~88
Chap~ee 2 Differentiation
Review Exercises for Chapter 2
1. f(x): x~ - 4x + 5
f’(x) : limf(X + Ax)- f(x)
= lim I(x+
Z~C)2 -- 4(x+
Ax--~ 0
Ax)+ 51- Ix~- 4x+ 51
z~C
= lira (x2 + 2x(Ax)+ (Ax)2- 4x- 4(Ax)+ 5)-(x~- 4x+ 5)
= lim2X(AX)+(Ax)2 -4(Ax)Axe0
= lim(2x+ Ax-4) = 2x-4
2. f(x) : "v/Tx + 1
= lim
&~---~ o N/~x
1
1
+ Ax + 4x
2~x
3. ~tx~_ x + 1
x-1
x + zDc+ 1 -x+l
t"’(x] = limf(X+ Ax)-f(x) = lim x+Ax- 1 x-1
~-~ o
Ax
~x ~ o
z~c
= lim(X+Ax+l)(x-1)-(x+As-1)(x+l)
kx(x + koc - 1)(x - 1)
= lim(X2 +xAx+x-x-Ax-1)-(x2 +xAx-x+x+Ax-1)
~x(x + ~x- 1)(, - 1)
f’(x) : lim f(x + kx)- f(x)
66
lim x + Ax x= lim6X-(6x+6kx) = lim -6Ax
~--,o ~(x + ~)~ ~-,o~(x +
= lim
-6
-6
~-,o(x + ~).~ -x~
5. fis differentiable for all x 3.
6. fis differentiable for all x ~: -1.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 2 189
12. (a) Using the limit definition,
f’(x) - -2
(a) Continuous at x = 2
(b) Not differentiable at x = 2 because of the sharp
turn in the graph. Also, the derivatives from the left
and right are not equal.
At x = 0, f’(0) = -2. The tangent line is
y - 2 = -2(x - o)
Y
y = -2x + 2.
7
6
54-
x
I 2 3 4 5 6N
IX2 q- 4X2,
+ 2,
X < --2
--2
8o f(x)= [1--4X--X
if if
X_>
(a) Nonremovable discontinuity at x = -2
(b) Not differentiable at x = -2 because the function is
discontinuous there.
Using the limit definition, you obtain g’(x) =~-X4 -- -~. At
X = --1,
g’(-1) -
4 1 __ 3
3
6
2"
10. Using the limit definition, you obtain h’(x) = 3-~_ 4x. At
X = --2,
13. g’(2)= limg(X) - g(2)
x-~2 x - 2
limX2(X - 1) - 4
x-~2 x - 2
X3 -- X2 -- 4
lim
x-+2
X -- 2
(x-2)(x2 +x+2)
= lim
x-,2
x-2
2
= lim(x
+ x+ 2] = 8
x--~2\
14. f’(3) = limf(x) - f(3)
x-~3 X -- 3
11
x
= lira + 4 7
x--,3 x - 3
7-x-4
= lim.
x~3(x- 3)(x + 4)7
-1
1
= lim.
=-~
x-~3(x + 4)7 49
15. y = 25
---.~2
h’(-2) = ’g -4(-2)11. (a) Using the limit definition, f’(x) = 3x2. At
x = -1, f’(-1) = 3. The tangent line is
y -(-2) = 3(x -(-1))
y = 3x+l.
(b)
4
(b)
12~_ 11" _-2-3-
(x+l)
.2
y’ = 0
16. y =-30
y =0
17. S(x) = x~
St(X) = gX7
18. g(x) : x2°
g’(x) = 20x19
© 2010 Brooks/Cole, Cengage Learning
Chapter 2 Differentiation
Y
20.
21l. f(x) : x3 - l lx2
f’(x) = 3x2 - 22x
22°
g(s) = 4S4 - 5S2
f’ > 0 where the slopes of tangent lines to the graph of
fare positive.
g’(s) = 16s3 - lOs
23.
htx) = + = + 6x|/~
3
3xl/3
32.
1
~4o f(x) = X112 - X-112
=2
25. g(t)= 3
-~x3--/~
2
f’ = 0 where the slopes of tangent lines to the graph of
fare 0.
t-2
g’(t) = -4t-~ - 43
3
3t
33.
26. h(x)= lOx-2
49
F = 200x/’~
100
(a) When T = 4, F’(4) = 50 vibrations/sec/lb.
h’(x)= --x-20 -3 _ 20
49
49x3
27. f(O) = 40 - 5 sin 0
f’(O) = 4-5cosO
(b) When T = 9, F’(9) = 33½ vibrations/sec/lb.
34. s = -16t2 + so
First ball:
-16t2 +100 = 0
28. g(a) = 4cos~ + 6
g’(a) = -4 sin a
~/ 16
10
- 2.5 seconds to hit ground
4
sin 0
29. t°(O] : 3 cos 0 - -4
cos 0
f’(O] = -3 sin 0 - -4
30. g(a)-
5 sin a
3
gqa~t 1 - 5cosa
3
Second ball:
-16t2 + 75 = 0
t2
=
7~
2a
~ 2.165 seconds to hit ground
4
Because the second ball was released one second after
the first ball, the first ball will hit the ground first.
The second ball will hit the ground
3.165 - 2.5 = 0.665 second later.
2
35.
s(t) =-16t2 + So
s(9.2) = -16(9.2)2 + So = 0
So = 1354.24
The building is approximately 1354 feet high (or 415 m).
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 2 191
36° s(t) = -16t2 + 14,400 = 0
16t2 = 14,400
38°
t = 30 sec
Because 600 miih = ~gmi/sec, in 30 seconds the bomb
will move horizontally (-~)(30) = 5miles.
128
37. (a)
322
(a) y = x __--Tx = x 1 -
15-
Vo
2
I0- ~
= 0ifx = 0orx = v°-32
5-
20
40
[ ~-x
Total horizontal distance: 50
(b) 0 = x - 0.02x2
o: o) mp.os
(c) Ball reaches maximum height when x = 25.
(d)
y = x - 0.02x~
y’ = 1-0.04x
y’(0) = 1
y’(10) = 0.6
y’(25) = 0
y’(30) = -0.2
Projectile strikes the ground when x = Vo2/32.
Projectile reaches its maximum height at
x = V02/64 (one-half the distance).
64
(b) y’ = 1 Tx
vo
When x = v°--~-~ y’= 1-64(v°2/ = 0.
64’
V02 ~ 64 )
(c) y = x ..-Tx = x 1-_.-~-x = 0
Vo
Vo
when x = 0 and x = So2/32. Therefore, the range
is x = Vo2/32. When the initial velocity is doubled
the range is
(2Vo)2 4Vo2
X --
/(50) -- -1
(e) y’(25) = 0
--
32
32
or four times the initial range. From part (a), the
maximum height occurs when x = Vo2/64. The
maximum height is
Y ~ 64 Vo2 ~, 64 ) 64 128 128"
If the initial velocity is doubled, the maximum
height is
F(2Vo)z] (2Vo)= .(V02 ~
128 = 4/1-~)
or four times the original maximum height.
(d) Vo = 70 ft/sec
V02 -- (70)2 - 153.125 it
32
32
2
2
Maximum height: yVo- (70)
38.28 it
128 128
Range: x
5O
© 2010 Brooks/Cole, Cengage Learning
192
Chap,er 2 Differentiation
39° x(t) = t2 - 3t + 2 = (t- 2)(t-1)
45° f(x) -
(a) v(t) = x’(t)= 2t- 3
X2÷X--I
x= -1
f’(x) = (x’ - 1)(2x + !)- (x2 + x - 1)(2x)
(b) v(t) < 0 for t < 2_3
(c) ,,(~) = o for ~ = _32
=-
1
4
(d) x(t) = 0fort = 1,2
6x - 5
46.
v(1) = 2(1)-3 = 1
f’(x) : (x2 + 1)(6) -(6x - 5)(2x)
~(2) =12t21 - 3 = ~
The speed is 1 when the position is 0.
40. (a) y = 0.14x2 -4.43x+ 58.4
(b) 320
47. f(x) : (9 - 4x2)-’
c
f’(x) : -(9 - 4x2)-2(-8x) -
60
o
8x
(C) 12
48. f(x)= 9(3x2- 2x)-1
18(1- 3x)
f’(x) = -9(3x~- 2x)-=(6x- 2)= (3x~ 2x)~
(d) If x = 65, y ~ 362 feet.
(e) As the speed increases, the stopping distance
increases at an increasing rate.
49. y-
COS X
y, : (COS x) 4x3 - x4(-sin x)
COS2 X
41. f(x)=
4x3 cos x 4- x4 sin x
s’(-) : (5x~ + 8)(2x - 4) + (x~ - 4x - 6)(10x)
---
IOx3 + 16x
= 4(5x3 -15x2 -11x- 8)
42. g(x):
COS2X
- 20x; - 32 + lOx~ - 40x~ - 60x
=20x~-60x~-44x-32
g’(x) =
X4
50. y- x4
sin x
Y’ : (x4) cos x -(sin x)(4x~)
x cos x - 4 sin x
(; + 7x)(x + 3)
(X34- 7X)(1)4-(X 4-
3)(3x2 + 7)
7x + 3x3 4- 9x2 4- 7x + 21
4x3 +9x2 +14x+21
X34-
43. h(x): ~x sin x = X1/2 sin x
1
h’(x) : ~ sin x + U;zcos ~<
x5
51. y = 3x= sec x
y’ = 3x= sec x tan x + 6x sec x
52. y = 2x-x2tanx
y’ = 2 - x2 sec2 x - 2x tan x
53. y = xcosx-sinx
y’ =-xsinx+cosx-cosx =-xsinx
44. f(t) = 2t5 cos t
s’(O : 2tS(-sin t) + cos/<(10’4)
--
-2/5sin t + lO/4cos t
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 2
54° g(x) = 3x sin x + x2cos x
g’(x) = 3xcosx + 3sinx-x2sinx+ 2xcosx
193
90t
4t + 10
a(t) = (4t + 10)90- 90t(4)
(4t + 10)2
900 _
225
2
(4t + 10) (2t + 5)2
6~}. v(t) -
= 5x cos x + (3 - x2)sin x
55. f(x) - x2 - 2x - x-2, (1,1)
f’(x) = 2 + 2x-3
90
(a) v(1) = -- ~ 6.43 ft/sec
14
f’(1) = 4
a(1) = 22___~5 ~ 4.59 ft/sec2
49
Tangent line: y - 1 = 4(x - 1)
y = 4x-3
(b) v(5)= 90(5) _ 15 ft/sec
30
a(5) - 225 _ 1 ft/sec2
152
56.
f’(x) = (x-1)-(x + 1)
(X-1)2
-2
(c) v(10) - 90(10) _ 1S ft/sec
50
a(10) - 225 _ 0.36 ft/sec2
252
_2_
= (1/4) s
Tangent line: y + 3 =-8IX - ~)
y = -8x + 1
61. g(t) = -8t3 - 5t + 12
g’(t) = -24t2 - 5
g"(0 = -48,
57. f(x) = -x tan x, (0,0)
f’(x) = -x sec2 x - tan x
s’(o) = o
62. h(x) = 21x-3 + 3x
h’(x) = -63x-4 + 3
Tangent line: y - 0 = 0(x - 0)
y=0
h"(x) = 252x-’ 63.
58.
f(x) - 1-cos
’
f’(x) = (1 - cos x)(-sin x) -(1 + cos x)(sin x)
(1 - cos x)2
-2 sin x
(1 - cos x)2
252
x5
f(x) = 15x5/2
f’(x) ="~-x75 3/2
f"(x)= 2254.. 1-1/2
= --~-’,4x225/’-
64. f(x) = 20s~x = 20x115
f’(x) = 4x-415
f"(x) = -16x-9/5 - 915
16
5
5x
65. f(O) = 3tan 0
f’(O) = 3 sec2 0
y = -2x+l +~r
59. v(t) = 36-t2, 0 < t < 6
a(t) = v’(,)= -2t
v(4) = 36 -16 = 20 rn!sec
a(4) = -8 rn!sec2
f"(O) = 6 sec O(sec 0 tan 0)
= 6sec20tan0
66. h(t) = 10 cos t - 15 sin t
h’(t) = -10 sin t - 15 cos t
h"(t) = -10 cos t + 15 sin t
© 2010 Brooks/Cole, Cengage Learning
~4
Chapter 2 Differentiation
y = 2sinx + 3cosx
y = 2cosx-3sinx
y = -2sinx-3cosx
y"+ y =-(2sinx+3cosx) + (2 sinx+ 3cosx)
=0
68.
y=
(10 - cos x)
75° y=
x
2
Y’-21
sin 2x
4
41
cos 2x(2)
- cos 2x)= sinZx
= ½(1
sec5 x
7
5
6
y’ = sec x(sec x tan x) - see4x(sec x tan x)
%° y-
sec7 x
x
= secSxtanx(sec2x -1)
= sec5 x tan3 x
xy + cos x = 10
xy’ + y - sin x = 0
xy + y = sin x
2. 3/2 x--sin2. 7/2 x
77° y =-sin3
7
y’= sin~/2x cos x - sin5/2x cos x
(
x+5)2
h’(x) :
= (cos x)si~(1- sin2x)
= cos3x.~sin x
3t(0-
2¢x+5
~x-~-~+ ~t (;+ ~)~
2(x + 5)(-x2- l Ox + 3)
(; +,3)3
70. S(x)=
x -t-
78. S(x)- ~3x+ 1
3(x2 + I)I/2- 3x#(xz + l)-I/Z(2x)
f’(x) =
x2 + 1
3(X2 ÷ 1)-3X2
f’(x) =
5(x2 + 1 4
(; ÷ 0~,~
_1
79. y-
71. f(s) =
y’
S’(~) :
80. y-
n. h(o)-
~’(o) =
yt ~
0
(1 - o)~
(I- O)~ - 013(I-
(1- 0)6
(1-0)~(1-0+30) _ 20+1
(1 - O)6
(1 i 0)4
73. y = 5 cos(9x + 1)
3
(; ÷ 0~,~
sin
x+2
(x + 2)~ cos ~x - sin ~x
(x+2)z
cos(x- 1)
x-1
-(x - 1)sin(x - 1) - cos(x - 1)(1)
(x-l)z
1_ F(x -1)sin(x -1) + cos(x - 1)l
(x 1)z
SL S(x)= ,/7- x3
i3X2
f’(x) : ~(1-x3)-l/2(-3xz) 2x/~- x3
-12
f’(-2)- 2(3)--2
y’= -5 sin(9x + 1)(9)= -45 sin (9x + 1)
74. y = 1 - cos 2x + 2 COS2X
y’ = 2 sin 2x - 4 cos x sin x
= 2[2 sin x cos x] - 4 sin x cos x
=0
82. f(x) = 3wr-~x2 -1
:,(x) : ~(x, - ,)-~’~(~x) :
2x
2(3)_ 1
f’(3)- 3(4) 2
©,2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 2
1
2
= -csc 2x cot 2x
y = - csc 2x
=0
84.
y = -4/-5-~x(x + 2)3
3(x + 2)2(7x + 2)
2-~x
y’ does not equal zero for any x in the domain. The
graph has no horizontal tangent lines.
Y=
Y = csc 3x + cot 3x
y’ = -3 csc 3x cot 3x - 3 CSC2 3x
75
= 0-3 = -3
8S. g(x) : 2x(x + 1)
x+2
g’(x) - (x + 1)3/~
89. f(t) = t2(t-1)5, (2, 4)
(a) f’(t) = t(t- 1)4(7t- 2)
f’(2) = 24
(b) y- 4 = 24(t- 2)
y = 24t - 44
-2
(c)
5"
4"
3"
g’ does not equal zero for any value ofx in the domain.
The graph ofg has no horizontal tangent lines.
(2, 4)
21"
86. f(x) = E(x - 2)(x + 4)~2 = (x2 + 2x - 8)2
f’(x) = 4(x3 + 3x2 - 6x - 8)
= 4(x- 2)(x + 1)(x + 4)
The zeros of f’ correspond to the points on the graph of
fwhere the tangent line is horizontal.
lOO
\
87. f(t) = ~,/7-~ ~-~/7 + 1
f(t) = (t + 1)~/2(t + 1)’13 = (t + 1)5/6
6(1 + 1)1/6
f’ does not equal zero for any x in the domain. The
graph off has no horizontal tangent lineg.
5
x2
(a) g’(x) = x/xz + 1 + ~
g’(3)- 19x/ri-6
lO
J
-6O
f’(t) - S
90. g(x)=x~, (3,3x/i-~)
(b) y - 3x/i-6 - 19~i~-(x - 3)
19-x/’i"d 27-~ti-6
10
10
(c)
108"
6"
4"
2"
© 2010 Brooks/Cole, Cengage Learning
196
Chapter 2 Differentiation
95. f(x) = cot x
91. f(x) = tan ~- x, (-2, tan
f’(x) = -csc2x
-1
(a) f’(x) = 2.,fi- x cos2xfi- x
f"(x) = -2 csc x(-csc x. cot x)
= 2 csc2x cot x
f’(-2) - 6 cos2x/~ ~ -11.1983
96. y = sin2x
(b) y - tan-,/-5 - 6 cos2
(x + 2)
y’ = 2sinxcosx = sin2x
y" = 2cos2x
(x + 2)-,/~
Y = 6 cos2-,/~ + tan-,!~
(c)
97.
y
4t2
f(O - (1- t)a
f’(,) _
8t
(1- ~)3
f"(t)- 8(21 + 1)
(1- 1)4
6x - 5
98. g(x) - ~ + 1
92. S(x)= 2 csc3(~x)- 2
g’(x) = 2(-3x2 + 5x + 3)
(1, 2 csc3(1))
(x2 +1)2
g"(X) = 2(6x3 - 15x2 -18x + 5)
(a) S’(x)= -4’7 sin4(~x)
f’(1) - -3 cos(l)
sin4(1)
99. g(O) = tan 30 - sin(O - 1)
g’(O) = 3 sec2 30 - cos(O - 1)
-3 cos(l).
(b) Y - 2 csc3(1)’" ~n-~ (x-1)
y-
x
-3 cos(l)
(c)
2
g"(O) = 18 see2 30 tan 30 + sin(O - 1)
3 cos(l)
sin 4(1)
100.
h (x) = 5 X4 X2 l 16
2
- 80
h’(x) = lOx
",,/x2 - 16
3 - 240x
4321-
h"(x) = lOx
93. y = 7x2 + COS 2x
y’ = 14x-2sin2x
y" = 14-4cos2x
94. y = x-1 + tan x
y’ = -x-2 + sec2 x
y" = 2x-3 + 2 sec x(sec x tan x)
2
= --3 + 2 sec2x tan x
x
© 20!0 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 2 197
700
101o T = a
t + 4t + 10
v : 2x/~ = N/r~-32)h =
4
dv
T = 700(t2 + 4t + 10)-1
Tr =
dv 4
-- dh
= -3ft/sec.
(a) Whenh = 9,
-1400(t + 2)
(b) When h = 4,--dv = 2 ft/sec.
dh
(a) When t = 1,
T’ = -1400(1 + 2) ~ -18.667 deg/h.
(1 + 4 + 10)2
103o
(b) Whent = 3,
-1400(3 + 2) ~ -7.284 deg/h.
Tr =
(9 + 12 + 10)2
X2 q- 3xy + y3 =
10
’
2x + 3xy’ + 3y + 3y2y = 0
3(x + y2)y, = -(2x + 3y)
y,= -(2x + By)
(c) Whent = 5,
-1400(5 + 2) ~ -3.240 deg/h.
T’=
(25 +20+10)2
y2 = x3 _x2y+xy_y2
104.
0 = X3 -- x2y
(d) Whent = 10,
+ xy - 2y2
0 = 3x2 - x2y’ - 2xy + xy’ + y - 4yy’
(x2 - x + 4y)y’ = 3x2 - 2xy + y
T’ -1400(10
=
+ 2) ~ -0.747 deg/h.
(100 + 40 + 10)2
3x2 - 2xy
+y
Y’ = x2 -x+4y
x - 4y
105.
xy’+ y =
24- - 84-gy’
x + 8x/-~y’ = 2~-~- - y
2~/’-~- y _ 2(x- 4y)- y 2x - 9y
x + 8~-~ x + 8(x - 4y) 9x - 32y
y-,/-~x- xx/"f = 25
106.
yl~x-i/2) + xl/2y’ -
_ yl/2 = 0
2~x
2xi"~ - x
2y~x- YX/-f
2xx/-f -
© 2010 Brooks/Cole, Cengage Learning
Chapter 2 Differentiation
cos(x + y)= u
xsiny = ycosx
197o
(x cos y)y’ + sin y = -y sin x + y’ cos x
-(1 + y’)sin(x + y)= 1
y’(x cos y - cos x) = -y sin x - sin y
ysinx+ siny
y=
cosx-xcosy
X2 q- y2
-y’sin(x + y)= 1 + sin(x + y)
Y’= l+sin(x~) =-csc(x~-in---~-
= 10
2x + 2yy’ = 0
--X
y
At (3, 1), y’ = -3
4
Tangent line:
y-l=-3(x-3)~3x+y-10=o
Normal line:
y-l= x- ___1( 3) ~- x 3y
= 0
3
112o Surface area = A = 6x2, x = length of edge
119. x2 - y2 = 20
2x - 2yy’ = 0
dx
x
dA
dt
3
At (6, 4), y’ = 2
Tangent line:
3 6)
y-4 = -~(x3
y = --x-5
2
2y-3x+10 = 0
Normal line:
12xdX
dt
12(6.5)(8) 624 cm2/sec
113. s 1/2h
2
dV
-1
dt
2 6) y - 4 = -~(x
2
y = --x + 8
3
3y+2x-24 = 0
’2
Width of water at depth h:
w=2+2s=2+2/¼h/= 4+h2
V= -~(2+--h=
4 -~ hI(8+
-~ h )h
111.
-: = 2 units/sec
dt
=> -- = 2x/Txdy = 4x/Tx
dt 2xitTx dt dt
dt
1 dx _ 2x/~ units/sec.
(a) Whenx - 2’ dt
dt (4 + dt
dh _ 2(dV/dt)
~t 5(4 + h)
1_
(b) When x = 1,--dx = 4 units/sec.
dt
dh 2dt 25
When
rn/min.
2
(c) When x = 4,--dx = 8 units/sec.
dt
© 2010 Brooks/Cole, Cengage Learning
+ 1)-1
Problem Solving for Chapter 2 199
tan0 = x
dO
-- = 3(2rc] rad/min
dt
114.
s = 35 = 60-4.9tz
4.9tz = 25
5
dt
dt
1
tan 30 - ~- - x(t)
When x = -~,
dr w/-~ds = x/~(-9.8)~-4.9 ~ -38.34 m/sec
dt dt
s(t)
30°
x(t)
Problem Solving for Chapter 2
1. (a) x2 +(y-r)2 = r2,Circle
x2 = y, Parabola
Substituting:
y2
3
(y- r)2 = r2 _ y
_ 2ry + r2 = r2 -
y
y2 -2ry+y = 0
y(y- 2r + 1) = 0
Because you want only one solution, let 1 - 2r = 0 ~ r = -. Graph y = x2 and X2 +
2
(b) Let (x, y) be a point oftangency:
x2 +(y-b)2 : 1 ~ 2x+ 2(y-b)y’ = 0 ~ y’ - x ,Circle
b-y
y = x~ ~ y’ = 2x, Parabola
Equating:
3
x
2(b - y) = 1
b-y=-~ b 1=y+- 1
2
2
Also, x2 + (y - b)~ = 1 and y = x2 imply:
1
3 5
= 1 ~ y+- = 1 ~ y =-andb =4
4 4
y+(y-b)~ =1 ~ y+ y- y+
Graphy = x2and x2 + y-
= 1.
© 2010 Brooks/Cole, Cengage Learning