Thomas Calculus 13th [Solutions]

Section 3.2 The Derivative as a Function 123
9.
t h t
2( t h) 1 2t
1
s r( t ) t and r( t h) t h ds lim
lim
2t
1
2( t h) 1 dt
h 0
h
h 0
lim ( t h )(2 t 1) t (2 t 2 h 1)
2 2
lim 2t t 2ht h 2t 2ht t lim h
h 0
(2t 2h 1)(2t 1) h
h 0
(2t 2h 1)(2t 1) h
h 0
(2t 2h 1)(2t 1) h
1
1
(2t 1)(2t 1)
2
(2t
1)
( t h)(2t 1) t(2t 2h
1)
(2t 2h 1)(2t
1)
h
lim 1
h 0
(2 t 2 h 1)(2 t 1)
10.
dv
dt
lim
h 0
1
t
1
t h t
( t h) ( )
h
1 1
t h t
h
lim lim
h 0
h
h 0
h( t h) t t ( t h)
( t h)
t
h
2 2
lim ht h t h
h 0
h( t h)
t
2
lim t ht 1
h 0
( t h)
t
2
t
t
1 1 1
t
2 2
11.
3/2 3/2 1/2 1/2
dp ( ) ( )( )
lim
q h q lim
q h q h q q
dq
h 0
h
h 0
h
1/2 1/2 1/2
[( ) ] ( )
lim q q h q h q h
h
h
h 0
lim
1/2 1/2 1/2 1/2 1/2
q[( q h) q ][( q h) q ] ( q h) 1/2 lim q[( q h) q]
( q h) 1/2 lim q ( q h)
1/2 q
q 1/2 3
q 1/2
1/2 1/2 1/2 1/2 1/2 1/2
[( ) ] [( ) ] ( )
2 2
h 0 h q h q h 0 h q h q h 0 q h q
12.
dz
dw
1 1 2 2 2 2
2 2
2 2
w w h w w h
( w h) 1 w 1 w 1 ( w h) 1
1 ( ) 1 1 ( ) 1
lim lim lim
2 2
h 0
h
h 0 0
2 2 2 2
h ( w h) 1 w 1 h h ( w h) 1 w 1 w 1 ( w h) 1
2 2 2
w 1 ( w 2wh h 1)
lim
lim
2w h
h 0
2 2 2 2
0
2 2 2
h ( w h) 1 w 1 w 1 ( w h) 1 h h ( w h) 1 w 1 w 1
w
2
( w h) 1 ( w 1)
2 3/2
13. f ( x)
x 9 and 9 f ( x h) f ( x)
( x h)
f ( x h) ( x h)
x
( x h)
h
3 2 2 3 2 2 2 2
2
x 2x h xh 9x x x h 9x 9h x h xh 9h h( x xh 9) x xh 9
x( x h) h x( x h) h x( x h) h x( x h) ;
m f ( 3) 0
9 9
x
( x h)
x
h
2 2
x( x h) 9 x x ( x h) 9( x h)
x( x h)
h
2 2
f ( x ) lim x xh 9 x 9 9
( )
1 ;
2 2
h 0
x x h x x
14. k( x ) 1 and 1
( ) ( )
k( x h)
k ( x) lim k x h k x
2 x
2 ( x h)
h 0
h
lim h lim 1
1 ; k (2) 1
h 0
h(2 x)(2 x h)
h 0
(2 x)(2 x h)
2
(2 x)
16
1 1
2 2
lim x h x
h
0
h
(2 x) (2 x h)
lim
h 0
h(2 x)(2 x h)
15.
16.
3 2 3 2
3 2 2 3 2 2 3 2
ds lim [( t h ) ( t h ) ] ( t t ) ( t 3t h 3 th h ) ( t 2 th h ) t t
lim
dt
h 0
h
h 0
h
2 2
(3 3 2 )
lim h t th h t h
2 2
2
lim (3t 3th h 2 t h)
3t
2 t;
m
h 0
h
h 0
ds
dt t
2 2 3 2
lim 3t h 3th h 2th h
h 0
h
( x h) 3 x 3
( x h 3)(1 x) ( x 3)(1 x h)
dy 1 ( x h) 1 x (1 x h)(1 x) 2 2
x h 3 x xh 3x x 3 x 3x xh 3h 4h
lim lim lim lim
dx
h 0
h
h 0
h
h 0
h(1 x h)(1 x) h 0
h(1 x h)(1 x)
lim 4 4
(1 )(1 ) ; 2
h 0
x h x (1 x)
dy
dx x
2
4 4
2
(3) 9
1
5
8 8
( x h) 2 x 2
17. f ( x ) 8 and 8 f ( x h) f ( x)
8 x 2 x h 2 x 2 x h 2
f ( x h)
x 2
( x h) 2 h
h h x h 2 x 2 x 2 x h 2
8[( x 2) ( x h 2)]
8h
f ( x) lim 8
h x h 2 x 2 x 2 x h 2 h x h 2 x 2 x 2 x h 2
h 0 x h 2 x 2 x 2 x h 2
8
4 ; m f (6) 4 1 the equation of the tangent line at (6,4) is
x 2 x 2 x 2 x 2 ( x 2) x 2 4 4 2
y 4 1 ( x 6) y 1 x 3 4 y 1 x 7.
2
2
2
Copyright
2014 Pearson Education, Inc.