Thomas Calculus 13th [Solutions]

132 Chapter 3 Derivatives
6.
y
3 2
x x
3 2
e
x
dy
dx
2
x x e
x
2
d y
x
2
2x 1 e NOTE: dy e x e x from Example 8(b).
dx
dx
7.
2 1
w 3z z
dw
dz
3 2
6z z 6 1
3 2
z z
2
d w
2
dz
4 3
18z 2z 18 2
4 3
z z
8.
1 2
s 2t 4t
ds
dt
2 3
2t
8t
2
2 8 d s
2
t
3
t
2
dt
3 4
4t 24t 4 24
3 4
t t
9.
2 2
y 6x 10x 5x
dy
dx
3
12x 10 10x 12x
10 10
3
x
2
d y
dx
2
4
12 0 30x 12 30
4
x
10.
y 4 2x x
3
dy
dx
2 3
4
3
x
x 2
4
2
d y
dx
2
0 12
5
12
x
x
5
11.
1 2 5 1
r s s
3 2
dr
ds
2 3 5 2
s s
3 2
2 5 d r
3
3s 2
2s 2
ds
2
4 3
2s 5s 2 5
4 3
s s
12.
r
12
1
4
3 4
24 48 20
3 5 6
dr
d
2 4 5
12 12 4
12 12 4 d r
2 4 5 2
d
2
3 5 6
24 48 20
13. (a)
(b)
14. (a)
(b)
15. (a)
(b)
2 3
2
y (3 x ) ( x x 1) (3 ) d 3
3
y x ( x x 1) ( 1) d 2
x x (3 x )
dx
dx
2 2 3
4 2
(3 x ) (3x 1) ( x x 1) ( 2 x) 5x 12x 2x
3
5 3 2
4 2
y x 4x x 3x
3 y 5x
12x
2x
3
y (2x 2
3)(5x 4 x) y (2x 3)(10 x 4)
2
(5x 4 x)(2) 2
30x 14x
12
y (2x 2
3)(5x 4 x) 3
10x 2
7x 12x y
2
30x 14x
12
2
y ( x 1) ( 5 1
2
x ) ( 1) d
x
y x 1
2
( x 5 ) ( x 5 1) d ( x 1)
dx x
x dx
2
2 1
2 2 2
2
( x 1) (1 x ) ( x 5 x ) (2 x) ( x 1 1 x ) (2x 10x 2) 3x
10x
2 1
2
x
3 2
5 2 5 1
2
y x x x y 3x 10x
2 1
x
2
x
16.
2 3/4 3
y (1 x )( x x )
2 1/4 4 3/4 3 7/4
(a) y (1 x ) 3 x 3 x ( x x )(2 x)
3 3 11 x 1
4 1/4 4 4
2
4x x x
3/4 3 11/4 1
(b) y x x x x y 3 3 11 7/4
x 1
1/4
4x x
4 4
2
x
17. y 2x
5
3x
2 ; use the quotient rule: u 2x 5 and v 3x 2 u 2 and 3
6x
4 6x
15 19
2
2
(3x
2) (3x
2)
vu uv
v
(3x
2)(2) (2x
5)(3)
v y
2
2
18. y 4 3x
; use the quotient rule: 4 3
2
u x and v 3x x u 3 and v 6x 1 y vu uv
2
2
3x
x
v
2
(3 x x)( 3) (4 3 x)(6x
1) 2 2
2
9x 3x 18x 21x
4 9x
24x
4
2 2
2 2
2 2
(3 x x)
(3 x x)
(3 x x)
(3x
2)
19.
2 4
g ( x ) x
x 0.5
; use the quotient rule:
2
( x 0.5)(2 x) ( x 4)(1) 2 2
2x
x x 4
2
2
( x 0.5)
( x 0.5)
2
u x 4 and v x 0.5 u 2x and 1
2
x x
( x 0.5)
4
2
v g x
2
( ) vu uv
v
Copyright
2014 Pearson Education, Inc.