Section 3.5 The Chain Rule and Parametric Equations 147 87. x 2t 5, y 4t 7, t 88. x 3 3t, y 2t, 0 t 1 t œ œ _ _ œ œ Ÿ Ÿ Ê œ y # x 5 2t 2(x 5) 4t x 3 3 2x 6 3y Ê œ Ê œ Ê œ Ê œ ˆ‰ y # y 2(x 5) 7 y 2x 3 y 2 x, x Ê œ Ê œ Ê œ !Ÿ Ÿ$ 2 3 89. x t, y 1 t, 1 t 0 90. x t 1, y t, t 0 œ œ Ÿ Ÿ œ œ È È È # y 1 x y t x y 1, y 0 Ê œ Ê œ Ê œ È È # # # 91. x sec t 1, y tan t, t 92. x sec t, y tan t, t œ œ œ œ # # # # # 1 1 1 1 sec t 1 tan t x y sec t tan t 1 x y 1 Ê œ Ê œ Ê œ Ê œ # # # # # # # 93. (a) x a cos t, y a sin t, 0 t 2 94. (a) x a sin t, y b cos t, t œ œ Ÿ Ÿ œ œ Ÿ Ÿ 1 1 1 # # 5 (b) x a cos t, y a sin t, 0 t 2 (b) x a cos t, y b sin t, 0 t 2 œ œ Ÿ Ÿ œ œ Ÿ Ÿ 1 1 (c) x a cos t, y a sin t, 0 t 4 (c) x a sin t, y b cos t, t œ œ Ÿ Ÿ œ œ Ÿ Ÿ 1 1 1 # # 9 (d) x a cos t, y a sin t, 0 t 4 (d) x a cos t, y b sin t, 0 t 4 œ œ Ÿ Ÿ œ œ Ÿ Ÿ 1 1 95. Using we create the parametric equations x at and y bt, representing a line which goes a b "ß $ œ " œ $ through at t . We determine a and b so that the line goes through when t . a b a b "ß $ œ! %ß " œ" Since a a . Since b b . Therefore, one possible parameterization is x t, % œ " Ê œ& " œ $ Ê œ% œ " & y t, 0 t . œ $ % Ÿ Ÿ"
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Section 3.5 The Chain Rule and Parametric Equations 147
87. x 2t 5, y 4t 7, t 88. x 3 3t, y 2t, 0 t 1 tœ � œ � �_ � � _ œ � œ Ÿ Ÿ Ê œy#
x 5 2t 2(x 5) 4t x 3 3 2x 6 3yÊ � œ Ê � œ Ê œ � Ê œ �ˆ ‰y#
y 2(x 5) 7 y 2x 3 y 2 x, xÊ œ � � Ê œ � Ê œ � ! Ÿ Ÿ $23
89. x t, y 1 t , 1 t 0 90. x t 1, y t, t 0œ œ � � Ÿ Ÿ œ � œ È È È#
y 1 x y t x y 1, y 0Ê œ � Ê œ Ê œ � È È# ##
91. x sec t 1, y tan t, t 92. x sec t, y tan t, tœ � œ � � � œ � œ � � �## # # #1 1 1 1
sec t 1 tan t x y sec t tan t 1 x y 1Ê � œ Ê œ Ê � œ Ê � œ# # # # # # #
93. (a) x a cos t, y a sin t, 0 t 2 94. (a) x a sin t, y b cos t, tœ œ � Ÿ Ÿ œ œ Ÿ Ÿ1 1 1
# #5
(b) x a cos t, y a sin t, 0 t 2 (b) x a cos t, y b sin t, 0 t 2œ œ Ÿ Ÿ œ œ Ÿ Ÿ1 1
(c) x a cos t, y a sin t, 0 t 4 (c) x a sin t, y b cos t, tœ œ � Ÿ Ÿ œ œ Ÿ Ÿ1 1 1
# #9
(d) x a cos t, y a sin t, 0 t 4 (d) x a cos t, y b sin t, 0 t 4œ œ Ÿ Ÿ œ œ Ÿ Ÿ1 1
95. Using we create the parametric equations x at and y bt, representing a line which goesa b�"ß �$ œ �" � œ �$ �
through at t . We determine a and b so that the line goes through when t .a b a b�"ß �$ œ ! %ß " œ "
Since a a . Since b b . Therefore, one possible parameterization is x t,% œ �" � Ê œ & " œ �$ � Ê œ % œ �" � &
y t, 0 t .œ �$ � % Ÿ Ÿ "
148 Chapter 3 Differentiation
96. Using we create the parametric equations x at and y bt, representing a line which goes througha b�"ß $ œ �" � œ $ �
at t . We determine a and b so that the line goes through when t . Since a a .a b a b�"ß $ œ ! $ß�# œ " $ œ �" � Ê œ %
Since b b . Therefore, one possible parameterization is x t, y t, 0 t .�# œ $ � Ê œ �& œ �" � % œ �$ � & Ÿ Ÿ "
97. The lower half of the parabola is given by x y for y . Substituting t for y, we obtain one possibleœ � " Ÿ !#
parameterization x t , y t, t 0œ � " œ Ÿ Þ#
98. The vertex of the parabola is at , so the left half of the parabola is given by y x x for x . Substitutinga b�"ß �" œ � # Ÿ �"#
t for x, we obtain one possible parametrization: x t, y t t, t .œ œ � # Ÿ �"#
99. For simplicity, we assume that x and y are linear functions of t and that the point x, y starts at for t and passesa b a b#ß $ œ !
through at t . Then x f t , where f and f .a b a b a b a b�"ß �" œ " œ ! œ # " œ �"
Since slope , x f t t t. Also, y g t , where g and g .œ œ œ �$ œ œ �$ � # œ # � $ œ ! œ $ " œ �"?
?
xt
�"�#"�! a b a b a b a b
Since slope 4. y g t t t.œ œ œ � œ œ �% � $ œ $ � %?
?
yt
3�"�"�! a b
One possible parameterization is: x t, y t, t .œ # � $ œ $ � % !
100. For simplicity, we assume that x and y are linear functions of t and that the point x, y starts at for t anda b a b�"ß # œ !
passes through at t . Then x f t , where f and f .a b a b a b a b!ß ! œ " œ ! œ �" " œ !
Since slope , x f t t t. Also, y g t , where g and g .œ œ œ " œ œ " � �" œ �" � œ ! œ # " œ !?
?
xt
!� �""�!a b a b a b a b a b a b
Since slope . y g t t t.œ œ œ �# œ œ �# � # œ # � #?
?
yt
!�#"�! a b
One possible parameterization is: x t, y t, t .œ �" � œ # � # !
101. t x 2 cos 2, y 2 sin 2; 2 sin t, 2 cos t cot tœ Ê œ œ œ œ œ � œ Ê œ œ œ �1 1 1
4 4 4 dt dt dx dx/dt 2 sin tdx 2 cos tdy dy dy/dtÈ È
�
cot 1; tangent line is y 2 1 x 2 or y x 2 2 ; csc tÊ œ � œ � � œ � � œ � � œ¹ Š ‹È È Èdy dydx 4 dt
tœ1
4
1w
#
2Ê œ œ œ � Ê œ �d y dy /dt d ydx dx/dt 2 sin t 2 sin t dx
csc t# w #
# $ #
#
�" ¹ È
tœ1
4
102. t x cos , y 3 cos ; sin t, 3 sin t 3œ Ê œ œ � œ œ � œ � œ � Ê œ œ2 2 2 dx3 3 3 dt dt dx sin t
3 3 sin tdy dy1 1 1"# # �
�È È ÈÈ È
3 ; tangent line is y 3 x or y 3 x; 0 0Ê œ � � œ � � œ œ Ê œ œ¹ Š ‹È È È� ‘ˆ ‰dy dy d ydx dt dx sin t
3 0
tœ 231
È# # �
" w #
#
0Ê œ¹d ydx
#
#
tœ 231
103. t x , y ; 1, 1; tangent line isœ Ê œ œ œ œ Ê œ œ Ê œ œ1 1 dx 14 4 dt dt dx dx/dt dx
dy dy dy/dt dyt 2 t
" " "# # #È È É¹
tœ 14
"
4
y 1 x or y x ; t t 2� œ � œ � œ � Ê œ œ � Ê œ �" " " " "#
�$Î# �$Î#† ˆ ‰ ¹4 4 dt 4 dx dx/dt 4 dx
dy d y dy /dt d yw # w #
# #
tœ 14
104. t 3 x 3 1 2, y 3(3) 3; (t 1) , (3t) œ Ê œ � � œ � œ œ œ � � œ Ê œÈ È dx 3dt dt dx
dy dy (3t)(t 1)
"# #
�"Î# �"Î#� �
ˆ ‰ˆ ‰
3#
�"Î#
"#
�"Î#
2; tangent line is y 3 2[x ( 2)] or y 2x 1;œ � œ œ œ � � œ � � � œ � �3 t 1 3 3 1
3tdydx 3(3)
È ÈÈ È� � �¹
t 3œ
dy d ydt 3t dx
3t (t 1) 3 t 1 (3t) 3 32t 3t t 1 t 3t
w ## #
�"Î# �"Î#
#œ œ Ê œ œ �È � ‘ � ‘È
È ÈÈ� � � �
�
3 3 Š ‹Š ‹
32t 3t t 1
12 t 1
È È
È
�
�
�
Ê œ �¹d ydx 3
#
#
t 3œ
"
Section 3.5 The Chain Rule and Parametric Equations 149
105. t 1 x 5, y 1; 4t, 4t t ( 1) 1; tangent line isœ � Ê œ œ œ œ Ê œ œ œ Ê œ � œdx 4tdt dt dx dx/dt 4t dx
dy dy dy/dt dy$ # #$ ¹t 1œ�
y 1 1 (x 5) or y x 4; 2t � œ � œ � œ Ê œ œ œ Ê œ†
dy d y dy /dt d ydt dx dx/dt 4t dx
2tw # w #
# #
" "# #¹
t 1œ�
106. t x sin , y 1 cos 1 ; 1 cos t, sin t œ Ê œ � œ � œ � œ � œ œ � œ Ê œ1 1 1 1 1
3 3 3 3 3 dt dt dx dx/dt3 dx dy dy dy/dtÈ
# # #" "
3 ; tangent line is y 3 xœ Ê œ œ œ � œ � �sin t1 cos t dx 3
dy sin1 cos
3� # #�
"¹ Š ‹È Ètœ1
3
ˆ ‰ˆ ‰
Š ‹ˆ ‰
È1
1
3
3
3È#
"
#
1
y 3x 2; Ê œ � � œ œ Ê œ œÈ 1È ˆ ‰33 dt (1 cos t) 1 cos t dx dx/dt 1 cos t
dy (1 cos t)(cos t) (sin t)(sin t) d y dy /dt1w # w
# #
�
�� �� � �
�1
1 cos t
4œ Ê œ ���
1(1 cos t) dx
d y# #
# ¹tœ1
3
107. t x cos 0, y 1 sin 2; sin t, cos t cot tœ Ê œ œ œ � œ œ � œ Ê œ œ �1 1 1
2 2 2 dt dt dx sin tdx cos tdy dy
�
cot 0; tangent line is y 2; csc t csc t 1Ê œ � œ œ œ Ê œ œ � Ê œ �¹ ¹dy dy d y d ydx dt dx sin t dx
csc t
t tœ œ
1 1
2 2
1
# �# $
w # #
# #
#
108. t x sec 1 1, y tan 1; 2 sec t tan t, sec tœ � Ê œ � � œ œ � œ � œ œ1 1 1
4 4 4 dt dtdx dy# # #ˆ ‰ ˆ ‰
cot t cot ; tangent line isÊ œ œ œ Ê œ � œ �dy dydx 2 sec t tan t 2 tan t dx 4
sec t#
#
" " " "# # #¹ ˆ ‰
tœ� 1
4
1
y ( 1) (x 1) or y x ; csc t cot t� � œ � � œ � � œ � Ê œ œ �" " " " "# # # #
# $�dy d ydt dx 2 sec t tan t 4
csc tw #
# #
"
#
#
Ê œ¹d ydx 4
#
#
tœ� 1
4
"
109. s A cos (2 bt) v A sin (2 bt)(2 b) 2 bA sin (2 bt). If we replace b with 2b to double theœ Ê œ œ � œ �1 1 1 1 1dsdt
frequency, the velocity formula gives v 4 bA sin (4 bt) doubling the frequency causes the velocity toœ � Ê1 1
double. Also v bA sin (2 bt) a 4 b A cos (2 bt). If we replace b with 2b in theœ �# Ê œ œ �1 1 1 1dvdt
# #
acceleration formula, we get a 16 b A cos (4 bt) doubling the frequency causes the acceleration toœ � Ê1 1# #
quadruple. Finally, a 4 b A cos (2 bt) j 8 b A sin (2 bt). If we replace b with 2b in the jerkœ � Ê œ œ1 1 1 1# # $ $dadt
formula, we get j 64 b A sin (4 bt) doubling the frequency multiplies the jerk by a factor of 8.œ Ê1 1$ $
110. (a) y 37 sin (x 101) 25 y 37 cos (x 101) cos (x 101) .œ � � Ê œ � œ �� ‘ � ‘ ˆ ‰ � ‘2 2 2 74 2365 365 365 365 3651 1 1 1 1w
The temperature is increasing the fastest when y is as large as possible. The largest value ofw
cos (x 101) is 1 and occurs when (x 101) 0 x 101 on day 101 of the year� ‘2 2365 3651 1� � œ Ê œ Ê
( April 11), the temperature is increasing the fastest.µ
(b) y (101) cos (101 101) cos (0) 0.64 °F/dayw œ � œ œ ¸74 2 74 74365 365 365 3651 1 1 1� ‘
111. s ( 4t) v (1 4t) (4) 2(1 4t) v(6) 2( 6) m/sec;œ " � Ê œ œ � œ � Ê œ " � % œ"Î# �"Î# �"Î# �"Î#"#
ds 2dt 5†
v 2( 4t) a 2(1 4t) (4) 4(1 4t) a(6) 4(1 4 6) m/secœ " � Ê œ œ � � œ � � Ê œ � � œ ��"Î# �$Î# �$Î# �$Î# #"# #
dv 4dt 1 5† †
112. We need to show a is constant: a and k s a vœ œ œ œ œ Ê œ œdv dv dv ds dv d k dv ds dvdt dt ds dt ds ds ds dt ds2 s
† † †ˆ ‰È È k s which is a constant.œ œk k
2 sÈ † È #
#
113. v proportional to v for some constant k . Thus, a v"È Ès sk dv k dv dv ds dv
ds dt ds dt ds2sÊ œ Ê œ � œ œ œ$Î# † †
acceleration is a constant times so a is inversely proportional to s .œ � œ � Êk k k2s s s s$Î#
#
# #† È #" " #ˆ ‰
114. Let f(x). Then, a f(x) f(x) (f(x)) f(x) f (x)f(x), as required.dx dv dv dx dv d dx ddt dt dx dt dx dx dt dxœ œ œ œ œ œ œ† † † †ˆ ‰ w
150 Chapter 3 Differentiation
115. T 2 2 . Therefore, kL 2 kœ Ê œ œ œ œ œ œ œ1 1 1É ÉL dT dT dT dL Lg dL g du dL du gg gL gL
k Lg† † † † †
" " "
# #É É È È ÈÈ
L Lg g
1 1 1 1
, as required.œ kT2
116. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f g is‰
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so there is no contradiction.
117. The graph of y (f g)(x) has a horizontal tangent at x 1 provided that (f g) (1) 0 f (g(1))g (1) 0œ ‰ œ ‰ œ Ê œw w w
either f (g(1)) 0 or g (1) 0 (or both) either the graph of f has a horizontal tangent at u g(1), or theÊ œ œ Ê œw w
graph of g has a horizontal tangent at x 1 (or both).œ
118. (f g) ( 5) 0 f (g( 5)) g ( 5) 0 f (g( 5)) and g ( 5) are both nonzero and have opposite signs.‰ � � Ê � � � Ê � �w w w w w†
That is, either f (g( 5)) 0 and g ( 5) 0 or f (g( 5)) 0 and g ( 5) 0 .c d c dw w w w� � � � � � � �
119. As h 0, the graph of yÄ œ sin 2(x h) sin 2xh
� �
approaches the graph of y 2 cos 2x becauseœ
lim (sin 2x) 2 cos 2x.h Ä !
sin 2(x h) sin 2xh dx
d� � œ œ
120. As h 0, the graph of yÄ œ cos (x h) cos xh
c d a b� �# #
approaches the graph of y 2x sin x becauseœ � a b# lim cos x 2x sin x .
h Ä !
cos (x h) cos xh dx
dc d a b� � # ## #
œ œ �c d a ba b
121. cos t and 2 cos 2t ; then 0 0dx 2 cos 2tdt dt dx dx/dt cos t cos t dx cos t
dy dy dy/dt dy2 2 cos t 1 2 2 cos t 1œ œ Ê œ œ œ œ Ê œa b a b# #� �
2 cos t 1 0 cos t t , , , . In the 1st quadrant: t x sin andÊ � œ Ê œ „ Ê œ œ Ê œ œ# "#È
È2 4 4 4 4 4 4
3 5 7 21 1 1 1 1 1
y sin 2 1 1 is the point where the tangent line is horizontal. At the origin: x 0 and y 0œ œ Ê ß œ œˆ ‰ Š ‹1
42È
#
sin t 0 t 0 or t and sin 2t 0 t 0, , , ; thus t 0 and t give the tangent lines atÊ œ Ê œ œ œ Ê œ œ œ1 1 11 1
# #3
the origin. Tangents at origin: 2 y 2x and 2 y 2x¹ ¹dy dydx dx
t 0 tœ œ
œ Ê œ œ � Ê œ �1
122. 2 cos 2t and 3 cos 3t dx 3 cos 3tdt dt dx dx/dt 2 cos 2t 2 2 cos t 1
dy dy dy/dt 3(cos 2t cos t sin 2t sin t)œ œ Ê œ œ œ ��a b#
œ œ œ3 2 cos t 1 (cos t) 2 sin t cos t sin t2 2 cos t 1 2 2 cos t 1 2 2 cos t
(3 cos t) 2 cos t 1 2 sin t (3 cos t) 4 cos t 3c da ba b a b a ba b a b#
# # #
# # #� �� �
� � ��1 ; then
Section 3.5 The Chain Rule and Parametric Equations 151
0 0 3 cos t 0 or 4 cos t 3 0: 3 cos t 0 t , anddydx 2 2 cos t 1
(3 cos t) 4 cos t 3 3œ Ê œ Ê œ � œ œ Ê œa ba b#
#
�� # #
# 1 1
4 cos t 3 0 cos t t , , , . In the 1st quadrant: t x sin 2## #� œ Ê œ „ Ê œ œ Ê œ œ
È È3 36 6 6 6 6 6
5 7 111 1 1 1 1 1ˆ ‰ and y sin 3 1 1 is the point where the graph has a horizontal tangent. At the origin: x 0œ œ Ê ß œˆ ‰ Š ‹1
63È
#
and y 0 sin 2t 0 and sin 3t 0 t 0, , , and t 0, , , , , t 0 and t giveœ Ê œ œ Ê œ œ Ê œ œ1 1 1 1 1 1
# #1 1 13 2 4 53 3 3 3
the tangent lines at the origin. Tangents at the origin: y x, and ¹ ¹dy dydx 2 cos 0 dx
3 cos 0 3 3
t 0 tœ œ
œ œ Ê œ# #1
y xœ œ � Ê œ �3 cos (3 )2 cos (2 )
3 31
1 # #
123. From the power rule, with y x , we get x . From the chain rule, y xœ œ œ"Î% �$Î%"dydx 4
ÉÈ x x , in agreement.Ê œ œ œdy
dx dx 4x x
dx
" " " "
# # #�$Î%
É ÉÈ È È† †ˆ ‰È
124. From the power rule, with y x , we get x . From the chain rule, y x xœ œ œ$Î% �"Î%dydx 4
3 É È x x x x xÊ œ Ê œ � œ œdy dy
dx dx dxx x x x x x 4 x x
d 3x
3 x" " " "
# # ## #É É É ÉÈ È È ÈÈÈ
† † † †ˆ ‰ ˆ ‰È È ÈŠ ‹ x , in agreement.œ œ
3 x
4 x x
34
ÈÈ ÈÉ
�"Î%
125. (a)
(b) 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14tdfdt œ � � �
(c) The curve of y approximates yœ œdfdt dt
dg
the best when t is not , , 0, , nor .� �1 11 1
# #
126. (a)
(b) 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)dhdt œ � � � �
152 Chapter 3 Differentiation
(c)
125-130. Example CAS commands: :Maple f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t) - 0.02546*cos(10*t) - 0.01299*cos(14*t); g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t<Pi/2, t, Pi-t ); plot( [f(t),g(t)], t=-Pi..Pi ); Df := D(f); Dg := D(g); plot( [Df(t),Dg(t)], t=-Pi..Pi ); : (functions, domains, and value for t0 may change):Mathematica To see the relationship between f[t] and f'[t] in 111 and h[t] in 112 Clear[t, f] f[t_] = 0.78540 0.63662 Cos[2t] 0.07074 Cos[6t] 0.02546 Cos[10t] 0.01299 Cos[14t]� � � �
f'[t] Plot[{f[t], f'[t]},{t, , }]�1 1
For the parametric equations in 113 - 116, do the following. Do NOT use the colon when defining tanline. Clear[x, y, t] t0 = p/4; x[t_]:=1 Cos[t]�
43. x y 16 3x 3y y 0 3y y 3x y ; we differentiate y y x to find y :$ $ # # w # w # w # w # ww� œ Ê � œ Ê œ � Ê œ � œ �xy
#
#
y y y 2y y 2x y y 2x 2y y y# ww w w # ww w ww# � � � � �� œ � Ê œ � � Ê œ œc d c d†
2x 2y 2x
y y
Š ‹x 2xy y
# %
# $
#
# #
2œ Ê œ œ �� � � �2xy 2x d yy dx 32
32 32$ % #
& # ¹(2 2)ß
44. xy y 1 xy y 2yy 0 y (x 2y) y y y ;� œ Ê � � œ Ê � œ � Ê œ Ê œ# w w w w ww�� �
� � � � �y(x 2y) (x 2y)
(x 2y) y ( y) 1 2ya b a bw w
#
since y we obtain yk kw ww" "#
� �(0 1) (0 1)� �
œ � œ œ �( 2) (1)(0)
4 4
ˆ ‰"#
45. y x y 2x at ( ) and ( 1) 2y 2x 4y 2 2y 4y 2 2x# # % $ $� œ � �#ß " �#ß� Ê � œ � Ê � œ � �dy dy dy dydx dx dx dx
2y 4y 2 2x 1 and 1Ê � œ � � Ê œ Ê œ � œdy dy dy dydx dx y y dx dx
xa b ¹ ¹$ �"# �$
( 2 1) ( 2 1)� ß � ß�
46. x y (x y) at( ) and ( 1) 2 x y 2x 2y 2(x y) 1a b a b Š ‹ Š ‹# # # # ##� œ � "ß ! "ß� Ê � � œ � �dy dy
dx dx
2y x y (x y) 2x x y (x y) 1Ê � � � œ � � � � Ê œ Ê œ �dy dy dydx dx 2y x y (x y) dx
2x x y (x y)c d a ba b ¹# # # # � � � �� � �
a ba b# #
# #
(1 0)ß
and 1¹dydx
(1 1)�
œ
47. x xy y 1 2x y xy 2yy 0 (x 2y)y 2x y y ;# # w w w w ��� � œ Ê � � � œ Ê � œ � � Ê œ 2x y
2y x
(a) the slope of the tangent line m y the tangent line is y 3 (x 2) y xœ œ Ê � œ � Ê œ �kw "#(2 3)ß
7 7 74 4 4
(b) the normal line is y 3 (x 2) y x� œ � � Ê œ � �4 4 297 7 7
156 Chapter 3 Differentiation
48. x y 25 2x 2yy 0 y ;# # w w� œ Ê � œ Ê œ � xy
(a) the slope of the tangent line m y the tangent line is y 4 (x 3)œ œ � œ Ê � œ �k ¹w(3 4)
(3 4)�
�
x 3 3y 4 4
y xÊ œ �3 254 4
(b) the normal line is y 4 (x 3) y x� œ � � Ê œ �4 43 3
49. x y 9 2xy 2x yy 0 x yy xy y ;# # # # w # w # wœ Ê � œ Ê œ � Ê œ � yx
(a) the slope of the tangent line m y 3 the tangent line is y 3 3(x 1)œ œ � œ Ê � œ �k ¸w( 1 3) ( 1 3)� ß � ß
yx
y 3x 6Ê œ �
(b) the normal line is y 3 (x 1) y x� œ � � Ê œ � �" "3 3 3
8
50. y 2x 4y 2yy 2 4y 0 2(y 2)y 2 y ;# w w w w "�#� � � " œ ! Ê � � œ Ê � œ Ê œ y
(a) the slope of the tangent line m y 1 the tangent line is y 1 1(x 2) y x 1œ œ � Ê � œ � � Ê œ � �kw ( 2 1)� ß
(b) the normal line is y 1 1(x 2) y x 3� œ � Ê œ �
51. 6x 3xy 2y 17y 6 0 12x 3y 3xy 4yy 17y 0 y (3x 4y 17) 12x 3y# # w w w w� � � � œ Ê � � � � œ Ê � � œ � �
y ;Ê œw � �� �12x 3y
3x 4y 17
(a) the slope of the tangent line m y the tangent line is y 0 (x 1)œ œ œ Ê � œ �k ¹w �" �� �( 1 0)
( 1 0)� ß
� ß
2x 3y3x 4y 17 7 7
6 6
y xÊ œ �6 67 7
(b) the normal line is y 0 (x 1) y x� œ � � Ê œ � �7 7 76 6 6
52. x 3xy 2y 5 2x 3xy 3y 4yy 0 y 4y 3x 3y 2x y ;# # w w w w �
�� � œ Ê � � � œ Ê � œ � Ê œÈ È È È ÈŠ ‹ È
È3y 2x4y 3x
(a) the slope of the tangent line m y 0 the tangent line is y 2œ œ œ Ê œk ¹w �
�Š ‹ÈŠ ‹È3 2
3 2ß
ß
ÈÈ3y 2x
4y 3x
(b) the normal line is x 3œ È53. 2xy sin y 2 2xy 2y (cos y)y 0 y (2x cos y) 2y y ;� œ Ê � � œ Ê � œ � Ê œ1 1 1 1w w w w �
�2y
2x cos y1
(a) the slope of the tangent line m y the tangent line isœ œ œ � Êk ¹w �� #ˆ ‰
ˆ ‰11
ß
ß
1
122
2y2x cos y1
1
y (x 1) y x� œ � � Ê œ � �1 1 1
# # # 1
(b) the normal line is y (x 1) y x� œ � Ê œ � �1 1
1 1 1# #2 2 2
54. x sin 2y y cos 2x x(cos 2y)2y sin 2y 2y sin 2x y cos 2x y (2x cos 2y cos 2x)œ Ê � œ � � Ê �w w w
sin 2y 2y sin 2x y ;œ � � Ê œw ��
sin 2y 2y sin 2xcos 2x 2x cos 2y
(a) the slope of the tangent line m y 2 the tangent line isœ œ œ œ Êk ¹w ��ˆ ‰
ˆ ‰1 1
1 14 24 2
ß
ß
sin 2y 2y sin 2xcos 2x 2x cos 2y
11
#
y 2 x y 2x� œ � Ê œ1 1
#ˆ ‰
4
(b) the normal line is y x y x� œ � � Ê œ � �1 1 1
# # #" "ˆ ‰
4 85
55. y 2 sin ( x y) y 2 [cos ( x y)] y y [1 2 cos ( x y)] 2 cos ( x y)œ � Ê œ � � Ê � � œ �1 1 1 1 1 1w w w† a b
y ;Ê œw ��# �2 cos ( x y)
1 cos ( x y)1 1
1
(a) the slope of the tangent line m y 2 the tangent line isœ œ œ Êk ¹w �� �(1 0)
(1 0)ß
ß
2 cos ( x y)1 2 cos ( x y)
1 1
11
y 0 2 (x 1) y 2 x 2� œ � Ê œ �1 1 1
(b) the normal line is y 0 (x 1) y� œ � � Ê œ � �" "# #1 1 1
x2
Section 3.6 Implicit Differentiation 157
56. x cos y sin y 0 x (2 cos y)( sin y)y 2x cos y y cos y 0 y 2x cos y sin y cos y# # # w # w w #� œ Ê � � � œ Ê � �c d 2x cos y y ;œ � Ê œ# w
�2x cos y
2x cos y sin y cos y
#
#
(a) the slope of the tangent line m y 0 the tangent line is yœ œ œ Ê œk ¹w�(0 )
(0 )ß
ß
1
1
2x cos y2x cos y sin y cos y
#
# 1
(b) the normal line is x 0œ
57. Solving x xy y 7 and y 0 x 7 x 7 7 and 7 are the points where the# # #� � œ œ Ê œ Ê œ „ Ê � ß ! ß !È È ÈŠ ‹ Š ‹ curve crosses the x-axis. Now x xy y 7 2x y xy 2yy 0 (x 2y)y 2x y# # w w w� � œ Ê � � � œ Ê � œ � �
y m the slope at 7 is m 2 and the slope at 7 isÊ œ � Ê œ � Ê � ß ! œ � œ � ß !w � �� �
�
�
2x y 2x yx 2y x 2y
2 77
Š ‹ Š ‹È ÈÈÈ
m 2. Since the slope is 2 in each case, the corresponding tangents must be parallel.œ � œ � �2 77
ÈÈ
58. x xy y 7 2x y x 2y 0 (x 2y) 2x y and ;# # � � �� � �� � œ Ê � � � œ Ê � œ � � Ê œ œdy dy dy dy 2x y x 2y
dx dx dx dx x 2y dy 2x ydx
(a) Solving 0 2x y 0 y 2x and substitution into the original equation givesdydx œ Ê � � œ Ê œ �
x x( 2x) ( 2x) 7 3x 7 x and y 2 when the tangents are parallel to the# # #� � � � œ Ê œ Ê œ „ œ …É É7 73 3
x-axis.
(b) Solving 0 x 2y 0 y and substitution gives x x 7 7dx x x x 3xdy 4œ Ê � œ Ê œ � � � � � œ Ê œ# # #
# #ˆ ‰ ˆ ‰ #
x 2 and y when the tangents are parallel to the y-axis.Ê œ „ œ …É É7 73 3
59. y y x 4y y 2yy 2x 2 2y y y 2x y ; the slope of the tangent line at% # # $ w w $ w w�œ � Ê œ � Ê � œ � Ê œa b x
y 2y$
is 1; the slope of the tangent line at Š ‹ ¹ Š ‹È È È3 3 34 y 2y 3 4
xß œ œ œ œ � ß# � #� #� �" "
$ "
#
"
#Œ �È È3 34 2ß
È
È È
34 4
3 6 38
34
is 3¹ Èxy 2y 4 2
2 3� ��$ "
#Œ �È34 2
1ß
œ œ œÈ3
428
È
60. y (2 x) x 2yy (2 x) y ( 1) 3x y ; the slope of the tangent line is# $ w # # w ��� œ Ê � � � œ Ê œ y 3x
2y(2 x)
# #
m 2 the tangent line is y 1 2(x 1) y 2x 1; the normal line isœ œ œ Ê � œ � Ê œ �¹y 3x2y(2 x)
4# #�� #
(1 1)ß
y 1 (x 1) y x� œ � � Ê œ � �" "# # #
3
61. y 4y x 9x 4y y 8yy 4x 18x y 4y 8y 4x 18x y% # % # $ w w $ w $ $ w � �� �� œ � Ê � œ � Ê � œ � Ê œ œa b 4x 18x 2x 9x
62. x y 9xy 0 3x 3y y 9xy 9y 0 y 3y 9x 9y 3x y$ $ # # w w w # # w � �� �� � œ Ê � � � œ Ê � œ � Ê œ œa b 9y 3x 3y x
3y 9x y 3x
# #
# #
(a) y and y ;k kw w(4 2) (2 4)ß ß
œ œ5 44 5
(b) y 0 0 3y x 0 y x 9x 0 x 54x 0w # $ ' $��
$
œ Ê œ Ê � œ Ê œ Ê � � œ Ê � œ3y xy 3x 3 3 3
x x x#
#
# # #Š ‹ Š ‹ x x 54 0 x 0 or x 54 3 2 there is a horizontal tangent at x 3 2 . To find theÊ � œ Ê œ œ œ Ê œ$ $ $ $ $a b È È È corresponding y-value, we will use part (c).
(c) 0 0 y 3x 0 y 3x ; y 3x x 3x 9x 3x 0dxdy 3y x
y 3xœ Ê œ Ê � œ Ê œ „ œ Ê � � œ#
#
��
# $$È È È ÈŠ ‹
x 6 3 x 0 x x 6 3 0 x 0 or x 6 3 x 0 or x 108 3 4 .Ê � œ Ê � œ Ê œ œ Ê œ œ œ$ $Î# $Î# $Î# $Î# $Î# $ $È È È ÈŠ ‹ È Since the equation x y 9xy 0 is symmetric in x and y, the graph is symmetric about the line y x.$ $� � œ œ
That is, if (a b) is a point on the folium, then so is (b a). Moreover, if y m, then y .ß ß œ œk kw w "(a b) (b a)ß ß m
Thus, if the folium has a horizontal tangent at (a b), it has a vertical tangent at (b a) so one might expectß ß
158 Chapter 3 Differentiation
that with a horizontal tangent at x 54 and a vertical tangent at x 3 4, the points of tangency areœ œ$ $È È 54 3 4 and 3 4 54 , respectively. One can check that these points do satisfy the equationŠ ‹ Š ‹È ÈÈ È$ $ $ $ß ß
x y 9xy 0.$ $� � œ
63. x 2tx 2t 4 2x 2x 2t 4t 0 (2x 2t) 2x 4t ;# # � �� �� � œ Ê � � � œ Ê � œ � Ê œ œdx dx dx dx 2x 4t x 2t
dt dt dt dt 2x 2t x t
2y 3t 4 6y 6t 0 ; thus ; t 2$ # # ��� œ Ê � œ Ê œ œ œ œ œ œdy dy dy dy/dt t(x t)
dt dt 6y y dx dx/dt y (x 2t)6t t# # � #
#
�
Š ‹ˆ ‰
ty
x 2tx t
x 2(2)x 2(2) 4 x 4x 4 0 (x 2) 0 x 2; t 2 2y 3(2) 4Ê � � œ Ê � � œ Ê � œ Ê œ œ Ê � œ# # # # $ #
2y 16 y 8 y 2; therefore 0Ê œ Ê œ Ê œ œ œ$ $ ��¹dy 2(2 2)
dx (2) (2 2(2))t 2œ
#
64. x 5 t 5 t t ; y(t 1) t y (t 1) tœ � Ê œ � � œ � � œ Ê � � œÉ È È Èˆ ‰ ˆ ‰dxdt dt4 t 5 t
dy" " " "# # #
�"Î# �"Î# �"Î#
�È ÈÉ
t 1 y ; thus Ê � œ � Ê œ œ œ œ œ †a b dy dy dydt dt t 1 dxt t t 2 t
y y t
"# # �
�
�"�#È È Èa b
È"
# # �
"�#
�"
�
È È ÈÈ
È ÈÉ
t t t 2 ty t
4 t 5 t
dydtdxdt
"�#
# �"
�
�"
y tt t
4 t 5 tÈÈ a bÈ ÈÉ
; t 4 x 5 4 3; t 4 y(3) 4 2œ œ Ê œ � œ œ Ê œ œ# "�# &�
"�
ˆ ‰È ÈÉy t t
tÉ È ÈÈ
therefore, ¹dydx 3
14
t 4œœ œ
2 2 2 4 4
4
Š ‹a bÈ ÈÉ"� &�
"�
65. x 2x t t 3x 2t 1 1 3x 2t 1 ; y t 1 2t y 4� œ � Ê � œ � Ê � œ � Ê œ � � œ$Î# # "Î# "Î# ��
dx dx dx dx 2t 1dt dt dt dt 1 3x
ˆ ‰ È È"Î#
t 1 y (t 1) 2 y 2t y 0 t 1 2 y 0Ê � � � � � œ Ê � � � � œdy dy dy y dydt dt dt y dt2 t 1
tÈ Èˆ ‰ ˆ ‰È È Š ‹" "# #
�"Î# �"Î#�È È
t 1 2 y ; thusÊ � � œ � Ê œ œŠ ‹È Èty dt dt
dy y dy2 t 1È È��
Š ‹ÈŠ ‹È
È ÈÈ È
�
�
y2 t 1
ty
È
È
�
� �
� � �
� � �
2 y
t 1
y y 4y t 1
2 y (t 1) 2t t 1
; t 0 x 2x 0 x 1 2x 0 x 0; t 0dy dy/dtdx dx/dtœ œ œ Ê � œ Ê � œ Ê œ œ
Œ �Š ‹� � �
� � �
�
�"Î#
y y 4y t 1
2 y (t 1) 2t t 1
2t 11 3x
È È
È È$Î# "Î#ˆ ‰
y 0 1 2(0) y 4 y 4; therefore 6Ê � � œ Ê œ œ œ �È È ¹dydx
t 0œ
Œ �Œ �
� � �
� � �
�
� "Î#
4 4 4(4) 0 1
2 4(0 1) 2(0) 0 1
2(0) 1
1 3(0)
È ÈÈ È
66. x sin t 2x t sin t x cos t 2 1 (sin t 2) 1 x cos t ;� œ Ê � � œ Ê � œ � Ê œdx dx dx dx 1 x cos tdt dt dt dt sin t 2
��
t sin t 2t y sin t t cos t 2 ; thus ; t x sin 2x� œ Ê � � œ œ œ Ê � œdy dydt dx
sin t t cos t 2� �ˆ ‰1 x cos tsin t 2�
�
1 1 1
x ; therefore 4Ê œ œ œ œ �1 1 1 1 1
1# �� � � �¹dy
dx 2sin cos 2 4 8
tœ1 – —1 cos
sin 2
�#
�
Š ‹1 1
1
67. (a) if f(x) x 3, then f (x) x and f (x) x so the claim f (x) x is falseœ � œ œ � œ33#
#Î$ w �"Î$ ww �%Î$ ww �"Î$"
(b) if f(x) x 7, then f (x) x and f (x) x is trueœ � œ œ9 310
&Î$ w #Î$ ww �"Î$#
(c) f (x) x f (x) x is trueww �"Î$ www �%Î$"œ Ê œ � 3
(d) if f (x) x 6, then f (x) x is truew #Î$ ww �"Î$#œ � œ3
68. 2x 3y 5 4x 6yy 0 y y and y ; also,# # w w w w� œ Ê � œ Ê œ � Ê œ � œ � œ � œ2x 2x 2 2x 23y 3y 3 3y 3k k¹ ¹(1 1) (1 1)
(1 1) (1 1)ß ß�
ß ß�
y x 2yy 3x y y and y . Therefore the# $ w # w w w# #œ Ê œ Ê œ Ê œ œ œ œ �3x 3x 3 3x 3
2y 2y 2y
# # #k k¹ ¹(1 1) (1 1)(1 1) (1 1)
ß ß�
ß ß�
tangents to the curves are perpendicular at (1 1) and (1 1) (i.e., the curves are orthogonal at these two points ofß ß �
intersection).
Section 3.6 Implicit Differentiation 159
69. x 2xy 3y 0 2x 2xy 2y 6yy 0 y (2x 6y) 2x 2y y the slope of the# # w w w w ��� � œ Ê � � � œ Ê � œ � � Ê œ Êx y
3y x
tangent line m y 1 the equation of the normal line at (1 1) is y 1 1(x 1)œ œ œ Ê ß � œ � �k ¹w ��(1 1)
(1 1)ß
ß
x y3y x
y x 2. To find where the normal line intersects the curve we substitute into its equation:Ê œ � �
x 2x(2 x) 3(2 x) 0 x 4x 2x 3 4 4x x 0 4x 16x 12 0# # # # # #� � � � œ Ê � � � � � œ Ê � � � œa b x 4x 3 0 (x 3)(x 1) 0 x 3 and y x 2 1. Therefore, the normal to the curveÊ � � œ Ê � � œ Ê œ œ � � œ �#
at (1 1) intersects the curve at the point (3 1). Note that it also intersects the curve at (1 1).ß ß � ß
70. xy 2x y 0 x y 2 0 ; the slope of the line 2x y 0 is 2. In order to be� � œ Ê � � � œ Ê œ � œ �dy dy dy y 2dx dx dx 1 x
��
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of�
the tangent is . Therefore, 2y 4 1 x x 3 2y. Substituting in the original equation," "# � #
�y 21 x œ Ê � œ � Ê œ � �
y( 3 2y) 2( 3 2y) y 0 y 4y 3 0 y 3 or y 1. If y 3, then x 3 and� � � � � � œ Ê � � œ Ê œ � œ � œ � œ#
y 3 2(x 3) y 2x 3. If y 1, then x 1 and y 1 2(x 1) y 2x 3.� œ � � Ê œ � � œ � œ � � œ � � Ê œ � �
71. y x . If a normal is drawn from (a 0) to (x y ) on the curve its slope satisfies 2y# "# �" " "
�œ Ê œ ß ß œ �dy y 0dx y x a
"
"
y 2y (x a) or a x . Since x 0 on the curve, we must have that a . By symmetry, theÊ œ � � œ � " " " " "" "# #
two points on the parabola are x x and x x . For the normal to be perpendicular,ˆ ‰ ˆ ‰È È" " " "ß ß �
1 1 x (a x ) x x x x and y .Š ‹Š ‹ ˆ ‰È Èx xx a a x (a x ) 4
x" "
" " "
"
#� � � # #" " " " " " "# " " "#
œ � Ê œ Ê œ � Ê œ � � Ê œ œ „
Therefore, and a .ˆ ‰" "#4 4
3ß „ œ
72. Ex. 6b.) y x has no derivative at x 0 because the slope of the graph becomes vertical at x 0.œ œ œ"Î#
Ex. 7a.) y 1 x has a derivative only on ( ) because the function is defined only on [ ] andœ � �"ß " �"ß "a b# "Î%
the slope of the tangent becomes vertical at both x 1 and x 1.œ � œ
73. xy x y 6 x 3y y x 2xy 0 3xy x y 2xy $ # # $ # # # $ � ��� œ Ê � � � œ Ê � œ � � Ê œŠ ‹ a bdy dy dy dy y 2xy
dx dx dx dx 3xy x
$
# #
; also, xy x y 6 x 3y y x y 2x 0 y 2xy 3xy xœ � � œ Ê � � � œ Ê � œ � �y 2xy3xy x dy dy dy
dx dx dx$
# #
��
$ # # $ # $ # #a b a bŠ ‹ ; thus appears to equal . The two different treatments view the graphs as functionsÊ œ �dx dx
dy y 2xy dy3xy x# #
$
��
"dydx
symmetric across the line y x, so their slopes are reciprocals of one another at the corresponding pointsœ
(a b) and (b a).ß ß
74. x y sin y 3x 2y (2 sin y)(cos y) (2y 2 sin y cos y) 3x $ # # # # ��� œ Ê � œ Ê � œ � Ê œdy dy dy dy
dx dx dx dx 2y 2 sin y cos y3x#
; also, x y sin y 3x 2y 2 sin y cos y ; thus œ � œ Ê � œ Ê œ3x dx dx dx2 sin y cos y 2y dy dy 3x dy
2 sin y cos y 2y#
#�$ # # # �
appears to equal . The two different treatments view the graphs as functions symmetric across the line"dydx
y x so their slopes are reciprocals of one another at the corresponding points (a b) and (b a).œ ß ß
160 Chapter 3 Differentiation
75. x 4y 1:% #� œ
(a) y y 1 x# � "#
%œ Ê œ „ �1 x4
% È 1 x 4x ;Ê œ „ � � œdy
dx 4x
1 x" „% $�"Î#
�a b a b $
% "Î#a b differentiating implicitly, we find, 4x 8y 0$ � œdy
dx
.Ê œ œ œdydx 8y
4x 4x x
8 1 x 1 x� � „
„ � �
$ $ $
"#
% % "Î#Š ‹È a b
(b)
76. (x 2) y 4:� � œ# #
(a) y 4 (x 2)œ „ � �È #
4 (x 2) ( 2(x 2))Ê œ „ � � � �dydx
"#
# �"Î#a b ; differentiating implicitly,œ „ �
� �
(x 2)
4 (x 2)c d# "Î#
2(x 2) 2y 0 � � œ Ê œdy dy 2(x 2)dx dx 2y
� �
.œ œ œ� � � � „ �
„ � � � �#
(x 2) (x 2) (x 2)y 4 (x 2) 4 (x )c d c d# #"Î# "Î#
(b)
77-84. Example CAS commands: :Maple q1 := x^3-x*y+y^3 = 7; pt := [x=2,y=1]; p1 := implicitplot( q1, x=-3..3, y=-3..3 ): p1; eval( q1, pt ); q2 := implicitdiff( q1, y, x ); m := eval( q2, pt ); tan_line := y = 1 + m*(x-2); p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ): p3 := pointplot( eval([x,y],pt), color=blue ): display( [p1,p2,p3], ="Section 3.6 #77(c)" ); : (functions and x0 may vary):Mathematica Note use of double equal sign (logic statement) in definition of eqn and tanline. <<Graphics`ImplicitPlot`
Section 3.7 Derivatives of Inverse Functions and Logarithms 161
87. y log 8t 3 ln t œ œ œ œ � Ê œ2ln 2a b ln 8 ln t
ln ln dt t3 ln 2 (ln 2)(ln t) dy�
# #� "ˆ ‰ln 2
Section 3.7 Derivatives of Inverse Functions and Logarithms 167
88. y t sin t sin t t cos tœ œ œ œ Ê œ �t ln e
ln 3 ln 3 ln 3 dtt ln 3 t(sin t)(ln 3) dyŠ ‹ˆ ‰ ˆ ‰ln 3 sin t
sin t
89. y (x 1) ln y ln (x 1) x ln (x 1) ln (x 1) x y (x 1) ln (x 1)œ � Ê œ � œ � Ê œ � � Ê œ � � �x x xyy (x 1) x 1
xw
†
"� �
w � ‘90. y x ln y ln x (x 1) ln x ln x (x 1) ln x 1œ Ê œ œ � Ê œ � � œ � �Ð �"Ñ Ð �"Ñx x y
y x x
w ˆ ‰" "
y x 1 ln xÊ œ � �w "Ð � Ñx 1 ˆ ‰x
91. y t t t ln y ln t ln t (ln t)œ œ œ Ê œ œ Ê œ � œ �ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰È t t t t"Î# Î## # # # #
" " " "Î# t t ln ty dt t
dy
tÊ œ �dydt
ln tˆ ‰ ˆ ‰È t
# #"
92. y t t ln y ln t t (ln t) t (ln t) tœ œ Ê œ œ Ê œ � œÈ ˆ ‰ ˆ ‰t t t"Î# "Î# ˆ ‰ ˆ ‰ ˆ ‰"Î# �"Î# "Î#" " " �#y dt t
dy ln t 22 tÈ
tÊ œdydt
ln t 22 t
Š ‹�È Èt
93. y (sin x) ln y ln (sin x) x ln (sin x) ln (sin x) x y (sin x) ln (sin x) x cot xœ Ê œ œ Ê œ � Ê œ �x x xyy sin x
cos xw ˆ ‰ c dw
94. y x ln y ln x (sin x)(ln x) (cos x)(ln x) (sin x)œ Ê œ œ Ê œ � œsin x sin x y sin x x (ln x)(cos x)y x x
w ˆ ‰" �
y xÊ œw �sin x ’ “sin x x(ln x)(cos x)x
95. y x , x 0 ln y (ln x) 2(ln x) y xœ � Ê œ Ê œ Ê œln x ln x# w"yy x x
ln xw #ˆ ‰ a b Š ‹96. y (ln x) ln y (ln x) ln (ln x) ln (ln x) (ln x) (ln x)œ Ê œ Ê œ � œ �ln x y ln (ln x)
y x ln x dx x xdw ˆ ‰ ˆ ‰" " "
y (ln x)Ê œw � "Š ‹ln (ln x)x
ln x
97. (g f)(x) x g(f(x)) x g (f(x))f (x) 1‰ œ Ê œ Ê œw w
98. lim 1 lim e for any x 0.1n nÄ _ Ä _
ˆ ‰ ” •Š ‹� œ œ ��xn
n 1n/x
n/xx
xa b
a b
99. y A sin ln x B cos ln x y A cos ln x B sin ln x A cos ln x B sin ln xœ � Ê œ † � † œ � †a b a b a b a b a ba b a bw 1 1 1x x x
y A cos ln x B sin ln x A sin ln x B cos ln xÊ œ � † � � † � † †ww �a b a b a ba b a b ˆ ‰1 1 1 1x x x x2
A cos ln x sin ln x B sin ln x cos ln xœ � � � � †a ba b a ba b a b a b a b 1x2
x y x y y A cos ln x sin ln x B sin ln x cos ln x A cos ln x B sin ln xÊ � � œ � � � � � �2 ww w a b a ba b a b a b a ba b a b a b a b A sin ln x B cos ln x� � œ !a ba b a b100. Suppose n 1. Then ln x 1 and so the base case is established. Now if the statement holds for n kœ œ œ � † œd 1 0!
dx x x0a b 1
we have that, for n k 1, the following holds:œ �
ln x ln x ln x 1 1 k 1 ! k xd d d d ddx dx dx xdx dx x x
k 1 k 1 ! k 1 1 k! 1 n 1 !k 1n k 1 k
n nk 1 k k k 1
k n 1a b a b a b a b a b a b a bŠ ‹ Š ‹œ œ œ � œ � � † � œ œ�
� �
�� � � � � �� �a b a b a b a b
Thus by mathematical induction the result is established for all n 1.
101-108. Example CAS commands: :Maple with( plots );#101 f := x -> sqrt(3*x-2);
168 Chapter 3 Differentiation
domain := 2/3 .. 4; x0 := 3; Df := D(f); # (a) plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#53(a) (Section 7.1)" ); q1 := solve( y=f(x), x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" ); (assigned function and values for a, b, and x0 may vary)Mathematica: If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. <<Miscellaneous `RealOnly` Clear[x, y] {a,b} = { 2, 1}; x0 = 1/2 ;�