Thomas Calculus 13th [Solutions]

Section 3.6 The Chain Rule 155
44.
g 1
( t ) 1 sin 3t 3 2t
(1 sin 3 t)( 2) (3 2 t)(3cos3 t)
g ( t)
2 2sin 3t 9cos3t 6t cos3t
3 2t
1 sin 3t
(1 sin 3 t)
2
(1 sin 3 t)
45.
r
2
sin( ) cos(2 ) dr 2
sin( )( sin 2 ) d
2 2
(2 ) cos(2 ) (cos( )) d ( )
d
d
d
2 2
2 2
sin( )( sin 2 )(2) (cos 2 )(cos ( ))(2 ) 2sin( )sin(2 ) 2 cos(2 ) cos( )
46. r sec tan 1
dr
sec sec 2 1
d 2
1 2
sec sec 1 1 tan 1 sec tan sec
2
2
1 tan 1 (sec tan ) 1
2
1 2 1
tan tan sec
2
2
47.
q
sin
t
t 1
dq
dt
cos t d t
t 1 dt t 1
cos
t
t
d
dt
2
t 1(1) t. t 1
1 t 1
cos
1
t 1
t
1
t
t
2 t 1
1
cos
t
t
2( t 1) t
1 2( t 1)
3/2
t 2
2( t 1)
3/2
cos
t
t
1
48. q cot( sin t )
t
dq
dt
2
csc sin t d sin t
t dt t
2
csc sin t
t cost sint
t
t 2
49.
2 2 2 2 2 2 2
dy 2
y cos e sin e d e sin e e d ( ) 2 e sin e
d d d
50.
51.
y 3 e 2 dy 2 2 3 2 3 2
cos5 (3 )( cos5 ) ( cos5 ) d ( 2 ) 5(sin 5 )( )
d
e e d
e
2 2
e (3cos5 2 cos5 5 sin 5 )
y 2
dy
sin ( t 2)
dt
2sin( t 2) d
dt
sin( t 2) 2sin( t 2) cos( t 2) d ( t 2)
dt
2 sin( t 2) cos( t 2)
52.
2 dy
y sec t (2sec t ) d (sec t) (2sec t)(sec t tan t)
dt
dt
d
2
( t) 2 sec t tan t
dt
53.
y
(1 cos 2 t)
4
dy
dt
5
4(1 cos 2 t) d (1 cos 2 t)
dt
5
4(1 cos 2 t) ( sin 2 t)
d (2 t)
8sin 2t
dt
5
(1 cos 2 t)
54.
y
1 cot t
2
2
dy
dt
2 1 cot
t
2
3
d 1 cot t 2 1 cot t
dt 2 2
3
2
csc t d t
2 dt 2
csc
2
1 cot
t
2
t
2
3
55.
10 dy
9 2
y ( t tan t) 10 ( t tan t) t sec t 1 tan t
dt
10 9 2 9 10
10t tan t sec t 10t tan t
9 9 2
10t tan t ( t sec t tan t)
56.
3/4 4/3
y ( t sin t)
1/3
(sin t) (4t cos t 3cos t)
2
3t
1 4/3
t (sin t)
dy
dt
t 1 4 1/3 2 4/3
(sin ) cos (sin )
3 t t t t 1/3 4/3
4(sin t) cos t (sin t)
3t
2
t
57.
2 2 2
cos ( t 1) dy cos ( t 1) cos t( 1)
y e e 2cos( t 1) ( sin( t 1)) 2 sin( t 1) cos( t 1) e
dt
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