Thomas Calculus 13th [Solutions]

122 Chapter 3 Derivatives
2.
2
2
F( x) ( x 1) 1 and F( x h) ( x h 1) 1
2 2 2
( x 2xh h 2x 2h 1 1) ( x 2x
1 1)
lim
h 0
h
F ( 1) 4, F (0) 2, F (2) 2
2 2
[( x h 1) 1] [( x 1) 1]
F ( x) lim
h 0
h
2
lim 2 xh h 2 h lim (2x h 2) 2( x 1);
h 0
h
h 0
3. Step 1: g( t)
1 and g( t h)
1
2
2
t
( t h)
1 1
( )
2 2
t
2
( t h)
2
2 2
2 2 2
g( t h) g( t) t h t ( t h)
t t ( t 2 th h )
2
Step 2:
2th h h( 2 t h)
h
h
h 2 2
( t h ) t h
2 2
2 2
( t h)
t h ( t h)
t h
Step 3: g ( t) lim 2 t h 2t
2 ; g ( 1) 2, g (2) 1
2 2 2 2 3
, g 3 2
h 0 ( t h)
t t t t
4 3 3
2t h
( t h)
t
2 2
1 ( z h) 1 z
4. ( ) 1 z
1 ( z h)
2( z h) 2z
(1 ) (1 )( )
k z and k( z h)
k ( z) lim
2z
2( z h)
h 0
h
lim z h z z z h
h 0
2( z h)
zh
lim h lim 1 1 ; k ( 1) 1 , k (1) 1 , k 2 1
h 0
2( z h) zh
h 0
2( z h)
z 2
2z
2 2 4
2 2
lim z z zh z h z zh
h 0
2( z h)
zh
5. Step 1: p( ) 3 and p( h) 3( h)
p( h) p( ) 3( h) 3 3 3h
3 3 3h
3 (3 3 h) 3
Step 2:
h h h 3 3h 3 h 3 3h
3
h
3h
3
3 3h
3 3 3h
3
Step 3: p ( ) lim 3
3 3 ; p (1) 3 , p (3) 1 2 3
h 0 3 3h
3 3 3 2 3 2 3 2 , p
3 2 2
6. r( s) 2s 1 and r( s h) 2( s h) 1
lim
h 0
2s h 1 2s 1 2s 2h 1 2s
1
lim 2h
h 0 h 2s 2h 1 2s
1
h 2s 2h 1 2s
1
r (0) 1, r (1) 1 , r 1 1
3 2 2
lim 2
h 0 2 s 2 h 1 2 s 1
2 2 1 2 1
r ( s) lim s h s
h 0
h
(2s 2h 1) (2s
1)
lim
h 0 h 2s 2h 1 2s
1
2 2 1 ;
2s 1 2s 1 2 2s 1 2s
1
7.
3
y f ( x) 2x and
2 2 3
lim 6 x h 6 xh 2 h
h 0
h
3
f ( x h) 2( x h)
2 2
(6 6 2 )
lim h x xh h
h 0
h
3 3
3 2 2 3 3
dy 2( x h) 2x
2( 3 3 ) 2
lim
lim x x h xh h x
dx
h 0
h
h 0
h
2 2 2
lim (6x 6xh 2 h ) 6x
h 0
8.
3 2
r s 2s
3
dr
ds
3 2 3 2
lim (( s h ) 2( s h ) 3) ( s 2 s 3)
h 0
h
2 2
3 2 2 3 2 2 3 2
lim s 3s h 3sh h 2s 4sh 2h 3 s 2s
3
h 0
h
2 2 3 2
3 3 4 2
(3 3 4 2 ) 2 2 2
lim s h sh h sh h h s sh h s h
lim lim (3s 3sh h 4s 2 h) 3s 2s
h 0
h
h 0
h
h 0
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