s s s= + 6 4 2
–7 15 – 9 1s s s= + +
5 3
( ) –42 60 –18f s s s s′′ = +
(2) –900f ′′ =
16.
2
2
( –1)2( 1) – ( 1)
( )
( –1)
x x x
f x
x
+ +
′ =
2
2
– 2 – 3
( –1)
x x
x
=
2 2
4
( –1) (2 – 2) – ( – 2 – 3)2( –1)
( )
( –1)
x x x x x
f x
x
′′ =
2
3 3
( –1)(2 – 2) – ( – 2 – 3)(2) 8
( –1) ( –1)
x x x x
x x
= =
3
8
(2) 8
1
f ′′ = =
17. –1
( )n n
xD x nx=
2 –2
( ) ( –1)n n
xD x n n x=
3 –3
( ) ( –1)( – 2)n n
xD x n n n x=
4 –4
( ) ( –1)( – 2)( – 3)n n
xD x n n n n x=
1
( ) ( –1)( – 2)( – 3)...(2)n n
xD x n n n n x−
=
0
( ) ( –1)( – 2)( – 3)...2(1)n n
xD x n n n n x= = n!
18. Let k < n.
( ) [ ( )] ( !) 0n k n k k k
x x x xD x D D x D k−
= = =
so –1
1 0[ ] 0n n
x nD a x a x a+…+ + =
19. a. 4 3
(3 2 –19) 0xD x x+ =
b. 12 11 10
(100 79 ) 0xD x x− =
c. 11 2 5
( – 3) 0xD x =

20.
2
1 1
–xD
x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
2 –2 –3
3
1 2
(– ) 2x xD D x x
x x
⎛ ⎞
= = =⎜ ⎟
⎝ ⎠
3 –3
4
1 3(2)
(2 ) –x xD D x
x x
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
4
5
1 4(3)(2)
xD
x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
1
1 ( 1) !n
n
x n
n
D
x x +
−⎛ ⎞
=⎜ ⎟
⎝ ⎠
21. 2
( ) 3 6 – 45 3( 5)( 3)f x x x x x′ = + = + −
3(x + 5)(x – 3) = 0
x = –5, x = 3
( ) 6 6f x x′′ = +
(–5) –24f ′′ =
(3) 24f ′′ =
22. ( ) 2g t at b′ = +
( ) 2g t a′′ =
(1) 2 4
2
g a
a
′′ = = −
= −
(1) 2 3g a b′ = + =
2(–2) + b = 3
b = 7
( )
( ) ( )
1 5
2 7 5
0
g a b c
c
c
= + + =
− + + =
=
23. a. ( ) 12 – 4
ds
v t t
dt
= =
2
2
( ) –4
d s
a t
dt
= =
b. 12 – 4t > 0
4t < 12
t < 3; ( ),3−∞
c. 12 – 4t < 0
t > 3; (3, )∞
d. a(t) = –4 < 0 for all t
e.
24. a. 2
( ) 3 –12
ds
v t t t
dt
= =
2
2
( ) 6 –12
d s
a t t
dt
= =
b. 2
3 –12 0t t >
3t(t – 4) > 0; ( ,0) (4, )−∞ ∪ ∞
c. 2
3 –12 0t t <
(0, 4)
d. 6t – 12 < 0
6t < 12
t < 2; ( ,2)−∞
e.
25. a. 2
( ) 3 –18 24
ds
v t t t
dt
= = +
2
2
( ) 6 –18
d s
a t t
dt
= =
b. 2
3 –18 24 0t t + >
3(t – 2)(t – 4) > 0
( ,2) (4, )−∞ ∪ ∞
c. 3t2
–18t +24 < 0
(2, 4)
d. 6t – 18 < 0
6t < 18
t < 3; ( ,3)−∞
e.
26. a. 2
( ) 6 – 6
ds
v t t
dt
= =
2
2
( ) 12
d s
a t t
dt
= =
b. 2
6 – 6 0t >
6(t + 1)(t – 1) > 0
( , 1) (1, )−∞ − ∪ ∞
c. 2
6 – 6 0t <
(–1, 1)
d. 12t < 0
t < 0
The acceleration is negative for negative t.

e.
27. a.
2
16
( ) 2 –
ds
v t t
dt t
= =
2
2 3
32
( ) 2
d s
a t
dt t
= = +
b.
2
16
2 – 0t
t
>
3
2
2 –16
0;
t
t
> (2, )∞
c. 2t –
16
t2
< 0; (0, 2)
d. 2 +
32
t3
< 0
2t3
+ 32
t3
< 0; The acceleration is not
negative for any positive t.
e.
28. a.
2
4
( ) 1–
ds
v t
dt t
= =
2
2 3
8
( )
d s
a t
dt t
= =
b.
2
4
1– 0
t
>
2
2
– 4
0;
t
t
> (2, )∞
c.
2
4
1– 0;
t
< (0, 2)
d.
3
8
0;
t
< The acceleration is not negative for
any positive t.
e.
29. 3 2
( ) 2 –15 24
ds
v t t t t
dt
= = +
2
2
2
( ) 6 – 30 24
d s
a t t t
dt
= = +
2
6 – 30 24 0t t + =
6(t – 4)(t – 1) = 0
t = 4, 1
v(4) = –16, v(1) = 11
30. 3 21
( ) (4 – 42 120 )
10
ds
v t t t t
dt
= = +
2
2
2
1
( ) (12 – 84 120)
10
d s
a t t t
dt
= = +
21
(12 – 84 120) 0
10
t t + =
12
( 2)( 5) 0
10
t t− − =
t = 2, t = 5
v(2) = 10.4, v(5) = 5
31. 1
1( ) 4 – 6
ds
v t t
dt
= =
2
2 ( ) 2 – 2
ds
v t t
dt
= =
a. 4 – 6t = 2t – 2
8t = 6
3
4
t = sec
b. 4 – 6 2 – 2 ;t t= 4 – 6t = –2t + 2
1 3
sec and
2 4
t t= = sec
c. 2 2
4 – 3 – 2t t t t=
2
4 – 6 0t t =
2t(2t – 3) = 0
3
0 sec and sec
2
t t= =
32. 21
1( ) 9 – 24 18
ds
v t t t
dt
= = +
22
2 ( ) –3 18 –12
ds
v t t t
dt
= = +
2 2
9 – 24 18 –3 18 –12t t t t+ = +
2
12 – 42 30 0t t + =
2
2 – 7 5 0t t + =
(2t – 5)(t – 1) = 0
t =1,
5
2

33. a. v(t) = –32t + 48
initial velocity = v0 = 48 ft/sec
b. –32t + 48 = 0
t =
3
2
sec
c. 2
–16(1.5) 48(1.5) 256 292s = + + = ft
d. 2
–16 48 256 0t t+ + =
2
–48 48 – 4(–16)(256)
–2.77, 5.77
–32
t
±
= ≈
The object hits the ground at t = 5.77 sec.
e. v(5.77) ≈ –137 ft/sec;
speed = 137 137ft/sec.− =
34. v(t) = 48 –32t
a. 48 – 32t = 0
t = 1.5
2
48(1.5) –16(1.5) 36s = = ft
b. v(1) = 16 ft/sec upward
c. 2
48 –16 0t t =
–16t(–3 + t) = 0
t = 3 sec
35. 0( ) – 32v t v t=
0 – 32 0v t =
0
32
v
t =
2
0 0
0 –16 5280
32 32
v v
v
⎛ ⎞ ⎛ ⎞
=⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2 2
0 0
– 5280
32 64
v v
=
2
0
5280
64
v
=
0 337,920 581v = ≈ ft/sec
36. 0( ) 32v t v t= +
0 32 140v t+ =
0 32(3) 140v + =
0 44v =
2
44(3) 16(3) 276s = + = ft
37. 2
( ) 3 – 6 – 24v t t t=
2
2
2
3 – 6 – 24
3 – 6 – 24 (6 – 6)
3 – 6 – 24
t td
t t t
dt t t
=
( – 4)( 2)
(6 – 6)
( – 4)( 2)
t t
t
t t
+
=
+
( – 4)( 2) (6 – 6)
0
( – 4)( 2)
t t t
t t
+
<
+
t < –2, 1 < t < 4; ( , 2)−∞ − ∪ (1, 4)
38. Point slowing down when
( ) 0
d
v t
dt
<
( ) ( )
( )
( )
v t a td
v t
dt v t
=
( ) ( )
0
( )
v t a t
v t
< when a(t) and v(t) have opposite
signs.
39. ( )xD uv uv u v′ ′= +
2
( )
2
xD uv uv u v u v u v
uv u v u v
′′ ′ ′ ′ ′ ′′= + + +
′′ ′ ′ ′′= + +
3
( ) 2( )xD uv uv u v u v u v u v u v′′′ ′ ′′ ′ ′′ ′′ ′ ′′ ′ ′′′= + + + + +
3 3uv u v u v u v′′′ ′ ′′ ′′ ′ ′′′= + + +
0
( ) ( ) ( )
n
n n k k
x x x
k
n
D uv D u D v
k
−
=
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
∑
where
n
k
⎛
⎝
⎜
⎞
⎠
⎟ is the binomial coefficient
!
.
( – )! !
n
n k k
40. 4 4 4 4 04
( sin ) ( ) (sin )
0
x x xD x x D x D x
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
3 4 1 2 4 24 4
( ) (sin ) ( ) (sin )
1 2
x x x xD x D x D x D x
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1 4 3 0 4 44 4
( ) (sin ) ( ) (sin )
3 4
x x x xD x D x D x D x
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
3 4
24sin 96 cos 72 sin
16 cos sin
x x x x x
x x x x
= + −
− +

41. a.
b. ′′′f (2.13) ≈ –1.2826
42. a.
b. (2.13) 0.0271f ′′′ ≈
2.7 Concepts Review
1.
3
9
– 3x
2. 2
3
dy
y
dx
3. 2 2 2
(2 ) 3 – 3
dy dy dy
x y y y x
dx dx dx
+ + =
4. –1 2 2/35
; ( – 5 ) (2 – 5)
3
p qp
x x x x
q
Problem Set 2.7
1. 2 – 2 0xy D y x =
2
2
x
x x
D y
y y
= =
2. 18 8 0xx y D y+ =
–18 9
–
8 4
x
x x
D y
y y
= =
3. 0xx D y y+ =
–x
y
D y
x
=
4. 2
2 2 0xx yD yα+ =
2 2
2
– –
2
x
x x
D y
y yα α
= =
5. 2
(2 ) 1xx y D y y+ =
2
1–
2
x
y
D y
xy
=
6. 2
2 2 4 3 3 0x xx x D y xy x D y y+ + + + =
2
(2 3 ) –2 – 4 – 3xD y x x x xy y+ =
2
–2 – 4 – 3
2 3
x
x xy y
D y
x x
=
+
7. 2 2 2
12 7 (2 ) 7 6x xx x y D y y y D y+ + =
2 2 2
12 7 6 –14x xx y y D y xyD y+ =
2 2
2
12 7
6 –14
x
x y
D y
y xy
+
=
8. 2 2
2 (2 )x xx D y xy y x y D y+ = +
2 2
– 2 – 2x xx D y xyD y y xy=
2
2
– 2
– 2
x
y xy
D y
x xy
=
9.
2 3
1
(5 5 ) 2
2 5
2 (3 )
x x
x x
x D y y D y
xy
y D y x y D y y
⋅ + +
= + +
2
3
5
2 – 2 – 3
2 5
5
–
2 5
x x x x
x
D y D y y D y xy D y
xy
y
y
xy
+
=
53
2 5
25
2 5
–
2 – 2 – 3
y
xy
x x
xy
y
D y
y xy
=
+
10.
1
1
2 1
x xx D y y x D y y
y
+ + = +
+
– – 1
2 1
x x
x
D y x D y y y
y
= +
+
2 1
– 1
–
x x
y
y y
D y
x
+
+
=

11. cos( )( ) 0x xx D y y xy x D y y+ + + =
cos( ) – – cos( )x xx D y x xy D y y y xy+ =
– – cos( )
–
cos( )
x
y y xy y
D y
x x xy x
= =
+
12. 2 2
–sin( )(2 ) 2 1x xxy xy D y y yD y+ = +
2 2 2
–2 sin( ) – 2 1 sin( )x xxy xy D y y D y y xy= +
2 2
2
1 sin( )
–2 sin( ) – 2
x
y xy
D y
xy xy y
+
=
13. 3 2 3 2
3 3 0x y x y y xy y′ ′+ + + =
3 2 2 3
( 3 ) –3 –y x xy x y y′ + =
2 3
3 2
–3 –
3
x y y
y
x xy
′ =
+
At (1, 3),
36 9
– –
28 7
y′ = =
Tangent line:
9
– 3 – ( –1)
7
y x=
14. 2 2
(2 ) 2 4 4 12x y y xy xy y y′ ′ ′+ + + =
2 2
(2 4 –12) –2 – 4y x y x xy y′ + =
2 2
2 2
–2 – 4 – – 2
2 4 –12 2 – 6
xy y xy y
y
x y x x y x
′ = =
+ +
At (2, 1), –2y′ =
Tangent line: –1 –2( – 2)y x=
15. cos( )( )xy xy y y′ ′+ =
[ cos( ) –1] – cos( )y x xy y xy′ =
– cos( ) cos( )
cos( ) –1 1– cos( )
y xy y xy
y
x xy x xy
′ = =
At ,1 , 0
2
y
π⎛ ⎞ ′ =⎜ ⎟
⎝ ⎠
Tangent line: –1 0 –
2
y x
π⎛ ⎞
= ⎜ ⎟
⎝ ⎠
y = 1
16. 2 2
[–sin( )][2 ] 6 0y xy xyy y x′ ′+ + + =
2 2 2
[1– 2 sin( )] sin( ) – 6y xy xy y xy x′ =
2 2
2
sin( ) – 6
1– 2 sin( )
y xy x
y
xy xy
′ =
At (1, 0),
6
– –6
1
y′ = =
Tangent line: y – 0 = –6(x – 1)
17. –1/3 –1/32 2
– – 2 0
3 3
x y y y′ ′ =
–1/3 –1/32 2
2
3 3
x y y
⎛ ⎞′= +⎜ ⎟
⎝ ⎠
–1/32
3
–1/ 32
3
2
x
y
y
′ =
+
At (1, –1),
2
3
4
3
1
2
y′ = =
Tangent line:
1
1 ( –1)
2
y x+ =
18. 21
2 0
2
y xyy y
y
′ ′+ + =
21
2 –
2
y xy y
y
⎛ ⎞
′ + =⎜ ⎟⎜ ⎟
⎝ ⎠
2
1
2
–
2
y
y
y
xy
′ =
+
At (4, 1),
17
2
–1 2
–
17
y′ = =
Tangent line:
2
–1 – ( – 4)
17
y x=
19. 2/3 1
5
2
dy
x
dx x
= +
20. –2/3 5/ 2 5/ 2
3 2
1 1
– 7 – 7
3 3
dy
x x x
dx x
= =
21. –2/3 –4/3
3 32 4
1 1 1 1
– –
3 3 3 3
dy
x x
dx x x
= =
22. –3/ 4
34
1 1
(2 1) (2)
4 2 (2 1)
dy
x
dx x
= + =
+
23. 2 –3/ 41
(3 – 4 ) (6 – 4)
4
dy
x x x
dx
=
2 3 2 34 4
6 – 4 3 – 2
4 (3 – 4 ) 2 (3 – 4 )
x x
x x x x
= =
24. 3 –2/3 21
( – 2 ) (3 – 2)
3
dy
x x x
dx
=
25. 3 2/3
[( 2 ) ]
dy d
x x
dx dx
−
= +
2
3 –5/3 2
3 53
2 6 4
– ( 2 ) (3 2)
3 3 ( 2 )
x
x x x
x x
+
= + + = −
+

26. –8/3 –8/35
– (3 – 9) (3) –5(3 – 9)
3
dy
x x
dx
= =
27.
2
1
(2 cos )
2 sin
dy
x x
dx x x
= +
+
2
2 cos
2 sin
x x
x x
+
=
+
28. 2
2
1
[ (–sin ) 2 cos ]
2 cos
dy
x x x x
dx x x
= +
2
2
2 cos – sin
2 cos
x x x x
x x
=
29. 2 –1/3
[( sin ) ]
dy d
x x
dx dx
=
2 –4/3 21
– ( sin ) ( cos 2 sin )
3
x x x x x x= +
2
2 43
cos 2 sin
–
3 ( sin )
x x x x
x x
+
=
30. –3/ 41
(1 sin5 ) (cos5 )(5)
4
dy
x x
dx
= +
34
5cos5
4 (1 sin5 )
x
x
=
+
31.
2 –3/ 4 2
[1 cos( 2 )] [–sin( 2 )(2 2)]
4
dy x x x x x
dx
+ + + +
=
2
2 34
( 1)sin( 2 )
2 [1 cos( 2 )]
x x x
x x
+ +
= −
+ +
32.
2 2 –1/ 2 2
(tan sin ) (2 tan sec 2sin cos )
2
x x x x x xdy
dx
+ +
=
2
2 2
tan sec sin cos
tan sin
x x x x
x x
+
=
+
33. 2 2
2 3 0
ds
s st t
dt
+ + =
2 2 2 2
– – 3 3
2 2
ds s t s t
dt st st
+
= = −
2 2
2 3 0
dt dt
s st t
ds ds
+ + =
2 2
( 3 ) –2
dt
s t st
ds
+ =
2 2
2
3
dt st
ds s t
= −
+
34. 2 2
1 cos( )(2 ) 6
dx dx
x x x
dy dy
= +
2 2
1
2 cos( ) 6
dx
dy x x x
=
+
35.
5
−5
5−5
y
x
(x + 2)2
+ y2
= 1
2 4 2 0
dy
x y
dx
+ + =
2 4 2
2
dy x x
dx y y
+ +
= − = −
The tangent line at 0 0( , )x y has equation
0
0 0
0
2
– ( – )
x
y y x x
y
+
= − which simplifies to
2 2
0 0 0 0 02 – – 2 – 0.x yy x xx y x+ + = Since
0 0( , )x y is on the circle, 2 2
0 0 0–3 – 4 ,x y x+ =
so the equation of the tangent line is
0 0 0– – 2 – 2 – 3.yy x x xx =
If (0, 0) is on the tangent line, then 0
3
– .
2
x =
Solve for 0y in the equation of the circle to get
0
3
.
2
y = ± Put these values into the equation of
the tangent line to get that the tangent lines are
3 0y x+ = and 3 – 0.y x =
36. 2 2
16( )(2 2 ) 100(2 – 2 )x y x yy x yy′ ′+ + =
3 2 2 3
32 32 32 32 200 – 200x x yy xy y y x yy′ ′ ′+ + + =
2 3 3 2
(4 4 25 ) 25 – 4 – 4y x y y y x x xy′ + + =
3 2
2 3
25 – 4 – 4
4 4 25
x x xy
y
x y y y
′ =
+ +
The slope of the normal line
1
–
y
=
′
2 3
3 2
4 4 25
4 4 – 25
x y y y
x xy x
+ +
=
+
At (3, 1),
65 13
slope
45 9
= =
Normal line:
13
–1 ( – 3)
9
y x=

37. a. 2
3 0xy y y y′ ′+ + =
2
( 3 ) –y x y y′ + =
2
–
3
y
y
x y
′ =
+
b. 2
2 2
2
2
– –
3
3 3
–
6 0
3
y y
xy y y
x y x y
y
y
x y
⎛ ⎞ ⎛ ⎞
′′ ′′+ + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
⎛ ⎞
+ =⎜ ⎟⎜ ⎟+⎝ ⎠
3
2
2 2 2
2 6
3 – 0
3 ( 3 )
y y
xy y y
x y x y
′′ ′′+ + =
+ +
3
2
2 2 2
2 6
( 3 ) –
3 ( 3 )
y y
y x y
x y x y
′′ + =
+ +
2
2 2
2
( 3 )
( 3 )
xy
y x y
x y
′′ + =
+
2 3
2
( 3 )
xy
y
x y
′′ =
+
38. 2
3 – 8 0x yy′ =
2
3
8
x
y
y
′ =
2
6 – 8( ( ) ) 0x yy y′′ ′+ =
22
3
6 – 8 – 8 0
8
x
x yy
y
⎛ ⎞
′′ =⎜ ⎟
⎜ ⎟
⎝ ⎠
4
2
9
6 – 8 – 0
8
x
x yy
y
′′ =
2 4
2
48 9
8
8
xy x
yy
y
−
′′=
2 4
3
48 – 9
64
xy x
y
y
′′ =
39. 2 2
2( 2 ) –12 0x y xy y y′ ′+ =
2 2
2 –12 –4x y y y xy′ ′ =
2 2
2
6 –
xy
y
y x
′ =
2 2 2
2( 2 2 2 ) –12[ 2 ( ) ] 0x y xy xy y y y y y′′ ′ ′ ′′ ′+ + + + =
2 2 2
2 12 8 4 24 ( )x y y y xy y y y′′ ′′ ′ ′− = − − +
2 2 3
2 2
2 2 2 2 2
16 96
(2 –12 ) – 4
6 – (6 )
x y x y
y x y y
y x y x
′′ = − +
−
4 2 3 5
2 2
2 2 2
12 48 144
(2 –12 )
(6 – )
x y x y y
y x y
y x
+ −
′′ =
5 4 2 3
2 2
2 2 2
72 6 24
(6 – )
(6 – )
y x y x y
y y x
y x
− −
′′ =
5 4 2 3
2 2 3
72 6 24
(6 – )
y x y x y
y
y x
− −
′′ =
At (2, 1),
120
15
8
y
−
′′ = = −
40. 2 2 0x yy′+ =
2
– –
2
x x
y
y y
′ = =
2
2 2[ ( ) ] 0yy y′′ ′+ + =
2
2 2 2 – 0
x
yy
y
⎛ ⎞
′′+ + =⎜ ⎟
⎝ ⎠
2
2
2
2 2
x
yy
y
′′ = − −
2 2 2
3 3
1 x y x
y
y y y
+
′′ = − − = −
At (3, 4),
25
64
y′′ = −
41. 2 2
3 3 3( )x y y xy y′ ′+ = +
2 2
(3 – 3 ) 3 – 3y y x y x′ =
2
2
–
–
y x
y
y x
′ =
At
3 3
, ,
2 2
⎛ ⎞
⎜ ⎟
⎝ ⎠
–1y′ =
Slope of the normal line is 1.
Normal line:
3 3
– 1 – ;
2 2
y x y x
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
This line includes the point (0, 0).
42. 0xy y′+ =
–
y
y
x
′ =
2 2 0x yy′− =
x
y
y
′ =
The slopes of the tangents are negative
reciprocals, so the hyperbolas intersect at right
angles.

43. Implicitly differentiate the first equation.
4 2 0x yy′+ =
2
–
x
y
y
′ =
Implicitly differentiate the second equation.
2 4yy′ =
2
y
y
′ =
Solve for the points of intersection.
2
2 4 6x x+ =
2
2( 2 – 3) 0x x+ =
(x + 3)(x – 1) = 0
x = –3, x = 1
x = –3 is extraneous, and y = –2, 2 when x = 1.
The graphs intersect at (1, –2) and (1, 2).
At (1, –2): 1 21, –1m m= =
At (1, 2): 1 2–1, 1m m= =
44. Find the intersection points:
2 2 2 2
1 1x y y x+ = → = −
( )
( ) ( )
2 2
2 2
2 2
1 1
1 1 1
1
2 1 1 1
2
x y
x x
x x x x
− + =
− + − =
− + + − = ⇒ =
Points of intersection:
1 3 1 3
, and , –
2 2 2 2
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Implicitly differentiate the first equation.
2 2 0x yy′+ =
–
x
y
y
′ =
Implicitly differentiate the second equation.
2( –1) 2 0x yy′+ =
1– x
y
y
′ =
At 1 2
1 3 1 1
, : – ,
2 2 3 3
m m
⎛ ⎞
= =⎜ ⎟⎜ ⎟
⎝ ⎠
( )( )
1 1 2
3 3 3
21 1
3
3 3
tan 3
31
θ θ
+ π
= = = → =
+ −
At 1 2
1 3 1 1
, – : , –
2 2 3 3
m m
⎛ ⎞
= =⎜ ⎟⎜ ⎟
⎝ ⎠
( )( )
1 1 2
3 3 3
21 1
3
3 3
tan 3
1 –
θ
− − −
= = = −
+
2
3
θ
π
=
45. 2 2
– (2 ) 2(2 ) 28x x x x+ =
2
7 28x =
2
4x =
x = –2, 2
Intersection point in first quadrant: (2, 4)
1 2y′ =
2 22 – – 4 0x xy y yy′ ′+ =
2 (4 – ) – 2y y x y x′ =
2
– 2
4 –
y x
y
y x
′ =
At (2, 4): 1 22, 0m m= =
–10 – 2
tan –2; tan (–2) 2.034
1 (0)(2)
θ θ= = = π + ≈
+
46. The equation is 2 2 2 2
0 0– – .mv mv kx kx=
Differentiate implicitly with respect to t to get
2 –2 .
dv dx
mv kx
dt dt
= Since
dx
v
dt
= this simplifies
to 2 –2
dv
mv kxv
dt
= or – .
dv
m kx
dt
=
47. 2 2
– 16x xy y+ = , when y = 0,
2
16x =
x = –4, 4
The ellipse intersects the x-axis at (–4, 0) and
(4, 0).
2 – – 2 0x xy y yy′ ′+ =
(2 – ) – 2y y x y x′ =
– 2
2 –
y x
y
y x
′ =
At (–4, 0), 2y′ =
At (4, 0), 2y′ =
Tangent lines: y = 2(x + 4) and y = 2(x – 4)

48. 2 2
2 – 2 – 0
dx dx
x xy xy y
dy dy
+ =
2 2
(2 – ) 2 – ;
dx
xy y xy x
dy
=
2
2
2 –
2 –
dx xy x
dy xy y
=
2
2
2 –
0
2 –
xy x
xy y
= if x(2y – x) = 0, which occurs
when x = 0 or .
2
x
y = There are no points on
2 2
– 2x y xy = where x = 0. If ,
2
x
y = then
2 3 3 3
2
2 – –
2 2 2 4 4
x x x x x
x x
⎛ ⎞ ⎛ ⎞
= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
so x = 2,
2
1.
2
y = =
The tangent line is vertical at (2, 1).
49. 2 2 0; –
dy dy x
x y
dx dx y
+ = =
The tangent line at 0 0( , )x y has slope 0
0
– ,
x
y
hence the equation of the tangent line is
0
0 0
0
– – ( – )
x
y y x x
y
= which simplifies to
2 2
0 0 0 0– ( ) 0yy xx x y+ + = or 0 0 1yy xx+ =
since 0 0( , )x y is on 2 2
1x y+ = . If (1.25, 0) is
on the tangent line through 0 0( , )x y , 0 0.8.x =
Put this into 2 2
1x y+ = to get 0 0.6,y = since
0 0.y > The line is 6y + 8x = 10. When x = –2,
13
,
3
y = so the light bulb must be
13
3
units high.
2.8 Concepts Review
1. ; 2
du
t
dt
=
2. 400 mi/hr
3. negative
4. negative; positive
Problem Set 2.8
1. 3
; 3
dx
V x
dt
= =
2
3
dV dx
x
dt dt
=
When x = 12, 2
3(12) (3) 1296
dV
dt
= = in.3/s.
2. 34
; 3
3
dV
V r
dt
= π =
2
4
dV dr
r
dt dt
= π
When r = 3, 2
3 4 (3)
dr
dt
= π .
1
0.027
12
dr
dt
= ≈
π
in./s
3. 2 2 2
1 ; 400
dx
y x
dt
= + =
2 2
dy dx
y x
dt dt
=
dy x dx
dt y dt
= mi/hr
When
5
5, 26, (400)
26
dy
x y
dt
= = =
≈ 392 mi/h.
4. 21 3 3
; ;
3 10 10
r h
V r h r
h
= π = =
2 3
1 3 3
; 3, 5
3 10 100
h h dV
V h h
dt
π⎛ ⎞
= π = = =⎜ ⎟
⎝ ⎠
2
9
100
dV h dh
dt dt
π
=
When h = 5,
2
9 (5)
3
100
dh
dt
π
=
4
0.42
3
dh
dt
= ≈
π
cm/s
5. 2 2 2
( 300) ; 300, 400,
dx dy
s x y
dt dt
= + + = =
2 2( 300) 2
ds dx dy
s x y
dt dt dt
= + +
( 300)
ds dx dy
s x y
dt dt dt
= + +
When x = 300, y = 400, 200 13s = , so
200 13 (300 300)(300) 400(400)
ds
dt
= + +
471
ds
dt
≈ mi/h

6. 2 2 2
(10) ; 2
dy
y x
dt
= + =
2 2
dy dx
y x
dt dt
=
When y = 25, x ≈ 22.9, so
25
(2) 2.18
22.9
dx y dy
dt x dt
= ≈ ≈ ft/s
7. 2 2 2
20 ; 1
dx
x y
dt
= + =
0 2 2
dx dy
x y
dt dt
= +
When x = 5, 375 5 15y = = , so
5
– – (1) –0.258
5 15
dy x dx
dt y dt
= = ≈ ft/s
The top of the ladder is moving down at
0.258 ft/s.
8. –4
dV
dt
= ft3/h; 2
; –0.0005
dh
V hr
dt
= π = ft/h
2 –1
,
V
A r Vh
h
= π = = so –1
2
–
dA dV V dh
h
dt dt dth
= .
When h = 0.001 ft, 2
(0.001)(250) 62.5V = π = π
and 1000(–4) –1,000,000(62.5 )(–0.0005)
dA
dt
= π
= –4000 + 31,250π ≈ 94,175 ft2/h.
(The height is decreasing due to the spreading of
the oil rather than the bacteria.)
9. 21
; , 2
3 4 2
d r
V r h h r h= π = = =
2 31 4
(2 ) ; 16
3 3
dV
V h h h
dt
= π = π =
2
4
dV dh
h
dt dt
= π
When h = 4, 2
16 4 (4)
dh
dt
= π
1
0.0796
4
dh
dt
= ≈
π
ft/s
10. 2 2 2
(90) ; 5
dx
y x
dt
= + =
2 2
dy dx
y x
dt dt
=
When y = 150, x = 120, so
120
(5) 4
150
dy x dx
dt y dt
= = = ft/s
11.
40
(20); , 8
2 5
hx x
V x h
h
= = =
2
10 (8 ) 80 ; 40
dV
V h h h
dt
= = =
160
dV dh
h
dt dt
=
When h = 3, 40 160(3)
dh
dt
=
1
12
dh
dt
= ft/min
12. 2
– 4; 5
dx
y x
dt
= =
2 2
1
(2 )
2 – 4 – 4
dy dx x dx
x
dt dt dtx x
= =
When x = 3,
2
3 15
(5) 6.7
53 – 4
dy
dt
= = ≈ units/s
13. 2
; 0.02
dr
A r
dt
= π =
2
dA dr
r
dt dt
= π
When r = 8.1, 2 (0.02)(8.1) 0.324
dA
dt
= π = π
≈ 1.018 in.2/s
14. 2 2 2
( 48) ; 30, 24
dx dy
s x y
dt dt
= + + = =
2 2 2( 48)
ds dx dy
s x y
dt dt dt
= + +
( 48)
ds dx dy
s x y
dt dt dt
= + +
At 2:00 p.m., x = 3(30) = 90, y = 3(24) = 72,
so s = 150.
(150) 90(30) (72 48)(24)
ds
dt
= + +
5580
37.2
150
ds
dt
= = knots/h

15. Let x be the distance from the beam to the point
opposite the lighthouse and θ be the angle
between the beam and the line from the
lighthouse to the point opposite.
tan ; 2(2 ) 4
1
x d
dt
θ
θ = = π = π rad/min,
2
sec
d dx
dt dt
θ
θ =
At –1 21 1 5
, tan and sec
2 2 4
x θ θ= = = .
5
(4 ) 15.71
4
dx
dt
= π ≈ km/min
16.
4000
tan
x
θ =
2
2
4000
sec
d dx
dt dtx
θ
θ = −
When
1
2
1 1 4000
, and 7322.
2 10 tan
d
x
dt
θ
θ = = = ≈
2
2 1 1 (7322)
sec
2 10 4000
dx
dt
⎡ ⎤⎛ ⎞
≈ −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
≈ –1740 ft/s or –1186 mi/h
The plane’s ground speed is 1186 mi/h.
17. a. Let x be the distance along the ground from
the light pole to Chris, and let s be the
distance from Chris to the tip of his shadow.
By similar triangles,
6 30
,
s x s
=
+
so
4
x
s =
and
1
.
4
ds dx
dt dt
= 2
dx
dt
= ft/s, hence
1
2
ds
dt
= ft/s no matter how far from the light
pole Chris is.
b. Let l = x + s, then
1 5
2
2 2
dl dx ds
dt dt dt
= + = + = ft/s.
c. The angular rate at which Chris must lift his
head to follow his shadow is the same as the
rate at which the angle that the light makes
with the ground is decreasing. Let θ be the
angle that the light makes with the ground at
the tip of Chris' shadow.
6
tan
s
θ = so 2
2
6
sec –
d ds
dt dts
θ
θ = and
2
2
6cos
– .
d ds
dt dts
θ θ
=
1
2
ds
dt
= ft/s
When s = 6, ,
4
θ
π
= so
( )
2
1
2
2
6
1 1
– – .
2 246
d
dt
θ ⎛ ⎞
= =⎜ ⎟
⎝ ⎠
Chris must lift his head at the rate of
1
24
rad/s.
18. Let θ be the measure of the vertex angle, a be the
measure of the equal sides, and b be the measure
of the base. Observe that 2 sin
2
b a
θ
= and the
height of the triangle is cos .
2
a
θ
21 1
2 sin cos sin
2 2 2 2
A a a a
θ θ
θ
⎛ ⎞⎛ ⎞
= =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
21 1
(100) sin 5000sin ;
2 10
d
A
dt
θ
θ θ= = =
5000cos
dA d
dt dt
θ
θ=
When
1
, 5000 cos 250 3
6 6 10
dA
dt
θ
π π⎛ ⎞⎛ ⎞
= = =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
2
433 cm min≈ .
19. Let p be the point on the bridge directly above
the railroad tracks. If a is the distance between p
and the automobile, then 66
da
dt
= ft/s. If l is the
distance between the train and the point directly
below p, then 88
dl
dt
= ft/s. The distance from the
train to p is 2 2
100 ,l+ while the distance from
p to the automobile is a. The distance between
the train and automobile is
2
2 2 2 2 2 2
100 100 .D a l a l⎛ ⎞= + + = + +⎜ ⎟
⎝ ⎠
2 2 2
1
2 2
2 100
dD da dl
a l
dt dt dta l
⎛ ⎞
= ⋅ +⎜ ⎟
⎝ ⎠+ +
2 2 2
.
100
da dl
dt dt
a l
a l
+
=
+ +
After 10 seconds, a = 660
and l = 880, so
2 2 2
660(66) 880(88)
110
660 880 100
dD
dt
+
= ≈
+ +
ft/s.

20. 2 21
( ); 20, 20,
3 4
h
V h a ab b a b= π ⋅ + + = = +
2
1
400 5 400 10 400
3 16
h
V h h h
⎛ ⎞
= π + + + + +⎜ ⎟
⎜ ⎟
⎝ ⎠
3
21
1200 15
3 16
h
h h
⎛ ⎞
= π + +⎜ ⎟
⎜ ⎟
⎝ ⎠
2
1 3
1200 30
3 16
dV h dh
h
dt dt
⎛ ⎞
= π + +⎜ ⎟
⎜ ⎟
⎝ ⎠
When h = 30 and 2000,
dV
dt
=
1 675 3025
2000 1200 900
3 4 4
dh dh
dt dt
π⎛ ⎞
= π + + =⎜ ⎟
⎝ ⎠
320
0.84
121
dh
dt
= ≈
π
cm/min.
21. 2
– ; –2, 8
3
h dV
V h r r
dt
⎡ ⎤
= π = =⎢ ⎥
⎣ ⎦
3 3
2 2
– 8 –
3 3
h h
V rh h
π π
= π = π
2
16 –
dV dh dh
h h
dt dt dt
= π π
When h = 3, 2
–2 [16 (3) – (3) ]
dh
dt
= π π
–2
–0.016
39
dh
dt
= ≈
π
ft/hr
22. 2 2 2
2 cos ;s a b ab θ= + −
a = 5, b = 4,
11
2 –
6 6
d
dt
θ π π
= π = rad/h
2
41– 40coss θ=
2 40sin
ds d
s
dt dt
θ
θ=
At 3:00, and 41
2
sθ
π
= = , so
11 220
2 41 40sin
2 6 3
ds
dt
π π π⎛ ⎞⎛ ⎞
= =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
18
ds
dt
≈ in./hr
23. Let P be the point on the ground where the ball
hits. Then the distance from P to the bottom of
the light pole is 10 ft. Let s be the distance
between P and the shadow of the ball. The height
of the ball t seconds after it is dropped is
2
64 –16 .t
By similar triangles,
2
48 10
64 –16
s
st
+
=
(for t > 1), so
2
2
10 – 40
.
1–
t
s
t
=
2 2
2 2
20 (1– ) – (10 – 40)(–2 )
(1– )
ds t t t t
dt t
=
2 2
60
–
(1– )
t
t
=
The ball hits the ground when t = 2,
120
– .
9
ds
dt
=
The shadow is moving
120
13.33
9
≈ ft/s.
24. 2
– ; 20
3
h
V h r r
⎛ ⎞
= π =⎜ ⎟
⎝ ⎠
2 2 3
20 – 20
3 3
h
V h h h
π⎛ ⎞
= π = π −⎜ ⎟
⎝ ⎠
2
(40 )
dV dh
h h
dt dt
= π − π
At 7:00 a.m., h = 15, 3,
dh
dt
≈ − so
2
(40 (15) (15) )( 3) 1125 3534.
dV
dt
= π − π − ≈ − π ≈ −
Webster City residents used water at the rate of
2400 + 3534 = 5934 ft3/h.
25. Assuming that the tank is now in the shape of an
upper hemisphere with radius r, we again let t be
the number of hours past midnight and h be the
height of the water at time t. The volume, V, of
water in the tank at that time is given by
( )3 22
( ) 2
3 3
V r r h r h
π
π= − − +
and so ( )216000
(20 ) 40
3 3
V h h
π
π= − − +
from which
( )2 2
(20 ) (20 ) 40
3 3
dV dh dh
h h h
dt dt dt
π π
= − − + − +
At 7t = , 525 1649
dV
dt
π≈ − ≈ −
Thus Webster City residents were using water at
the rate of 2400 1649 4049+ = cubic feet per
hour at 7:00 A.M.
26. The amount of water used by Webster City can
be found by:
usage beginning amount added amount
remaining amount
= +
−
Thus the usage is
2 2 3
(20) (9) 2400(12) (20) (10.5) 26,915 ftπ π≈ + − ≈
over the 12 hour period.

Instructor's Resource Manual Section 2.8 137
27. a. Let x be the distance from the bottom of the wall to the end of the ladder on the ground, so 2
dx
dt
= ft/s. Let y
be the height of the opposite end of the ladder. By similar triangles,
2
18
,
12 144
y
x
=
+
so
2
216
.
144
y
x
=
+
2 3/ 2
216
– 2
2(144 )
dy dx
x
dt dtx
=
+ 2 3/ 2
216
–
(144 )
x dx
dtx
=
+
When the ladder makes an angle of 60° with the ground, 4 3x = and
3/ 2
216(4 3)
– 2 –1.125
(144 48)
dy
dt
= ⋅ =
+
ft/s.
b.
2
2 2 3/ 2
216
–
(144 )
d y d x dx
dt dtdt x
⎛ ⎞
= ⎜ ⎟⎜ ⎟+⎝ ⎠
2
2 3/ 2 2 3/ 2 2
216 216
– –
(144 ) (144 )
d x dx x d x
dt dtx x dt
⎛ ⎞
= ⋅⎜ ⎟⎜ ⎟+ +⎝ ⎠
Since
2
2
2, 0,
dx d x
dt dt
= = thus
( )2 3/ 2 232
2
2 2 3
–216(144 ) 216 144 (2 )
(144 )
dx dx
dt dt
x x x xd y dx
dtdt x
⎡ ⎤+ + +
⎢ ⎥=
⎢ ⎥+⎢ ⎥⎣ ⎦
22 2
2 5/ 2
–216(144 ) 648
(144 )
x x dx
dtx
+ + ⎛ ⎞
= ⎜ ⎟
⎝ ⎠+
22
2 5/ 2
432 – 31,104
(144 )
x dx
dtx
⎛ ⎞
= ⎜ ⎟
⎝ ⎠+
When the ladder makes an angle of 60° with the ground,
2
2
2 5/ 2
432 48 – 31,104
(2) –0.08
(144 48)
d y
dt
⋅
= ≈
+
ft/s2
28. a. If the ball has radius 6 in., the volume of the
water in the tank is
33
2 4 1
8 – –
3 3 2
h
V h
π ⎛ ⎞
= π π⎜ ⎟
⎝ ⎠
3
2
8 – –
3 6
h
h
π π
= π
2
16 –
dV dh dh
h h
dt dt dt
= π π
This is the same as in Problem 21, so
dh
dt
is
again –0.016 ft/hr.
b. If the ball has radius 2 ft, and the height of
the water in the tank is h feet with 2 3h≤ ≤ ,
the part of the ball in the water has volume
2
3 24 4 – (6 – )
(2) – (4 – ) 2 – .
3 3 3
h h h
h
π⎡ ⎤
π π =⎢ ⎥
⎣ ⎦
The volume of water in the tank is
3 2
2 2(6 – )
8 – – 6
3 3
h h h
V h h
π π
= π = π
12
dV dh
h
dt dt
= π
1
12
dh dV
dt h dt
=
π
When h = 3,
1
(–2) –0.018
36
dh
dt
= ≈
π
ft/hr.
29. 2
(4 )
dV
k r
dt
= π
a. 34
3
V r= π
2
4
dV dr
r
dt dt
= π
2 2
(4 ) 4
dr
k r r
dt
π = π
dr
k
dt
=
b. If the original volume was 0V , the volume
after 1 hour is 0
8
.
27
V The original radius
was 3
0 0
3
4
r V=
π
while the radius after 1
hour is 3
1 0 0
8 3 2
.
27 4 3
r V r= ⋅ =
π
Since
dr
dt
is
constant, 0
1
–
3
dr
r
dt
= unit/hr. The snowball
will take 3 hours to melt completely.

30. PV = k
0
dV dP
P V
dt dt
+ =
At t = 6.5, P ≈ 67, –30,
dP
dt
≈ V = 300
300
– – (–30) 134
67
dV V dP
dt P dt
= = ≈ in.3/min
31. Let l be the distance along the ground from the
brother to the tip of the shadow. The shadow is
controlled by both siblings when
3 5
4l l
=
+
or
l = 6. Again using similar triangles, this occurs
when
6
,
20 3
y
= so y = 40. Thus, the girl controls
the tip of the shadow when y ≥ 40 and the boy
controls it when y < 40.
Let x be the distance along the ground from the
light pole to the girl. –4
dx
dt
=
When y ≥ 40,
20 5
–y y x
= or
4
.
3
y x=
When y < 40,
20 3
– ( 4)y y x
=
+
or
20
( 4).
17
y x= +
x = 30 when y = 40. Thus,
4
if 30
3
20
( 4) if 30
17
x x
y
x x
⎧
≥⎪⎪
= ⎨
⎪ + <
⎪⎩
and
4
if 30
3
20
if 30
17
dx
x
dy dt
dxdt
x
dt
⎧
≥⎪⎪
= ⎨
⎪ <
⎪⎩
Hence, the tip of the shadow is moving at the rate
of
4 16
(4)
3 3
= ft/s when the girl is at least 30 feet
from the light pole, and it is moving
20 80
(4)
17 17
= ft/s when the girl is less than 30 ft
from the light pole.
2.9 Concepts Review
1. ( )f x dx′
2. ;y dyΔ
3. Δx is small.
4. larger ; smaller
Problem Set 2.9
1. dy = (2x + 1)dx
2. 2
(21 6 )dy x x dx= +
3. –5 –5
–4(2 3) (2) –8(2 3)dy x dx x dx= + = +
4. 2 –3
–2(3 1) (6 1)dy x x x dx= + + +
2 –3
–2(6 1)(3 1)x x x dx= + + +
5. 2
3(sin cos ) (cos – sin )dy x x x x dx= +
6. 2 2
3(tan 1) (sec )dy x x dx= +
2 2
3sec (tan 1)x x dx= +
7. 2 –5/ 23
– (7 3 –1) (14 3)
2
dy x x x dx= + +
2 5 23
(14 3)(7 3 1)
2
x x x dx−
= − + + −
8. 10 9 1
2( sin 2 )[10 (cos 2 )(2)]
2 sin 2
x x x x
x
dy dx+ + ⋅=
9 10cos2
2 10 ( sin 2 )
sin 2
x
x x x dx
x
⎛ ⎞
= + +⎜ ⎟
⎝ ⎠
9. 2 1/ 2 23
( – cot 2) (2 csc )
2
ds t t t t dt= + +
2 23
(2 csc ) – cot 2
2
t t t t dt= + +
10. a. 2 2
3 3(0.5) (1) 0.75dy x dx= = =
b. 2 2
3 3(–1) (0.75) 2.25dy x dx= = =
11.
12. a.
2 2
0.5
– – –0.5
(1)
dx
dy
x
= = =
b.
2 2
0.75
– – –0.1875
(–2)
dx
dy
x
= = =

13.
14. a. 3 3
(1.5) – (0.5) 3.25yΔ = =
b. 3 3
(–0.25) – (–1) 0.984375yΔ = =
15. a.
1 1 1
– –
1.5 1 3
yΔ = =
b.
1 1
–0.3
–1.25 2
yΔ = + =
16. a. 2 2
[(2.5) – 3] –[(2) – 3] 2.25yΔ = =
dy = 2xdx = 2(2)(0.5) = 2
b. 2 2
[(2.88) – 3] –[(3) – 3] –0.7056yΔ = =
dy = 2xdx = 2(3)(–0.12) = –0.72
17. a. 4 4
[(3) 2(3)] –[(2) 2(2)] 67yΔ = + + =
3 3
(4 2) [4(2) 2](1) 34dy x dx= + = + =
b. 4 4
[(2.005) 2(2.005)] –[(2) 2(2)]yΔ = + +
≈ 0.1706
3 3
(4 2) [4(2) 2](0.005) 0.17dy x dx= + = + =
18.
1
; ; 400, 2
2
y x dy dx x dx
x
= = = =
1
(2) 0.05
2 400
dy = =
402 400 20 0.05 20.05dy≈ + = + =
19.
1
; ; 36, –0.1
2
y x dy dx x dx
x
= = = =
1
(–0.1) –0.0083
2 36
dy = ≈
35.9 36 6 – 0.0083 5.9917dy≈ + = =
20. –2/33
3 2
1 1
; ;
3 3
y x dy x dx dx
x
= = =
x = 27, dx = –0.09
23
1
(–0.09) –0.0033
3 (27)
dy = ≈
3 3
26.91 27 3 – 0.0033 2.9967dy≈ + = =
21. 34
; 5, 0.125
3
V r r dr= π = =
2 2
4 4 (5) (0.125) 39.27dV r dr= π = π ≈ cm3
22. 3 3
; 40, 0.5V x x dx= = =
2 23
3 3( 40) (0.5) 17.54dV x dx= = ≈ in.3
23. 34
; 6ft 72in., –0.3
3
V r r dr= π = = =
2 2
4 4 (72) (–0.3) –19,543dV r dr= π = π ≈
3
3 3
4
(72) –19,543
3
1,543,915 in 893 ft
V ≈ π
≈ ≈
24. 2
; 6ft 72in., 0.05,V r h r dr= π = = = −
8ft 96in.h = =
3
2 2 (72)(96)( 0.05) 2171in.dV rhdr= π = π − ≈ −
About 9.4 gal of paint are needed.
25. 2C rπ= ; r = 4000 mi = 21,120,000 ft, dr = 2
2 2 (2) 4 12.6dC dr ftπ π π= = = ≈
26. 2 ; 4, –0.03
32
L
T L dL= π = =
32
2 1
32 322 L
dT dL dL
L
π π
= ⋅ ⋅ =
(–0.03) –0.0083
32(4)
dT
π
= ≈
The time change in 24 hours is
(0.0083)(60)(60)(24) ≈ 717 sec
27. 3 34 4
(10) 4189
3 3
V r= π = π ≈
2 2
4 4 (10) (0.05) 62.8dV r dr= π = π ≈ The
volume is 4189 ± 62.8 cm3.
The absolute error is ≈ 62.8 while the relative
error is 62.8/ 4189 0.015 or 1.5%≈ .

28. 2 2
(3) (12) 339V r h= π = π ≈
24 24 (3)(0.0025) 0.565dV rdr= π = π ≈
The volume is 339 ± 0.565 in.3
error is 0.565/339 0.0017 or 0.17%≈ .
29. 2 2
– 2 coss a b ab θ= +
2 2
151 151 – 2(151)(151)cos0.53= + ≈ 79.097
45,602 – 45,602coss θ=
1
45,602sin
2 45,602 – 45,602cos
ds dθ θ
θ
= ⋅
22,801sin
45,602 – 45,602cos
d
θ
θ
θ
=
22,801sin 0.53
(0.005)
45,602 – 45,602cos0.53
= ≈ 0.729
s ≈ 79.097 ± 0.729 cm
error is 0.729/ 79.097 0.0092 or 0.92%≈ .
30.
1 1
sin (151)(151)sin 0.53 5763.33
2 2
A ab θ= = ≈
22,801
sin ; 0.53, 0.005
2
A dθ θ θ= = =
22,801
(cos )
2
dA dθ θ=
22,801
(cos0.53)(0.005) 49.18
2
= ≈
A ≈ 5763.33 ± 49.18 cm2
error is 49.18/5763.33 0.0085 or 0.85%≈ .
31. 2
3 – 2 11; 2, 0.001y x x x dx= + = =
dy = (6x – 2)dx = [6(2) – 2](0.001) = 0.01
2
2
6,
d y
dx
= so with Δx = 0.001,
21
– (6)(0.001) 0.000003
2
y dyΔ ≤ =
32. Using the approximation
( ) ( ) '( )f x x f x f x x+ Δ ≈ + Δ
we let 1.02x = and 0.02xΔ = − . We can rewrite
the above form as
( ) ( ) '( )f x f x x f x x≈ + Δ − Δ
which gives
(1.02) (1) '(1.02)( 0.02)
10 12(0.02) 10.24
f f f≈ − −
= + =
33. Using the approximation
( ) ( ) '( )f x x f x f x x+ Δ ≈ + Δ
we let 3.05x = and 0.05xΔ = − . We can rewrite
the above form as
( ) ( ) '( )f x f x x f x x≈ + Δ − Δ
which gives
(3.05) (3) '(3.05)( 0.05)
1
8 (0.05) 8.0125
4
f f f≈ − −
= + =
34. From similar triangles, the radius at height h is
2
.
5
h Thus, 2 31 4
,
3 75
V r h h= π = π so
24
.
25
dV h dh= π h = 10, dh = –1:
34
(100)( 1) 50cm
25
dV = π − ≈ −
The ice cube has volume 3 3
3 27cm ,= so there is
room for the ice cube without the cup
overflowing.
35. 2 34
3
V r h r= π + π
2 34
100 ; 10, 0.1
3
V r r r dr= π + π = =
2
(200 4 )dV r r dr= π + π
(2000 400 )(0.1) 240 754= π + π = π ≈ cm3
36. The percent increase in mass is .
dm
m
–3/ 22
0
2 2
2
– 1– –
2
m v v
dm dv
c c
⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
–3/ 22
0
2 2
1–
m v v
dv
c c
⎛ ⎞
= ⎜ ⎟
⎜ ⎟
⎝ ⎠
–12 2
2 2 2 2 2
1–
dm v v v c
dv dv
m c c c c v
⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
2 2
v
dv
c v
=
−
v = 0.9c, dv = 0.02c
2 2
0.9 0.018
(0.02 ) 0.095
0.190.81
dm c
c
m c c
= = ≈
−
The percent increase in mass is about 9.5.

37. 2
( ) ; '( ) 2 ; 2f x x f x x a= = =
The linear approximation is then
( ) (2) '(2)( 2)
4 4( 2) 4 4
L x f f x
x x
= + −
= + − = −
38. 2 2
( ) cos ; '( ) sin 2 cos
/ 2
g x x x g x x x x x
a π
= = − +
=
2
2 3
( ) 0
2 2
4 8
L x x
x
π π
π π
⎛ ⎞ ⎛ ⎞
= + − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= − +
2
2 3
( ) 0
4 2
4 8
L x x
x
π π
π π
⎛ ⎞
= + − −⎜ ⎟
⎝ ⎠
= − +
39. ( ) sin ; '( ) cos ; 0h x x h x x a= = =
( ) 0 1( 0)L x x x= + − =
40. ( ) 3 4; '( ) 3; 3F x x F x a= + = =
( ) 13 3( 3) 13 3 9
3 4
L x x x
x
= + − = + −
= +

41. ( )
( ) ( )
2
1/ 2
2
2
1 ;
1
1 ( 2 )
2
, 0
1
f x x
f x x x
x
a
x
−
= −
′ = − −
−
= =
−
( ) ( )1 0 0 1L x x= + − =
42. ( ) 2
1
x
g x
x
=
−
;
( )
( ) ( )
( ) ( )
2 2
2 22 2
1 2 1 1
' ,
21 1
x x x x
g x a
x x
− − − +
= = =
− −
( )
9
4
9
20
2
1
9
20
3
2
−=⎟
⎠
⎞
⎜
⎝
⎛
−+= xxxL
43. ( ) ( ) 0,tansecsec;sec =+=′= axxxxxhxxxh
( ) ( ) xxxL =−+= 010
44. ( ) ( ) 2/,2cos21;2sin π=+=′+= axxGxxxG
( ) ( ) π
ππ
+−=⎟
⎠
⎞
⎜
⎝
⎛
−−+= xxxL
2
1
2
45. ( ) ( ) mxfbmxxf =′+= ;
( ) ( )
( ) ( )xLxfbmx
mamxbamaxmbmaxL
=+=
−++=−++=
46. ( ) ( ) ( )
( )
2
1
2
2
22 2
0
2
L x f x a x a x
a
x a x a x a
x
a a
x a
a
− = + − −
− +
= − + =
−
= ≥
47. The linear approximation to ( )f x at a is
2
2
( ) ( ) '( )( )
2 ( )
2
L x f a f a x a
a a x a
ax a
= + −
= + −
= −
Thus,
( )2 2
2 2
2
( ) ( ) 2
2
( )
0
f x L x x ax a
x ax a
x a
− = − −
= − +
= −
≥
48. ( ) ( ) ( ) ( ) 0,1,1 1
=+=′+= −
axxfxxf αα
α
( ) ( ) 11 +=+= xxxL αα
5
−5
−5 5 x
y
2α = −
5
−5
−5 5 x
y
1α = −

5
−5
−5 5 x
y
0.5α = −
5
−5
−5 5 x
y
0α =
5
−5
−5 5 x
y
0.5α =
5
−5
−5 5 x
y
1α =
5
−5
−5 5 x
y
2α =
49. a. ( ) ( ) ( ) ( )( )
( ) ( ) ( )
0 0
lim lim
0 0
h h
h f x h f x f x h
f x f x f x
ε
→ →
′= + − −
′= − − =
b.
( ) ( ) ( )
( )
( ) ( )
0
lim lim
0
h
h f x h f x
f x
h h
f x f x
ε
→
+ −⎡ ⎤
′= −⎢ ⎥
⎣ ⎦
′ ′= − =
2.10 Chapter Review
Concepts Test
1. False: If 3 2
( ) , '( ) 3f x x f x x= = and the
tangent line 0 at 0y x= = crosses the
curve at the point of tangency.
2. False: The tangent line can touch the curve
at infinitely many points.
3. True: 3
tan 4 ,m x= which is unique for each
value of x.
4. False: tan –sin ,m x= which is periodic.
5. True: If the velocity is negative and
increasing, the speed is decreasing.
6. True: If the velocity is negative and
decreasing, the speed is increasing.
7. True: If the tangent line is horizontal, the
slope must be 0.
8. False: 2 2
( ) , ( ) ,f x ax b g x ax c= + = +
b c≠ . Then ( ) 2 ( ),f x ax g x′ ′= = but
f(x) ≠ g(x).
9. True: ( ( )) ( ( )) ( );xD f g x f g x g x′ ′= since
g(x) = x, ( ) 1,g x′ = so
( ( )) ( ( )).xD f g x f g x′=
10. False: 0xD y = because π is a constant, not
a variable.
11. True: Theorem 3.2.A
12. True: The derivative does not exist when the
tangent line is vertical.
13. False: ( ) ( ) ( ) ( ) ( ) ( )f g x f x g x g x f x′ ′ ′⋅ = +
14. True: Negative acceleration indicates
decreasing velocity.

15. True: If 3
( ) ( ),f x x g x= then
3 2
( ) ( ) 3 ( )xD f x x g x x g x′= +
2
[ ( ) 3 ( )].x xg x g x′= +
16. False: 2
3 ;xD y x= At (1, 1):
2
tan 3(1) 3m = =
Tangent line: y – 1 = 3(x – 1)
17. False: ( ) ( ) ( ) ( )xD y f x g x g x f x′ ′= +
2
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
xD y f x g x g x f x
g x f x f x g x
′′ ′ ′= +
′′ ′ ′+ +
( ) ( ) 2 ( ) ( ) ( ) ( )f x g x f x g x f x g x′′ ′ ′ ′′= + +
18. True: The degree of 3 8
( )y x x= + is 24, so
25
0.xD y =
19. True: –1
( ) ; ( )n n
f x ax f x anx′= =
20. True:
2
( ) ( ) ( ) – ( ) ( )
( ) ( )
x
f x g x f x f x g x
D
g x g x
′ ′
=
21. True: ( ) ( ) ( ) ( ) ( )h x f x g x g x f x′ ′ ′= +
( ) ( ) ( ) ( ) ( )h c f c g c g c f c′ ′ ′= +
= f(c)(0) + g(c)(0) = 0
22. True:
( )2
22
sin – sin
lim
2 –x
x
f
x
π
ππ→
π⎛ ⎞
′ =⎜ ⎟
⎝ ⎠
22
sin –1
lim
–x
x
x ππ→
=
23. True: 2 2
( )D kf kD f= and
2 2 2
( )D f g D f D g+ = +
24. True: ( ) ( ( )) ( )h x f g x g x′ ′ ′= ⋅
( ) ( ( )) ( ) 0h c f g c g c′ ′ ′= ⋅ =
25. True: ( ) (2) ( (2)) (2)f g f g g′ ′ ′= ⋅
(2) (2) 2 2 4f g′ ′= ⋅ = ⋅ =
26. False: Consider ( ) .f x x= The curve
always lies below the tangent.
27. False: The rate of volume change depends
on the radius of the sphere.
28. True: 2c rπ= ; 4
dr
dt
=
2 2 (4) 8
dc dr
dt dt
= π = π = π
29. True: (sin ) cos ;xD x x=
2
(sin ) –sin ;xD x x=
3
(sin ) – cos ;xD x x=
4
(sin ) sin ;xD x x=
5
(sin ) cosxD x x=
30. False: (cos ) –sin ;xD x x=
2
(cos ) – cos ;xD x x=
3
(cos ) sin ;xD x x=
4 3
(cos ) [ (cos )] (sin )x x x xD x D D x D x= =
Since 1 3 1
(cos ) (sin ),x xD x D x+
=
3
(cos ) (sin ).n n
x xD x D x+
=
31. True:
0 0
tan 1 sin
lim lim
3 3 cosx x
x x
x x x→ →
=
1 1
1
3 3
= ⋅ =
32. True: 2
15 6
ds
v t
dt
= = + which is greater
than 0 for all t.
33. True: 34
3
V r= π
2
4
dV dr
r
dt dt
= π
If 3,
dV
dt
= then
2
3
4
dr
dt r
=
π
so
0.
dr
dt
>
2
2 3
3
–
2
d r dr
dtdt r
=
π
so
2
2
0
d r
dt
<
34. True: When h > r, then
2
2
0
d h
dt
>
35. True: 34
,
3
V r= π 2
4S r= π
2
4dV r dr S dr= π = ⋅
If Δr = dr, then dV S r= ⋅Δ
36. False: 4
5 ,dy x dx= so dy > 0 when dx > 0,
but dy < 0 when dx < 0.
37. False: The slope of the linear approximation
is equal to
'( ) '(0) sin(0) 0f a f= = − = .

Sample Test Problems
1. a.
3 3
0
3( ) – 3
( ) lim
h
x h x
f x
h→
+
′ =
2 2 3
0
9 9 3
lim
h
x h xh h
h→
+ +
= 2 2 2
0
lim (9 9 3 ) 9
h
x xh h x
→
= + + =
b.
5 5
0
[2( ) 3( )] – (2 3 )
( ) lim
h
x h x h x x
f x
h→
+ + + +
′ =
4 3 2 2 3 4 5
0
10 20 20 10 2 3
lim
h
x h x h x h xh h h
h→
+ + + + +
=
4 3 2 2 3 4
0
lim (10 20 20 10 2 3)
h
x x h x h xh h
→
= + + + + + 4
10 3x= +
c.
1 1
3( ) 3
0 0
– 1
( ) lim lim –
3( )
x h x
h h
h
f x
h x h x h
+
→ →
⎡ ⎤
′ = = ⎢ ⎥
+⎣ ⎦ 20
1 1
lim – –
3 ( ) 3h x x h x→
⎛ ⎞
= =⎜ ⎟
+⎝ ⎠
d.
2 20
1 1 1
( ) lim –
3( ) 2 3 2h
f x
hx h x→
⎡ ⎤⎛ ⎞
′ = ⎢ ⎥⎜ ⎟⎜ ⎟+ + +⎢ ⎥⎝ ⎠⎣ ⎦
2 2
2 20
3 2 – 3( ) – 2 1
lim
(3( ) 2)(3 2)h
x x h
hx h x→
⎡ ⎤+ +
= ⋅⎢ ⎥
+ + +⎢ ⎥⎣ ⎦
2
2 20
–6 – 3 1
lim
(3( ) 2)(3 2)h
xh h
hx h x→
⎡ ⎤
= ⋅⎢ ⎥
+ + +⎢ ⎥⎣ ⎦
2 2 2 20
–6 – 3 6
lim –
(3( ) 2)(3 2) (3 2)h
x h x
x h x x→
= =
+ + + +
e.
0
3( ) – 3
( ) lim
h
x h x
f x
h→
+
′ =
0
( 3 3 – 3 )( 3 3 3 )
lim
( 3 3 3 )h
x h x x h x
h x h x→
+ + +
=
+ +
0 0
3 3
lim lim
( 3 3 3 ) 3 3 3h h
h
h x h x x h x→ →
= =
+ + + +
3
2 3x
=
f.
0
sin[3( )] – sin3
( ) lim
h
x h x
f x
h→
+
′ =
0
sin(3 3 ) – sin3
lim
h
x h x
h→
+
=
0
sin3 cos3 sin3 cos3 – sin3
lim
h
x h h x x
h→
+
=
0 0
sin3 (cos3 –1) sin3 cos3
lim lim
h h
x h h x
h h→ →
= +
0 0
cos3 –1 sin3
3sin3 lim cos3 lim
3h h
h h
x x
h h→ →
= +
0
sin3
(3sin3 )(0) (cos3 )3 lim
3h
h
x x
h→
= + (cos3 )(3)(1) 3cos3x x= =
g.
2 2
0
( ) 5 – 5
( ) lim
h
x h x
f x
h→
+ + +
′ =
2 2 2 2
0 2 2
( ) 5 – 5 ( ) 5 5
lim
( ) 5 5h
x h x x h x
h x h x→
⎛ ⎞⎛ ⎞+ + + + + + +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠=
⎛ ⎞+ + + +⎜ ⎟
⎝ ⎠
2
0 2 2
2
lim
( ) 5 5h
xh h
h x h x→
+
=
⎛ ⎞+ + + +⎜ ⎟
⎝ ⎠
2 2 2 20
2 2
lim
( ) 5 5 2 5 5h
x h x x
x h x x x→
+
= = =
+ + + + + +
h.
0
cos[ ( )] – cos
( ) lim
h
x h x
f x
h→
π + π
′ =
0
cos( ) – cos
lim
h
x h x
h→
π + π π
=
0
cos cos – sin sin – cos
lim
h
x h x h x
h→
π π π π π
=
0 0
1– cos sin
lim – cos lim sin
h h
h h
x x
h h→ →
π π⎛ ⎞ ⎛ ⎞
= π π − π π⎜ ⎟ ⎜ ⎟
π π⎝ ⎠ ⎝ ⎠
(– cos )(0) – ( sin ) – sinx x xπ= π π π = π π
2. a.
2 2
2 – 2 2( – )( )
( ) lim lim
– –t x t x
t x t x t x
g x
t x t x→ →
+
′ = =
2 lim( ) 2(2 ) 4
t x
t x x x
→
= + = =

b.
3 3
( ) – ( )
( ) lim
–t x
t t x x
g x
t x→
+ +
′ =
2 2
( – )( ) ( – )
lim
–t x
t x t tx x t x
t x→
+ + +
=
2 2
lim( 1)
t x
t tx x
→
= + + + 2
3 1x= +
c.
1 1– –
( ) lim lim
– ( – )
t x
t x t x
x t
g x
t x tx t x→ →
′ = =
2
–1 1
lim –
t x tx x→
= =
d.
2 2
1 1 1
( ) lim –
–1 1t x
g x
t xt x→
⎡ ⎤⎛ ⎞⎛ ⎞
′ = ⎢ ⎥⎜ ⎟⎜ ⎟
⎝ ⎠+ +⎝ ⎠⎣ ⎦
2 2
2 2
–
lim
( 1)( 1)( – )t x
x t
t x t x→
=
+ +
2 2
–( )( – )
lim
( 1)( 1)( – )t x
x t t x
t x t x→
+
=
+ +
2 2 2 2
–( ) 2
lim –
( 1)( 1) ( 1)t x
x t x
t x x→
+
= =
+ + +
e.
–
( ) lim
–t x
t x
g x
t x→
′ =
( – )( )
lim
( – )( )t x
t x t x
t x t x→
+
=
+
– 1
lim lim
( – )( )t x t x
t x
t x t x t x→ →
= =
+ +
1
2 x
=
f.
sin – sin
( ) lim
–t x
t x
g x
t x→
π π
′ =
Let v = t – x, then t = v + x and as
, 0.t x v→ →
0
sin – sin sin ( ) – sin
lim lim
–t x v
t x v x x
t x v→ →
π π π + π
=
0
sin cos sin cos – sin
lim
v
v x x v x
v→
π π + π π π
=
0
sin cos –1
lim cos sin
v
v v
x x
v v→
π π⎡ ⎤
= π π + π π⎢ ⎥π π⎣ ⎦
cos 1 sin 0 cosx x x= π π ⋅ + π π ⋅ = π π
Other method:
Use the subtraction formula
( ) ( )
sin – sin 2cos sin
2 2
t x t x
t x
π + π −
π π =
g.
3 3
–
( ) lim
–t x
t C x C
g x
t x→
+ +
′ =
3 3 3 3
3 3
–
lim
( – )t x
t C x C t C x C
t x t C x C→
⎛ ⎞⎛ ⎞+ + + + +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠=
⎛ ⎞+ + +⎜ ⎟
⎝ ⎠
3 3
3 3
–
lim
( – )t x
t x
t x t C x C→
=
⎛ ⎞+ + +⎜ ⎟
⎝ ⎠
2 2 2
3 3 3
3
lim
2t x
t tx x x
t C x C x C→
+ +
= =
+ + + +
h.
cos2 – cos2
( ) lim
–t x
t x
g x
t x→
′ =
Let v = t – x, then t = v + x and as
, 0.t x v→ →
0
cos2 – cos2 cos2( ) – cos2
lim lim
–t x v
t x v x x
t x v→ →
+
=
0
cos2 cos2 – sin 2 sin 2 – cos2
lim
v
v x v x x
v→
=
0
cos2 –1 sin 2
lim 2cos2 – 2sin 2
2 2v
v v
x x
v v→
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
2cos2 0 – 2sin 2 1 –2sin 2x x x= ⋅ ⋅ =
Other method:
Use the subtraction formula
cos2 cos2 2sin( )sin( ).t x t x t x− = − + −
3. a. f(x) = 3x at x = 1
b. 3
( ) 4f x x= at x = 2
c. 3
( )f x x= at x = 1
d. f(x) = sin x at x π=
e.
4
( )f x
x
= at x
f. f(x) = –sin 3x at x
g. f(x) = tan x at
4
x
π
=
h.
1
( )f x
x
= at x = 5
4. a.
3
(2) –
4
f ′ ≈
b.
3
(6)
2
f ′ ≈

c.
3
2
avg
6 – 9
7 – 3 8
V = =
d. 2 2
( ) ( )(2 )
d
f t f t t
dt
′=
At t = 2,
2 8
4 (4) 4
3 3
f
⎛ ⎞′ ≈ =⎜ ⎟
⎝ ⎠
e. 2
[ ( )] 2 ( ) ( )
d
f t f t f t
dt
′=
At t = 2,
2 (2) (2)f f ′
3
2(2) – –3
4
⎛ ⎞
≈ =⎜ ⎟
⎝ ⎠
f. ( ( ( ))) ( ( )) ( )
d
f f t f f t f t
dt
′ ′=
At t = 2, ( (2)) (2) (2) (2)f f f f f′ ′ ′ ′=
3 3 9
– –
4 4 16
⎛ ⎞⎛ ⎞
≈ =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
5. 5 4
(3 ) 15xD x x=
6. 3 2 –2 2 –3
( – 3 ) 3 – 6 (–2)xD x x x x x x+ = +
2 –3
3 – 6 – 2x x x=
7. 3 2 2
( 4 2 ) 3 8 2zD z z z z z+ + = + +
8.
2
2 2 2
3 – 5 ( 1)(3) – (3 – 5)(2 )
1 ( 1)
x
x x x x
D
x x
+⎛ ⎞
=⎜ ⎟
+ +⎝ ⎠
2
2 2
3 10 3
( 1)
x x
x
− + +
=
+
9.
2
2 2 2
4 5 (6 2 )(4) – (4 – 5)(12 2)
6 2 (6 2 )
t
t t t t t
D
t t t t
− + +⎛ ⎞
=⎜ ⎟
+ +⎝ ⎠
2
2 2
24 60 10
(6 2 )
t t
t t
− + +
=
+
10. 2/3 –1/32
(3 2) (3 2) (3)
3
xD x x+ = +
–1/3
2(3 2)x= +
2 2/3 –4/32
(3 2) – (3 2) (3)
3
xD x x+ = +
–4/3
–2(3 2)x= +
11.
2 3 2 2
3 3 2
4 – 2 ( )(8 ) – (4 – 2)(3 1)
( )
d x x x x x x
dx x x x x
⎛ ⎞ + +
=⎜ ⎟
⎜ ⎟+ +⎝ ⎠
4 2
3 2
4 10 2
( )
x x
x x
− + +
=
+
12.
1
( 2 6) (2) 2 6
2 2 6
tD t t t t
t
+ = + +
+
2 6
2 6
t
t
t
= + +
+
13.
2 –1/ 2
2
2 –3/ 2
1
( 4)
4
1
– ( 4) (2 )
2
d d
x
dx dxx
x x
⎛ ⎞
⎜ ⎟ = +
⎜ ⎟
+⎝ ⎠
= +
2 3
–
( 4)
x
x
=
+
14.
2
1 2
3 3 2
–1 1 1
– 2
d x d d
x
dx dx dxxx x x
−
= = = −
15. 3 2
(sin cos ) cos 3cos (–sin )Dθ θ θ θ θ θ+ = +
2
cos – 3sin cosθ θ θ=
2 3
3
(sin cos )
–sin – 3[sin (2)(cos )(–sin ) cos ]
Dθ θ θ
θ θ θ θ θ
+
= +
2 3
–sin 6sin cos – 3cosθ θ θ θ= +
16. 2 2 2
[sin( ) – sin ( )] cos( )(2 ) – (2sin )(cos )
d
t t t t t t
dt
=
2
2 cos( ) – sin(2 )t t t=
17. 2 2 2
[sin( )] cos( )(2 ) 2 cos( )Dθ θ θ θ θ θ= =
18.
3 2
2
(cos 5 ) (3cos 5 )(–sin5 )(5)
–15cos 5 sin5
d
x x x
dx
x x
=
=

19. 2
[sin (sin( ))] 2sin(sin( ))cos(sin( ))(cos( ))( )
d
d
θ θ θ θ
θ
π = π π π π 2 sin(sin( ))cos(sin( ))cos( )θ θ θ= π π π π
20. ( )2
[sin (cos4 )] 2sin(cos4 ) cos(cos4 ) (–sin 4 )(4)
d
t t t t
dt
= –8sin(cos4 )cos(cos4 )sin 4t t t=
21. 2 2
tan3 (sec 3 )(3) 3sec 3Dθ θ θ θ= =
22.
2 2
2 2 2
sin3 (cos5 )(cos3 )(3) – (sin3 )(–sin5 )(10 )
cos5 cos 5
d x x x x x x
dx x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
2 2
2 2
3cos5 cos3 10 sin3 sin5
cos 5
x x x x x
x
+
=
23. 2 2 2 3 2
( ) ( –1) (9 – 4) (3 – 4 )(2)( –1)(2 )f x x x x x x x′ = + 2 2 2 2 3
( –1) (9 – 4) 4 ( –1)(3 – 4 )x x x x x x= +
(2) 672f ′ =
24. ( ) 3cos3 2(sin3 )(cos3 )(3)g x x x x′ = + 3cos3 3sin 6x x= +
( ) –9sin3 18cos6g x x x′′ = +
(0) 18g′′ =
25.
2 2 2 2
2 2 2
cot (sec )(– csc ) – (cot )(sec )(tan )(2 )
sec sec
d x x x x x x x
dx x x
⎛ ⎞
=⎜ ⎟
⎝ ⎠
2 2
2
– csc – 2 cot tan
sec
x x x x
x
=
26.
2
4 sin (cos – sin )(4 cos 4sin ) – (4 sin )(–sin – cos )
cos – sin (cos – sin )
t
t t t t t t t t t t t
D
t t t t
+⎛ ⎞
=⎜ ⎟
⎝ ⎠
2 2 2
2
4 cos 2sin 2 – 4sin 4 sin
(cos – sin )
t t t t t t
t t
+ +
=
2
2
4 2sin 2 – 4sin
(cos – sin )
t t t
t t
+
=
27. 3 2 2
( ) ( –1) 2(sin – )( cos –1) (sin – ) 3( –1)f x x x x x x x x′ = π π π + π
3 2 2
2( –1) (sin – )( cos –1) 3(sin – ) ( –1)x x x x x x x= π π π + π
(2) 16 4 3.43f ′ = − π ≈
28. 4
( ) 5(sin(2 ) cos(3 )) (2cos(2 ) – 3sin(3 ))h t t t t t′ = +
4 3 2
( ) 5(sin(2 ) cos(3 )) ( 4sin(2 ) – 9cos(3 )) 20(sin(2 ) cos(3 )) (2cos(2 ) – 3sin(3 ))h t t t t t t t t t′′ = + − + +
4 3 2
(0) 5 1 ( 9) 20 1 2 35h′′ = ⋅ ⋅ − + ⋅ ⋅ =
29. 2
( ) 3(cos 5 )(–sin5 )(5)g r r r′ = 2
–15cos 5 sin5r r=
2
( ) –15[(cos 5 )(cos5 )(5) (sin5 )2(cos5 )(–sin5 )(5)]g r r r r r r′′ = + 3 2
–15[5cos 5 –10(sin 5 )(cos5 )]r r r=
2 2
( ) –15[5(3)(cos 5 )(–sin5 )(5) (10sin 5 )( sin5 )(5) (cos5 )(20sin5 )(cos5 )(5)]g r r r r r r r r′′′ = − − −
2 3
–15[ 175(cos 5 )(sin5 ) 50sin 5 ]r r r= − +
(1) 458.8g′′′ ≈
30. ( ) ( ( )) ( ) 2 ( ) ( )f t h g t g t g t g t′ ′ ′ ′= +
31. ( ) ( ( ) ( ))( ( ) ( )) ( )G x F r x s x r x s x s x′ ′ ′ ′ ′= + + +
( ) ( ( ) ( ))( ( ) ( )) ( ( ) ( )) ( ( ) ( ))( ( ) ( )) ( )G x F r x s x r x s x r x s x F r x s x r x s x s x′′ ′ ′′ ′′ ′ ′ ′′ ′ ′ ′′= + + + + + + +
2
( ( ) ( ))( ( ) ( )) ( ( ) ( )) ( ( ) ( )) ( )F r x s x r x s x r x s x F r x s x s x′ ′′ ′′ ′ ′ ′′ ′′= + + + + + +

32. 2
( ) ( ( )) ( ) 3[ ( )] (–sin )F x Q R x R x R x x′ ′ ′= =
2
–3cos sinx x=
33. 2
( ) ( ( )) ( ) [3cos(3 ( ))](9 )F z r s z s z s z z′ ′ ′= =
2 3
27 cos(9 )z z=
34. 2( – 2)
dy
x
dx
=
2x – y + 2 = 0; y = 2x + 2; m = 2
1
2( – 2) –
2
x =
7
4
x =
2
7 1 7 1
– 2 ; ,
4 16 4 16
y
⎛ ⎞ ⎛ ⎞
= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
35. 34
3
V r= π
2
4
dV
r
dr
= π
When r = 5, 2
4 (5) 100 314
dV
dr
= π = π ≈ m3 per
meter of increase in the radius.
36. 34
; 10
3
dV
V r
dt
= π =
2
4
dV dr
r
dt dt
= π
When r = 5, 2
10 4 (5)
dr
dt
= π
1
0.0318
10
dr
dt
= ≈
π
m/h
37.
1 6 3
(12); ;
2 4 2
b h
V bh b
h
= = =
23
6 9 ; 9
2
h dV
V h h
dt
⎛ ⎞
= = =⎜ ⎟
⎝ ⎠
18
dV dh
h
dt dt
=
When h = 3, 9 18(3)
dh
dt
=
1
0.167
6
dh
dt
= ≈ ft/min
38. a. v = 128 – 32t
v = 0, when 4t s=
2
128(4) –16(4) 256s = = ft
b. 2
128 –16 0t t =
–16t(t – 8) = 0
The object hits the ground when t = 8s
v = 128 – 32(8) = –128 ft/s
39. 3 2
– 6 9s t t t= +
2
( ) 3 –12 9
ds
v t t t
dt
= = +
2
2
( ) 6 –12
d s
a t t
dt
= =
a. 2
3 –12 9 0t t + <
3(t – 3)(t – 1) < 0
1 < t < 3; (1,3)
b. 2
3 –12 9 0t t + =
3(t – 3)(t – 1) = 0
t = 1, 3
a(1) = –6, a(3) = 6
c. 6t – 12 > 0
t > 2; (2, )∞
40. a. 20 19 12 5
( 100) 0xD x x x+ + + =
b. 20 20 19 18
( ) 20!xD x x x+ + =
c. 20 21 20
(7 3 ) (7 21!) (3 20!)xD x x x+ = ⋅ + ⋅
d. 20 4
(sin cos ) (sin cos )x xD x x D x x+ = +
= sin x + cos x
e. 20 20
(sin 2 ) 2 sin 2xD x x=
= 1,048,576 sin 2x
f.
20
20
21 21
1 (–1) (20!) 20!
xD
x x x
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
41. a. 2( –1) 2 0
dy
x y
dx
+ =
–( –1) 1–dy x x
dx y y
= =
b. 2 2
(2 ) (2 ) 0
dy dy
x y y y x x
dx dx
+ + + =
2 2
(2 ) –( 2 )
dy
xy x y xy
dx
+ = +
2
2
2
2
dy y xy
dx x xy
+
= −
+

c. 2 2 3 2 2 3
3 3 (3 ) 3
dy dy
x y x y x y
dx dx
+ = +
2 3 2 2 3 2
(3 – 3 ) 3 – 3
dy
y x y x y x
dx
=
2 3 2 2 3 2
2 3 2 2 3 2
3 – 3 –
3 – 3 –
dy x y x x y x
dx y x y y x y
= =
d. cos( ) sin( ) 2
dy
x xy x y xy x
dx
⎡ ⎤
+ + =⎢ ⎥
⎣ ⎦
2
cos( ) 2 – sin( ) – cos( )
dy
x xy x xy xy xy
dx
=
2
2 – sin( ) – cos( )
cos( )
dy x xy xy xy
dx x xy
=
e. 2
sec ( ) tan( ) 0
dy
x xy x y xy
dx
⎛ ⎞
+ + =⎜ ⎟
⎝ ⎠
2 2 2
sec ( ) –[tan( ) sec ( )]
dy
x xy xy xy xy
dx
= +
2
2 2
tan( ) sec ( )
–
sec ( )
dy xy xy xy
dx x xy
+
=
42. 2
12 12yy x′ =
2
1
6x
y
y
′ =
At (1, 2): 1 3y′ =
24 6 0x yy′+ =
2
2
–
3
x
y
y
′ =
At (1, 2): 2
1
–
3
y′ =
Since 1 2( )( ) –1y y′ ′ = at (1, 2), the tangents are
perpendicular.
43. [ cos( ) 2 ]dy x x dxπ π= + ; x = 2, dx = 0.01
[ cos(2 ) 2(2)](0.01) (4 )(0.01)dy π π π= + = +
≈ 0.0714
44. 2 2
(2 ) 2 [2( 2)] ( 2) (2) 0
dy dy
x y y y x x
dx dx
+ + + + + =
2 2
[2 2( 2) ] –[ 2 (2 4)]
dy
xy x y y x
dx
+ + = + +
2
2
–( 4 8 )
2 2( 2)
dy y xy y
dx xy x
+ +
=
+ +
2
2
4 8
–
2 2( 2)
y xy y
dy dx
xy x
+ +
=
+ +
When x = –2, y = ±1
a.
2
2
(1) 4(–2)(1) 8(1)
– (–0.01)
2(–2)(1) 2(–2 2)
dy
+ +
=
+ +
= –0.0025
b.
2
2
(–1) 4(–2)(–1) 8(–1)
– (–0.01)
2(–2)(–1) 2(–2 2)
dy
+ +
=
+ +
= 0.0025
45. a. 2 3
[ ( ) ( )]
d
f x g x
dx
+
2
2 ( ) ( ) 3 ( ) ( )f x f x g x g x′ ′= +
2
2 (2) (2) 3 (2) (2)f f g g′ ′+
2
2(3)(4) 3(2) (5) 84= + =
b. [ ( ) ( )] ( ) ( ) ( ) ( )
d
f x g x f x g x g x f x
dx
′ ′= +
(2) (2) (2) (2) (3)(5) (2)(4) 23f g g f′ ′+ = + =
c. [ ( ( ))] ( ( )) ( )
d
f g x f g x g x
dx
′ ′=
( (2)) (2) (2) (2) (4)(5) 20f g g f g′ ′ ′ ′= = =
d. 2
[ ( )] 2 ( ) ( )xD f x f x f x′=
2 2
[ ( )] 2[ ( ) ( ) ( ) ( )]xD f x f x f x f x f x′′ ′ ′= +
2
2 (2) (2) 2[ (2)]f f f′′ ′= +
2
2(3)(–1) 2(4) 26= + =
46. 2 2 2
(13) ; 2
dx
x y
dt
= + =
0 2 2
dx dy
x y
dt dt
= +
–
dy x dx
dt y dt
=
When y= 5, x = 12, so
12 24
– (2) – –4.8
5 5
dy
dt
= = = ft/s
47. sin15 , 400
y dx
x dt
° = =
sin15y x= °
sin15
dy dx
dt dt
= °
400sin15
dy
dt
= ° ≈ 104 mi/hr
48. a.
2 2
2 2( ) 2
( ) 2 2x
x x x
D x x x
x x x
= ⋅ = = =
b. 2
2 2
0
x
x
x x
x x
x x x
D x D
x x x
⎛ ⎞ −⎜ ⎟⎛ ⎞ −⎝ ⎠= = = =⎜ ⎟
⎝ ⎠

Instructor’s Resource Manual Review and Preview 151
c. 3 2
( ) (0) 0x x x xD x D D x D= = =
d.
22
( ) (2 ) 2x xD x D x= =
49. a.
sin
sin cos cot sin
sin
Dθ
θ
θ θ θ θ
θ
= =
b.
cos
cos ( sin ) tan cos
cos
Dθ
θ
θ θ θ θ
θ
= − = −
50. a. ( ) 1/ 2
1/ 2
1
( ) 1; '( ) 1 ; 3
2
( ) (3) '(3)( 3)
1
4 (4) ( 3)
2
1 3 1 11
2
4 4 4 4
f x x f x x a
L x f f x
x
x x
−
−
= + = − + =
= + −
= + − −
= − + = − +
b. ( ) cos ; '( ) sin cos ; 1
( ) (1) '(1)( 1)
cos1 ( sin1 cos1)( 1)
cos1 (sin1) sin1 (cos1) cos1
(cos1 sin1) sin1
0.3012 0.8415
f x x x f x x x x a
L x f f x
x
x x
x
x
= = − + =
= + −
= + − + −
= − + + −
= − +
≈ − +
Review and Preview Problems
1. ( )( )2 3 0x x− − <
( )( )2 3 0
2 or 3
x x
x x
− − =
= =
The split points are 2 and 3. The expression on
the left can only change signs at the split points.
Check a point in the intervals ( ),2−∞ , ( )2,3 ,
and ( )3,∞ . The solution set is { }| 2 3x x< < or
( )2,3 .
−2 76 853−1 10 2 4
2.
( )( )
2
6 0
3 2 0
x x
x x
− − >
− + >
( )( )3 2 0
3 or 2
x x
x x
− + =
= = −
The split points are 3 and 2− . The expression on
the left can only change signs at the split points.
Check a point in the intervals ( ), 2−∞ − , ( )2,3− ,
and ( )3,∞ . The solution set is
{ }| 2 or 3x x x< − > , or ( ) ( ), 2 3,−∞ − ∪ ∞ .
−5 −3−4 −2 53−1 10 2 4
3. ( )( )1 2 0x x x− − ≤
( )( )1 2 0x x x− − =
0, 1 or 2x x x= = =
The split points are 0, 1, and 2. The expression
on the left can only change signs at the split
points. Check a point in the intervals ( ),0−∞ ,
( )0,1 , ( )1,2 , and ( )2,∞ . The solution set is
{ }| 0 or 1 2x x x≤ ≤ ≤ , or ( ] [ ],0 1,2−∞ ∪ .
−5 −3−4 −2 53−1 10 2 4
4.
( )
( )( )
3 2
2
3 2 0
3 2 0
1 2 0
x x x
x x x
x x x
+ + ≥
+ + ≥
+ + ≥
( )( )1 2 0
0, 1, 2
x x x
x x x
+ + =
= = − = −
The split points are 0, 1− , and 2− . The
expression on the left can only change signs at
the split points. Check a point in the intervals
( ), 2−∞ − , ( )2, 1− − , ( )1,0− , and ( )0,∞ . The
solution set is { }| 2 1 or 0x x x− ≤ ≤ − ≥ , or
[ ] [ )2, 1 0,− − ∪ ∞ .
−5 −3−4 −2 53−1 10 2 4
5.
( )
( )
( )( )
2
2
0
4
2
0
2 2
x x
x
x x
x x
−
≥
−
−
≥
− +
The expression on the left is equal to 0 or
undefined at 0x = , 2x = , and 2x = − . These
are the split points. The expression on the left can
only change signs at the split points. Check a
point in the intervals: ( ), 2−∞ − , ( )2,0− , ( )0,2 ,
and ( )2,∞ . The solution set is
{ }| 2 or 0 2 or 2x x x x< − ≤ < > , or
( ) [ ) ( ), 2 0,2 2,−∞ − ∪ ∪ ∞ .
−5 −3−4 −2 53−1 10 2 4

152 Review and Preview Instructor’s Resource Manual
6.
( )( )
2
2
2
9
0
2
3 3
0
2
x
x
x x
x
−
>
+
− +
>
+
The expression on the left is equal to 0 at 3x = ,
and 3x = − . These are the split points. The
expression on the left can only change signs at
the split points. Check a point in the intervals:
( ), 3−∞ − , ( )3,3− , and ( )3,∞ . The solution set
is { }| 3 or 3x x x< − > , or ( ) ( ), 3 3,−∞ − ∪ ∞ .
−5 −3−4 −2 53−1 10 2 4
7. ( ) ( ) ( ) ( )3 3
' 4 2 1 2 8 2 1f x x x= + = +
8. ( ) ( ) ( )' cos cosf x x xπ π π π= ⋅ =
9. ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
2
2
' 1 sin 2 2 cos 2 2
2 1 sin 2 2 cos 2
f x x x x x
x x x x
= − ⋅− ⋅ + ⋅
= − − +
10. ( )
( )
2
2
sec tan sec 1
'
sec tan 1
x x x x
f x
x
x x x
x
⋅ − ⋅
=
−
=
11. ( ) ( )
( )( )
2
2
' 2 tan3 sec 3 3
6 sec 3 tan3
f x x x
x x
= ⋅ ⋅
=
12. ( ) ( ) ( )( )
1/ 22
2
1
' 1 sin 2sin cos
2
sin cos
1 sin
f x x x x
x x
x
−
= +
=
+
13. ( ) ( ) 1/ 21 cos
' cos
2 2
x
f x x x
x
−
= ⋅ =
(note: you cannot cancel the x here because it
is not a factor of both the numerator and
denominator. It is the argument for the cosine in
the numerator.)
14. ( ) ( ) 1/ 21 cos2
' sin 2 cos2 2
2 sin 2
x
f x x x
x
−
= ⋅ ⋅ =
15. The tangent line is horizontal when the derivative
is 0.
2
' 2tan secy x x= ⋅
2
2tan sec 0
2sin
0
cos
x x
x
x
=
=
The tangent line is horizontal whenever
sin 0x = . That is, for x kπ= where k is an
integer.
16. The tangent line is horizontal when the derivative
is 0.
' 1 cosy x= +
The tangent line is horizontal whenever
cos 1x = − . That is, for ( )2 1x k π= + where k is
an integer.
17. The line 2y x= + has slope 1, so any line parallel
to this line will also have a slope of 1.
For the tangent line to siny x x= + to be parallel
to the given line, we need its derivative to equal 1.
' 1 cos 1y x= + =
cos 0x =
The tangent line will be parallel to 2y x= +
whenever ( )2 1
2
x k
π
= + .
18. Length: 24 2x−
Width: 9 2x−
Height: x
Volume: ( )( )
( )( )
24 2 9 2
9 2 24 2
l w h x x x
x x x
⋅ ⋅ = − −
= − −
19. Consider the diagram:
1
4 x−
x
His distance swimming will be
2 2 2
1 1x x+ = + kilometers. His distance
running will be 4 x− kilometers.
Using the distance traveled formula, d r t= ⋅ , we
solve for t to get
d
t
r
= . Andy can swim at 4
kilometers per hour and run 10 kilometers per
hour. Therefore, the time to get from A to D will
be
2
1 4
4 10
x x+ −
+ hours.

Instructor’s Resource Manual Review and Preview 153
20. a. ( ) ( )
( ) ( ) ( )
0 0 cos 0 0 1 1
cos 1 1
f
f π π π π π
= − = − = −
= − = − − = +
Since cosx x− is continuous, ( )0 0f < ,
and ( ) 0f π > , there is at least one point c
.in the interval ( )0,π where ( ) 0f c = .
(Intermediate Value Theorem)
b. cos
2 2 2 2
f
π π π π⎛ ⎞ ⎛ ⎞
= − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )' 1 sinf x x= +
' 1 sin 1 1 2
2 2
f
π π⎛ ⎞ ⎛ ⎞
= + = + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
The slope of the tangent line is 2m = at the
point ,
2 2
π π⎛ ⎞
⎜ ⎟
⎝ ⎠
. Therefore,
2
2 2
y x
π π⎛ ⎞
− = −⎜ ⎟
⎝ ⎠
or 2
2
y x
π
= − .
c. 2 0
2
x
π
− = .
2
2
4
x
x
π
π
=
=
The tangent line will intersect the x-axis at
4
x
π
= .
21. a. The derivative of 2
x is 2x and the
derivative of a constant is 0. Therefore, one
possible function is ( ) 2
3f x x= + .
b. The derivative of cos x− is sin x and the
derivative of a constant is 0. Therefore, one
possible function is ( ) ( )cos 8f x x= − + .
c. The derivative of 3
x is 2
3x , so the
derivative of 31
3
x is 2
x . The derivative of
2
x is 2x , so the derivative of 21
2
x is x .
The derivative of x is 1, and the derivative of
a constant is 0. Therefore, one possible
function is 3 21 1
2
3 2
x x x+ + + .
22. Yes. Adding 1 only changes the constant term in
the function and the derivative of a constant is 0.
Therefore, we would get the same derivative
regardless of the value of the constant.

51543 0131469657 ism-2

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51543 0131469657 ism-2