130
Chapter 2
Differentiation
- f'g)'
111. (a) Ug'
= fg"
(b) (fgY=
- f'g' - f"g
= fg" + f'g'
-
f"g
True
Ug' + f'g)'
= fg" + f'g' + f'g' + f"g
= fg" + 2f'g' + f"g
4=fg" + f"g
False
The Chain Rule
Section 2.4
= g(x)
y
= f(u)
u=6x-5
y
= if
u=x+l
y
= U-I/2
3. y =.P-=t
u=x2-1
y= Ju
4. y = tan(= + 1)
u=m+l
y=tanu
5.y=csc3x
u=cscx
y = U3
y = f(g(x»
u
1. Y = (6x - 5)4
2.y=
1
x+ 1
3x
3x
6.y=cos2
u=2
y=cosu
7. y = e-2x
u=-2x
y = e"
8. y = (Inx)3
u = Inx
y = U3
9. y = (2x -
10. y = (3x2+ 1)4
7)3
y' = 3(2x - 7)2(2)= 6(2x - 7)2
y' = 4(3x2+ 1)3(6x)= 24x(3x2+ 1)3
12. f(x) = 2(1 - x2)3
11. g(x) = 3(4 - 9X)4
g'(x) = 12(4- 9x)3(-9) = -108(4 - 9x)3
13. f(x) = (9 - x2)2/3
14. f(t) = (91+ 2)2/3
f'(x) = ~(9 - x2)-1/3(-2x) = - 3(9 ~xx2)173
15. f(t) = (1 -
= ~(1 -
t)-1/2( -1)
6
f'(t) = ~(9t + 2)-1/3(9)= V9t+2
16. g(x) = (3 -
t)I/2
1
f'(t)
f'(x) = 6(1 - x2)2(-2x) = -12x(1 - x2)2
=-
2""1
-
t
2x)1/2
1
g'(x) = ~(3 - 2x)-1/2(-2) = - ""3 - ~
Section 2.4
17. Y = (9xl + 4)1/3
18. g(x) = ../xl- 2x+ 1 = ../(x-I)2=
6x
y/ = ~(9xl + 4)-2/3(18x)= (9xl + 4)2/3
19. y = 2(4 - xl)1/2
= (x -
= (t - 3)-2
+ 2) 3/2 = --
1
2(x + 2)3/2
27. lex) = xl(x - 2)4
= (1-
- -(2t+3)
- (P + 3t - 1)2
8
y/ = 8(t + 2)-3 = (t + 2)3
1
t
g/(t) = -"2(t2 - 2)-3/2(2t) = - (P - 2)372
= x(3x -
9)3
f'(x) = x[3(3x - 9)2(3)] + (3x - 9)3(1)
2)3[2x+ {x - 2)]
= (3x - 9)2[9x+ 3x - 9]
2)3(3x - 2)
= 27(x -
= x(1 - xl)1/2
y/ = x[k(1 - xl)-1/2(-2x)]
= -xl(1
s/(t) = -1(t2 + 3t - 1)-2(2t + 3)
28. lex)
f'(x) = xl[4(x - 2)3(1)]+ (x - 2)4(2x)
29. y = x.JI="?
f'(x) = -~2 - 9x)-3/4(- 9) = 4(2 ~~x)3/4
26. get) = (P - 2)-1/2
25. y = (x + 2)-1/2
= 2x(x = 2x(x -
I, x > 1
{ -1, x < 1
24. y = -4(t + 2)-2
-2
f'(t) = - 2(t - 3)-3 = (t - 3)3
dy
1
- = --(x
dx
2
'ex)=
22. set)= (t2 + 3t - 1)-1
2)-1
-1
y/ = -I(x - 2)-2(1) = (x - 2)2
23. Jet)
g
Ix-II
20. lex) = - 3(2 - 9X)I/4
2x
y/ = (4 - xl)-1/2(-2x) = - ../4 - xl
21. y
The Chain Rule
+ (1 - X2)1/2(1)
30. y = x2../9 - xl = xl(9 - X2)1/2
y/ = xl[~9 - xl)-1/2(-2x)]
= - x3(9 -
- X2)-1/2+ (1 - xl)1/2
xl)-1/2[-xl
3)2(4x- 3)
+ (9 - xl)I/2(2x)
xl)-1/2 + 2x(9 - xl)1/2
= x(9 - xl)-1/2[-xl + 2(9 - xl)]
+ (1 - X2)]
1-2r
- x(18 - 3X2)
- ../9 - xl
- .JI="?
- 3x(6 - xl)
- ../9 - xl
31. y =
x
r;--;-, = x(xl + 1)-1/2
...;xl+ 1
X2
32. y
= (X9
+ 9)1/2
(X9+ 9)1/2(2x) - xl(1/2)(X9 + 9)-1/2 9x8
y/ = x[ -k(xl + 1)-3/2(2x)]+ (xl + 1)-1/2(1)
= - xl(xl
=
+ 1)-3/2 + (xl + 1)-1/2
y/ =
X9+ 9
- (X9+ 9)-1/2x[2(X9+ 9) - (9/2)X9]
x9+9
(xl + 1)-3/2[- xl + (X2 + 1)]
1
- (xl + 1)372
- x[I8 - (5/2)X9]- x(36 - 5X9)
(X9+ 9)3/2 - 2(x9+ 9)3/2
131
132
Chapter 2
Differentiation
~
33. y=Jx+1
Xl + I
34. y = "";-;+t
I - 3x2- 4x3/2
y' = 2Jx(X2 + 1)2
I
y' = y'2X(X+ 1)3/2
The zero of y' corresponds to the point on the graph of
y where the tangent line is horizontal.
y' has no zeros.
7
2
J~5
~~. -1
~-2
3t2
35. g(t)
= ~t
- I
36. f(x) = Jx(2 - x)2
3t(P + 3t - 2)
g'(t) = (P + 2t - 1)3/2
f'(x) = (x - 2)(5x - 2)
2Jx
The zeros of g' correspond to the points on the graph of
g where the tangent lines are horizontal.
The zeros off' correspond to the points on the graph of
f where the tangent lines are horizontal.
4
FfD
--00.
_5~3
-2
38. y = (t2 -
37. y=JX;1
v'(x + I)/x
2x(x + 1)
y'=
-2
y' has no zeros.
,
Y
15
D2[]
-0llliJ
Bjd
39. s(t)
5P + 8t - 9
= 2.Jt+2
The zero of y' corresponds to the point on the graph of y
where the tangent line is horizontal.
4
,10,
9).Jt+2
-2
= -2(2 - 3t).Jl+t
t
'
-15
3
CIa
s'(t) = v'1 + t
The zero of s'(t) corresponds to the point on the graph of s(t) where the
tangent line is horizontal.
-3~6 -3
The Chain Rule
Section 2.4
6
40. g(x)=~+~
'
g (x )
-2~
1
+
1
2~
-2
g' has no zeros.
~
-2
41. Y = COS TTX
x + 1
3
dy - -1Txsin TTX- COS1TX- 1
dx XZ
,~.
TTX sin TTX + cos TTX + 1
-3
XZ
The zeros of y , correspond
42.
to the points
on the graph
of y where
the tangent
1
y=x2tan- x
~
= 2xtan.!. dx
x
6
secz!
x
_.~,
The zeros of y' correspond to the points on the graph of
y where the tangent lines are horizontal.
43. (a)
(b)
y = sin X
-6
y=sin2x
y' = 2cos 2x
y'=cosx
44. (a)
lines are horizontal.
y'(O) = 1
y'(O) = 2
1 cycle in [0. 21T]
2 cycles in [0, 21T]
(b)
y=sin3x
Y = Sin()
y'=3cos3x
y'(O) = 3
y' = ()
cos()
1
3 cycles in [0. 21T]
y'(O)= -2
Half cycle in [0, 27T]
45. Y = cos 3x
dy = -3 sin3x
dx
46.
y = sin 1TX
dy
dx = 1TCOS
1TX
g'
10
133
134
Chapter 2
47. g(x)
Differentiation
= 3 tan 4x
48. hex)= sec(x2)
g '(x) = 12 sec24x
49. 1(8)
=~
sin2 28
h'(x) = 2xsec(x2)tan(x2)
= ~(sin
50. get)
28)2
g '(t)
1'(8) = 2W(sin 28)(cos 28)(2)
= sin 28 cos 28 =
51.
Y=
! sin 48
h + ~ sin(2x)2
= 5 COS2
= 5(cos
7Tt
=
10 cos 7Tt( - sin 7Tt)( 7T)
=
-1O7T(sin
52. Y = 3x - 5 COS(7TX)2
7Tt)(COS 7Tt)
53.
= 3x - 5 cos(7T2X2)
= h + ~ sin(4x2)
dy
dx = 3 + 5 sin(7T2x2)(27T2X)
dy - 1 -1/2 1
dx - 2x
+ 4 cos(4x2)(8x)
=
3 +
7Tt)2
y
:
=-
57T sin 2m
= sin(cosx)
= cos(cosx)
=
-sinxcos(cosx)
1O7T2x sine 7T.x-)2
1
= 2h + 2xCOS(2x)2
54.
y
= sinh
+ ~sinx
.!X-I/2
+ .!(sinx)-1/2
. cosx
2
2
cosx
2~sinx
dy = cosh.
dx
cosh
=-+
2h
55. lex) = e2x
56. y=e
f'(x) = 2e2x
58. y
-;i'
57. Y= e../x
dy - e../x
dx - 2h
dy = -2xe-;i'
dx
= x2e-x
59. get) = (e-t + e')3
60. get) = e-I/t'
g '(t) = 3(e-t + et )2(et - e-t)
dy = - x2e-x+ 2xe-X
dx
g'(t) = 2e-I/t'
t3
= xe-x (2 - x)
2
61. Y =
= 2(eX + e-x ) -I
eX + e-X
dy
dx
= -2(eX
=
63. y
dy
dx
eX - e-X
62. y
=
2
dy - eX + e-X
dx2
+ e-x)-2(eX - e-x)
-2(eX - e-x)
(eX+ e-x)2
= x2eX= eX(2x-
2xeX+ 2eX = eX(x2- 2x + 2)
2) + eX(X2- 2x + 2)
65. g(x) = Inx2 = 21nx
2
g'(x) = -x
= x2eX
64. y
= xeX -
eX
= eX(x- I)
dy = eX+ eX(x- 1) = xeX
dx
66. hex) = In(x2+ 3)
h'(x) = x2~ 3
. (-sinx)
Section 2.4
67. y
= (In X)4
dy
dx
= 4(Inxp
68.
y=xlnx
:
x
(x )= 4(Inxp
1.
2x
dy
1
1
dx =
~ + "2 X2 -
(
1
)=
:
2r - 1
- 1)
X(x2
= x(~) + lnx = 1 + lnx
1
70. y = In~.x2- 4 = "2ln(.x2- 4)
1
= In x + "2ln(.x2- 1)
69. y = lnx~
= ~(.x2
~4)= .x2~4
x
x
= In~x + 1 = In x - In(.x2+ 1)
71. f(x)
1
2x
-
= In x + 1 = In x - In(x + 1)
72. f(x)
1
1_~
f'(x) = ~ - x + 1 - .x2+ x
1-.x2
f'(x) = ~ - .x2+ 1 - x(xl + 1)
(;+11
73. Y = In-V ~
:
~l
= "2[In(x+
-R+1
74. y = In-V
:
1
.x2
+ R+1
+
(
+ x + R+1
1
= .x2R+1
1
+ R+1
-#+4
2r
1
dx =
Note that:
)(
x +
R+1
R+1+x
)(
1+
)
R+1
2+
(
#+4
x
+ 4x#+4
2r
)- -#+4
-
1.ln(2
+ ~X2 + 4) + 1.lnx
4
4
1
- 1.
x
+
4 (2 + ~.x2 + 4 )(~.x2 + 4 ) 4x
4X4
1
1
.2-#+4_2-#+4
2 + ../.x2+ 4 - 2 + ../.x2+ 4 2 - ../.x2+ 4 Hence dy =
, dx
-1
2x../.x2 + 4
+
#+4
X3
= - 1 + (1/2)(2 - ff+4)
2x../.x2+ 4
4
4x../.x2 + 4
In t
g'(t)
-.x2
+ #+4
X3
4x
(
x
~X2 + 4
2tln t
1-21nt
t3
)+
.!.
4x
+ .!.
4x
../.x2+ 4
X3
78. h(t)
= t2
= t2(1/t) t4-
X3
.!.
-.x2
- 1.(2 - ff+4)
1
= -~.x2 + 4 + ../xl + 4 +-=-
77. g(t)
)
x
R+1
1) - In(x + 1)]
= ~[x ~ 1 - x ~ 1] = .x2~ 1
1+.x2
R+1
= .x2R+1 = ~
- 4"ln
- 2r(x/ ff+4)
1
(
1
= .x2R+1
dy
= "2[In(x -
~
dx
=
m
+ In(x + ...;.x2+ 1)
x
dy = -x(x/R+1)
76. y
1) - In(x- 1)]
= Ux ~ 1 - x ~.1] = 1 ~ X2
75. Y =
The Chain Rule
h'(t)
= Int t
= t(1/t) t 2-lnt
- 1 -lnt
-t2
135
136
79.
Chapter 2
y
dy
dx
Differentiation
= Inlsinxl
80. y = Inlsecxl
= cosx = cotx
dy = see x tan x = tan x
dx
see x
sin x
cosx
81.
y
= In cosx -
I
I
82. y
I
dy - seex tanx + see2x
dx secx + tanx
= Inleosxl - Inlcosx - II
dy
dx
83.
y
=
-sin x -
-sin x
eosx
= In
1
cosx - I
= -tan
x +
= seex(secx + tanx) = see x
sin x
eosx - I
- I + sin X
2 + sin x I
see x + tan x
=
cosx
-
-I+sinx
-
= In"'1
84. y
= Inj-I + sinxl -lnl2 + sinxl
dy
dx
= InIsee x + tan xl
+ sin2x = ~ In(I + sin2x)
cos x = sin x cos x
I + sin2x
(2) I + sin2x
= 1. 2 sinx
dy
dx
eosx
2+sinx
3cosx
- (sinx - 1)(sinx + 2)
85.
y
1 + e"
86. y=lnl-e"
(
= Ine"> = x2
dy=2x
dx
)
= In(I + ex) - In(I - ex)
dy
e"
-=-+dx I+e"
e"
l-e"
2e"
=1-e2x
87. f(x) = e-X Inx
f'(x) = e-x(~) - e-XInx = e-x(~ - Inx)
89.
y
:
= e"(sinx + cosx)
= e"(cos
=
x
-
e"(2 cos x)
90. y=lne"=x
~=
dx
sin x) + (sin x + cos x)(e")
=
1
s'(t) = -(t2
2 + 2t + 8)-1/2(2t+ 2)
t+ I
-ff+2i+8
1
2e" cos x
91. s(t) = (t2 + 2t + 8)1/2, (2,4)
3
s'(2) = -4
88. f(x) = e3Inx
e3
f'(x) = -x
92.
y
= (3x3 +
4X)I/5, (2,2)
I
y' = -(3x3
+ 4X)-4/5(9x2 + 4)
5
9x2 + 4
- 5(3x3 + 4X)4/5
1
y'(2)= -2
Section 2.4
93.
lex) = x3
/'(x) = f
~
4 = 3(X3 -
3(X3 -
4)-2(3r) =
(x3- 4)2
9
'(-I) = -- 25
94. lex) = (r ~ 3x)2= (X2- 3X)-2, (4, I~)
/'(x) = -2(r - 3x)-3(2x- 3) = -2(2x - 3)
(r - 3x)3
/,(4) = -2-32
3t + 2
95. fer)= _t- 1 '
97.
4)-1, (-1, -~)
9r
The Chain Rule
x+ 1
96. lex) = 2x - 3' (2,3)
(0, -2)
/'(t) = (t - 1)(3)- (3t + 2)(1)-' - 5
(t - 1)2
- (t - 1)2
'
f ( ) = (2x - 3)(1) - (x + 1)(2)= ---=1
/,(0) = -5
/,(2) = - 5
x
(2x- 3)2
2!:~
98. y = .!.+
.,/cosx '2"7T
x
( )
y = 37 - sec3(2x), (0, 36)
y' = - 3 sec2(2x)[2sec(2x)tan(2x)]
1
y' = -
= - 6 sec3(2x)tan(2x)
y'(O) = 0
r
sinx
- 2"/cosx
y'( '7T/2) is undefined.
99. (a) lex) = "/3r - 2, (3,5)
1
/'(x) = -(3r
- 2)-1/2(6x)
2
3x
- "/3X2- 2
~) _.~.
-3
/,(3) = ~
5
Tangent line:
y
9
- 5 = -(x
5
1
100. (a) lex) = 3x"/
r
9x
-
5y
- 2 =0
+ 5, (2, 2)
/'(x) = .!.X
+
3 [ .!.(X2
2
=
- 3) ~
5)-1/2(2x)
3
] + .!.(r
+ 5)1/2
~)
~58.
-6
X2
1
3../r + 5 + 3"/r + 5
/,(2) = 4
1
3{3)+ 3(3) = Q9
Tangent line:
13
y
- 2 = -(x
9
- 2) ~
I3x
-
9y
- 8 =0
101.(a) lex) = sin2x, ('7T,
0)
/'(x) = 2 cas 2x
/'('7T)= 2
Tangent line:
y = 2(x - '7T)
~ 2x - y - 2'7T= 0
(b).~~
-2
(2x - 3)2
137
138
Differentiation
Chapter 2
)
102. (a) f(x) = tan2x, (~, 1
f'(x) = 2 tan x sec2x
(b)",m
f'(~) = 2(1)(2)= 4
",
-4
Tangent line:
y -
1
= 4(X - ~) => 4x - y + (1 - 7T)= 0
103. (a) y = 4 - X2 - InG x + 1),
dy-
1
1
dx - -2x - (I/2)x + 1
= -2x--
(0,4)
(2)
~) _.~.
1
-4
x+2
dy -_!
Whenx=O,dx2.
Tangent line: y - 4
= -~(x
- 0)
1
y= --x+4
2
104. (a) y = 2el-x2
y' = 2el-X2(-2x) = -4xel-x2
At (1, 2), y' = -4
Tangent
line: y -
~)L1J.
2 = -4(x - 1)
-1
Y= -4x + 6
106. g(x) = 2-x
105. f(x) = 4X
f'(x) = (In 4) 4X
107. Y = 5x-2
g '(x) = - (In 2) 2-x
108. y = X(7-3x)
109. g(t)
dy
dx = xC- 3(In 7) 7-3x] + 7-3x
= 7-3x[ -3x(In 7) + 1]
dy = (In 5) 5x-2
dx
= t22'
110. f(t) = 32'
t
g'(t) = t2 (In2) 2' + (2t)2'
f'(t) = t(2ln 3)t2
32' - 32'
= t2'(tln2 + 2)
= 2't(2 + tin 2)
- 32'(2t In 3 - 1)
= 7-3x (I - 3x In 7)
111.
h(8)
=
2-9COS 7T8
h '(8) = 2-9(-
7Tsin 7T8) - (In 2)2-9 COS7T8
= -2-9((In
2) cos 7T8+ 7Tsin 7T8]
-
t2
112. g(o:) = 5-a/2 sin 20:
g '(0:) = 5-a/2 2 COS20: - !(In 5) 5-a/2 sin 20:
The Chain Rule
Section 2.4
113. Y
139
x2
= log3 X
114. y = 10gIO(2x) = 10gIO2 + 10gIOx
dy=~
dx xln3
dy=O+~=~
dx
xlnlO
115. f(x)
= log2 X = 210g2 X -
xlnlO
I
log2(x - I)
2
I
f'(x) = x In 2 - (x - I) In 2
x-2
= (In2)x(x - I)
116. h(x)
= log3
xh-=t
I
2
117.
y
I
g 3 (X - I) - log32
- Iog3x + -lo
2
h'(x) = X~3 +~.
=
~
logs
dy-.!..
dx - 2
=
2 logs
(X2
- I)
2x
x
(x2 - I)ln 5 - (X2- I)ln 5
(x- ~)ln3-0
I I
1
= In 3 ~ + 2(x - 1)
[
]
1
3X-2
- 1)]
= In 3[ 2x(x
x2 - 1
118. y = 10gIO~
119. g(t) = 1010&1
t t = JQ.
In4
= 10glO(x2- I) - 10gIOX
dy2x
dx-(x2-1)lnlO
g'(t) = JQ.
In4[
ln t
(t )
t(1/t) - In t
t2
]
-~
=~
t21nil
xlnlO
2x
1
=lnlOx2-1-~
1
5
t In 2
-Int] = ~(1
-Int )
[
]
1
x2+1
= In 10[x(x2 - 1)]
120. f (t) = t3/210g2 ~
= t3/2.!.
+ I)
2 In(tIn2
1
f'(t) = -21n2
3
+ -tl/2In(t
+ 1)
2
1
t312t+l
[
121.
]
-2
-3
The zeros off' correspond to the points where the graph
offhas horizontal tangents.
122.
123.
-3
\
-2 -I
I
2
3
-1
-2
-3
tis decreasing on (- 00, -1) sof' must be negative
there.fis increasing on (1,00) sof' must be postive
there.
The zeros off' correspond to the points where the graph
off has horizontal tangents.
140
Chapter 2
Differentiation
124.
125. f(x)
f'(x)
= 2(x2 = 6(x2 =
1)2(2x)
12x{x4 - 2x2 + 1)
= 12x5r(x)
1)3
24x3 + 12x
= 6OX4 - 72x2+ 12
=
12(5x2 - 1)(x2 - 1)
The zeros off' correspond to the points where the graph
off has horizontal tangents.
126. f(x)
= (x - 2)-1
-1
f'(x) = - (x - 2)-2 = (x - 2)2
2
r(x) = 2(x- 2)-3 = (x - 2)3
127. f(x) = sin X2
f'(x) = 2x cos X2
r(x)
=
= 2x[2x(-sinx2)] + 2cosx2
128. f(x)
f'(x)
r(x)
2[cosx2 - 2x2sinx2)
= sec2 'TTX
=
2 see 'TTx('TTsee 'TTXtan 'TTx)
=
2'TT sec2
'TTX tan
=
2'TT sec2
'TTX(sec2
'TTX
1TX)( 'TT)
+
2'TT tan
'TTx(2'TT sec2
= 2'TT2sec4
'TTX + 4'TT2 sec2 'TTXtan2 'TTX
=
2'TT2 sec2
'TTx(sec2
=
2'TT2 sec2
'TTx(3 sec2
'TTX
'TTX tan
'TTx)
+ 2 tan2 'TTx)
'TTX -
2)
129. f(x)
= (3 +
2x)e-3x
f'(x)
= (3 +
2x)( - 3e-3x)+ 2e-3x
130. g(x) = .Jx + eXInx
g'(x) =
= (-7 - 6x)e-3x
1
eX
r + - + eX1nx
2..;x x
1
g"(x) --4--V2+
x-
j(x) = (-7 - 6x)(-3e-3x) - 6e-3x.
XeX-eX
eX
x2
+-+eXi
x
nx
= 3(6x + 5)e-3x
= --1-
+
4x.Jx
131. (a) f(x)
= g(x)h(x)
(b) f(x)
eX(2x
-
1)
x2
+ eXInx
= g(h(x»
f'(x) = g(x)h'(x) + g'(x)h(x)
f'(x) = g'(h(x»h'(x)
f'(5) = (-3)(-2)
f'(5) = g '(3)(- 2) = - 2g '(3)
+ (6)(3)= 24
Need g '(3) to findf'(5).
(c) f(x) = g(x)
h(x)
(d) f(x)
f'(x)
f'(x) = h(x)g'(x) - g(x)h'(x)
[h(x»)2
f'(5) = (3)(6) - (-3)(-2) - 12
(3)2
- "9
f'(5)
4
= "3
= [g(x»)3
= 3(g(x»)2g'(x)
= 3(-3)2(6) = 162
Section 2.4
= 1 =>g/(x)= 0
g /(x) = 2 sinx cosx + 2 cosx( - sin x) = 0
tan2x + 1 = sec2x
g(x) + 1 = f(x)
132. (a) g(x)
(b)
= sin2x
+ cos2 x
133. (a) f = 132,400(331-
.
V)-l
- 132,400
- (331 - v)2
Whenv = 30,f' = 1.461.
(b) f = 132,400(331+ v)-l
g /(x) = I'(x).
g /(x) = 2 tan x
141
l' = (-1)(132,400)(331- v)-2(_1)
Taking derivatives of both sides,
Equivalently,f'(x)
The Chain Rule
.
l' = (-1)(132,400)(331+ v)-2(1)
.
= 2 see x sec x tan x and
sec2 x, which are the same.
- -132,400
- (331 + V)2
Whenv = 30,f' = -1.016.
134. Y
v
= 1cos 12t -
135. fJ= 0.2 cos 8t
sin 12t
~
= y/ = H-12 sin 12t]- H12 cos 12t]
= -4 sin 12t - 3 cos 12t
The maximum angular displacement is fJ = 0.2 (since
-1 ~ cos 8t ~ 1).
When t = '11/8,Y = 0.25 feet and v = 4 feet per second.
~~ = 0.2[- 8 sin 8t] = -1.6 sin 8t
Whent = 3, dfJ/dt = -1.6 sin 24 = 1.4489 radians per
second.
136. y
= A COSwt
137.
(2R dRdt - 2r drdt )
=C
= 325 = 1.75
dS
dt
=
Since r is constant, we have dr / dt
(a) Amplitude: A
y
S = C(R2 - r2)
1.75 cos CJJt
217
Period: 10 => w =
17
10 = "5
~~
= (1.76 x
= 4.224
17t
Y
= 0 and
1()5)(2)(1.2x 10-2)(10-5)
X 10-2
= 0.04224.
= 1.75 cos"'5
(b) v = y/ = 1.75[-~sin ~t]
= -0.35
17t
17sin "'5
138. (a) Using a graphing utility, or by trial and error, you
obtain a model of the form
T(t)
= 64.18 -
22.15 sin( ~t + 1)
(c) T'(t) = - 22.15cos(~t + 1)(~)
= -11.60 cos(~t + 1)
20
.~"
-20
0
(d) The temperature changes most rapidly when t
(April) and t
=
= 4.1
10.1 (October). The temperature
changes most slowly (T'(t) = 0) when t = 1.1
(January) and t = 7.1 (July).
142
Chapter 2
Differentiation
139. (a) g(x) = j(x)-2
(b) h(x) = 2j(x)
~
x
h'(x) = 2f'(x)
~
(c) r(x) = j(-3x)
g'(x) = f'(x)
~
= -3f'(-3x)
8
h'(x)
s'(x)
= 12
0
I
-3
-3
-3
I
-I
1
2
-3
12
r'(x)
r'(O) = - 3j'(0) = (- 3)(-f) = I
2
3
2
3
4
3
4
g'(x)
Hence, you need to know 1'( - 3x).
-I
4
f'(x)
r'(x) = f'(-3x)(-3)
r'(-I) = -31'(3) = (-3)(-4)
-2
I
2
3
-I
-2
-4
-I
-2
-4
-2
-4
-8
I
-2
-4
(d) s(x)= j(x + 2) ~ s'(x)= f'(x + 2)
Hence, you need to know f'(x + 2).
s'(-2)
=1'(0) = -~,etc.
140. V = 15,OOOe-o.6286t,
0 ~ t ~ 10
dV
(b)
(a)
~D"
dV
When t = I, dt = - 5028.84.
dV
When t = 5, dt = -406.89.
0
141. (a)
-dt = -942ge-O.6286t
350
(b)
T'(p)
= 34.96 +
p
(0)
3.955
,jp
T'(IO)= 4.75deg/lb/in2
100
0
0
0"
T'(70) = 0.97deg/lb/in2
(oj
.0..
0
Hm T'(p)
p->OO
142. V(t) = 20,OOO(~)t
(a)
( 4~)(~'y
4)
(b) dV = 20 ooo ln
v
dt
'
dV
When t = I: dt = -4315.23
16.000
(c)
=0
V'(x)
-1000
-2000
dV
8,000
When t
4.000
= 4: dt = -1820.49
-6000
246810
Horizontal asymptote: v' = 0
V(2)
143. C(t)
= 20,OOO(~J= $11,250
As the car ages, it is worth
less each year and depreciates
less each year, but the value of
the car will never reach $0.
= P(1.05)t
(a) C(IO) = 24.95(1.05)10
=$40.64
(b)
:~ = p(In
1.05)(1.05)t
dC
Whent = I: dt = 0.051P
dC
When t
= 8: dt = O.O72P
dC
(c) dt
= (In 1.05)[P(1.05)t]
=
(In 1.05)C(t)
The constant of proportionality
is In 1.05.
Section 2.4
144. f(x)
=
The Chain Rule
145. f(x + p) = f(x) for allx.
sin {3x
(a) Yes,f'(x + p) = f'(x), whichshowsthatf'is
periodicas well.
(a) f'(x) = {3cos {3x
- {32sin {3x
= - {33cos {3x
f"(x) =
f"'(x)
(b) Yes, let g(x) = f(2x), so g '(X) = 2f'(2x).
SinceI' is periodic, so is g '.
f(4) = {34sin {3x
(b) f"(x)
+ {32f(x)
= - {32sin {3x + {32(sin {3x)
=
0
f(2Ic)(X) = (-l)k {321c
sin {3x
(c)
f(2Ic-l)(X)
146. Iff(-x)
=
(-l)k+
1 {321c-l cos {3x
147. g(x) = ../x(x + n)
= -f(x), then
d
-[f(-x)]
dx
f'(-x)(-l)
= ../xl + nx
dg I
-dx = _(X2+
nx)-1/2(2x+ n)
2
d
= -[
x -f(x)]
= -f'(x)
2x + n
- 2../X2 + nx
1'( - x) = f'(x).
Thus,f'(x) is even.
- (2x + n)/2
- ../x(x + n)
- [x + (x + n)]/2
../x(x + n)
a
g
148.
lul=R
~[luJ] = ~[ R]
149. g(x) = 12x-
= ~(U2)-1/2(2UU~
=
(
= ull:1'
u oF 0
150. f(x) = Ixl - 41
31
2x-3
g'(x)=212x-31'
;;
)
3
xoF2
151. h(x) = Ixlcosx
x2-4
f'(x) = 2x(Ixl - 41),x oF:1:2
152. f(x) = Isinxl
sin x
1
h '(X)= -Ixl sinx + 1;1COSX,x oF0
f(x)
= cosx
( )
Isinxl ,x oFk1r
143
144
Chapter 2
153. (a) lex)
Differentiation
I(Tr) = sin.z!:=
1
2
= sin~
1
f'(x) =
1.
2
1
Tr
f'(Tr) = -cos2
2=0
x
2cos 2
II(
(b)
x
"
)
1 (Tr=
I 'x) = --sm4
2
1.
Tr
0
1
8
1
f
Pz
0
--sm4
2 = --4
6.28
P1(x)= f'( Tr)(x- Tr)+ I( Tr)= f( Tr)= 1
1
pix) = 2f"(Tr)(x - Tr)2+ f'(Tr)(x- Tr)+ I(Tr)
1
1
1
8
( )
= -2 --4
(x - Tr)2+ 1 = 1 - -(x - Tr)2
(d) The accuracy worsens as you move away
from x = Tr,
(c) Pz is a better approximation than PI'
154. (a) lex)
= sec(2x)
(b)
rex) = 2[2(sec 2x)(tan 2x)] tan 2x + 2(sec 2x)(secz2x)(2)
= 4[(sec 2x)(tanZ2x) + sec32x]
~
6
f'(x) = 2(sec 2x)(tan2x)
0
Pz
PI
0.78
I(~) = sec(~)= 2
I{~) = 2 sec(~)
tan(~) = 4J3
r(~) = 4[2(3)+ 23]= 56
PI(x)= 4J3(X -~) + 2
pz(x)= ~(56)(X- ~)z + 4J3(X -~) + 2
= 28(X- ~)z + 4J3(x -~) + 2
(c) Pz is a better approximation than PI'
(d) The accuracy worsens as you move away
from x = Tr/6,
2
155. lex) = e-x2/Z,f(O)= 1
f'(x) = - xe-x2/Z,f'(O)= 0
/'(x) = x2e-x2/Z- e-x2/Z= e-x2/Z(xZ- 1),/'(0) = -1
PI(x) = 1 + O(x- 0) = 1,PI(O)= 1
-,~,
-2
PI '(x) = 0, PI'(0) = 0
1
x2
Pix) = 1 + O(x- 0) - 2(x - 0)2 = 1 - 2' Pz(O)= 1
Pz'(x) = -x,Pz'(O) = 0
Pz"(x)= -1,Pz"(O) = -1
The values off, PI' Pz and their first derivatives agree at x
land Pz agree at x = 0,
= 0, The values of the second derivatives of