Thomas Calculus 13th [Solutions]

194 Chapter 3 Derivatives
x x x x
32. y xe dy ( xe e ) dx (1 x)
e dx
33. y ln(1 x 2 ) dy 2 x dx
2
1 x
34. y ln x 1 ln( x 1) 1 1 1 1
3
2 ln( x 1) dy x
1 2 1
2
x 1 x x
dx dx 2( x 1)
35.
2 2
1 x 1 x
y tan e dy e 2x dx 2xe
dx
2
2
2x
2
x
1 e
1 e
x
2
1 1
36. y cot 1 cos (2 x ) Note: d 1 1d
cot ,
2
2
x
d
1
so that
1
2
1
2
x
3
x
3
2 1
x
4
1
4
x
4
x
4
d 1
cot 1
2x
dx x 1 x 1
Note:
d
d
1
cos 1 d , so that
2
1
d 1
(cos (2 )) 1 2 2 .
dx
1 4x
1 4x
2x
2
x 1 1 4x
x Thus dy
2 2
4 2
dx
37.
1 x
sec ( ) 1 x
y e dy ( e ) dx 1 dx e dx
e e e
x x 2 2 2x
( ) 1 1
1
1
e
x
x
38.
1 2 1 2 tan
1
x
2
1
tan x 1 tan x 1 2 1/2
y e dy e 1 1 ( x 1) 2x dx xe dx
2 2
2
2 2
1 x 1
( x 2) x 1
39.
40.
41.
42.
2
f ( x) x 2 x,
x 0 1, dx 0.1 f ( x) 2x
2
(a) f f ( x0 dx) f ( x 0 ) f (1.1) f (1) 3.41 3 0.41
(b) df f ( x0
) dx [2(1) 2](0.1) 0.4
(c) | f df | |0.41 0.4| 0.01
2
f ( x) 2x 4x 3, x 0 1, dx 0.1 f ( x) 4x
4
(a) f f ( x0 dx) f ( x 0 ) f ( .9) f ( 1) .02
(b) df f ( x0
) dx [4( 1) 4](.1) 0
(c) | f df | |.02 0| .02
3
2
f ( x) x x,
x0
1, dx 0.1 f ( x) 3x
1
(a) f f ( x0 dx) f ( x 0 ) f (1.1) f (1) .231
2
(b) df f ( x0
) dx [3(1) 1](.1) .2
(c) | f df | |.231 .2| .031
4
3
f ( x) x , x0
1, dx 0.1 f ( x) 4x
(a) f f ( x0 dx) f ( x 0 ) f (1.1) f (1) .4641
3
(b) df f ( x0
) dx 4(1) (.1) .4
(c) | f df | |.4641 .4| .0641
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