Calculus 12 – Ch. 2 Derivatives
Lesson 6: The Chain Rule Cont’d
1) Find f !(x) for each of the following:
2
a) f (x) = ( x 2 + 3) ( x 3 ! 2x )
1
3
3
2
2
f !(x) = 2 ( x 2 + 3) ( 2x ) ( x 3 " 2x ) + ( x 2 + 3) ( 3) ( x 3 " 2x ) ( 3x 2 " 2 )
3
2
2
= 4x ( x 2 + 3) ( x 3 " 2x ) + 3 ( x 2 + 3) ( x 3 " 2x ) ( 3x 2 " 2 )
2
= ( x 2 + 3) ( x 3 " 2x ) #$4x ( x 3 " 2x ) + 3 ( x 2 + 3) ( 3x 2 " 2 )%&
2
= ( x 2 + 3) ( x 3 " 2x ) #$4x 4 " 8x 2 + 3 ( 3x 4 " 2x 2 + 9x 2 " 6 )%&
2
= ( x 2 + 3) ( x 3 " 2x ) #$4x 4 " 8x 2 + 9x 4 " 6x 2 + 27x 2 "18%&
2
= ( x 2 + 3) ( x 3 " 2x ) (13x 4 +13x 2 "18 )
6
" 2x !1 %
b) f (x) = $
'
# x+2 &
# ( x + 2 ) ( 2 ) " ( 2x "1) (1) &
%
(
2
%
(
x
+
2
(
)
$
'
5
# 2x "1 & # 2x + 4 " 2x +1 &
(
= 6%
( %
(
$ x + 2 ' %$ ( x + 2 )2
'
5
# 2x "1 & # 5 &
(
= 6%
( %
$ x + 2 ' %$ ( x + 2 )2 ('
5
# 2x "1 &
f !(x) = 6 %
(
$ x+2 '
=
30 ( 2x "1)
( x + 2)
7
5
3
c) f (x) = ( x 2 !1) ( 2x 3 + 5x )
2
2
2
3
1
f !(x) = 3 ( x 2 "1) ( 2x ) ( 2x 3 + 5x ) + ( x 2 "1) ( 2 ) ( 2x 3 + 5x ) ( 6x 2 + 5 )
2
= ( x 2 "1) ( 2x 3 + 5x ) #$6x ( 2x 3 + 5x ) + 2 ( x 2 "1) ( 6x 2 + 5 )%&
2
= ( x 2 "1) ( 2x 3 + 5x ) #$12x 4 + 30x 2 + 2 ( 6x 4 " x 2 " 5 )%&
2
= ( x 2 "1) ( 2x 3 + 5x ) ( 24x 4 + 28x 2 "10 )
2
= 2 ( x 2 "1) ( 2x 3 + 5x ) (12x 4 +14x 2 " 5 )
d) f (x) =
3
x2 + 5
x!3
2
1
"
#1& 2
2
3
( x " 3) % ( ( x + 5) (2x ) " ( x + 5) 3 (1)
$ 3'
f !(x) =
2
( x " 3)
2
"
#1& 2
3 ) x " 3 2x " 3 x 2 + 5 +
x
+
5
% ((
)( ) (
)
),
*(
$ 3'
=
2
( x " 3)
=
2x 2 " 6x " 3x 2 "15
2
3 ( x 2 + 5 ) 3 ( x " 3)
=
2
"x 2 " 6x "15
2
3 ( x 2 + 5 ) 3 ( x " 3)
2
e) f (x) = 2x 2 +1 ( 3x 2 ! 5 )
4
1
1
"
4
3
1
2
2
2
2
f !(x) = ( 2x +1) ( 4x ) ( 3x " 5 ) + ( 2x +1) 2 ( 4 ) ( 3x 2 " 5 ) ( 6x )
2
= ( 2x ) ( 2x 2 +1)
"
1
2
= ( 2x ) ( 2x 2 +1)
"
1
2
=
=
(2x ) ( 3x 2 " 5)
3
4
( 3x
2
3
" 5 ) #$3x 2 " 5 +12 ( 2x 2 +1)%&
#$3x 2 " 5 + 24x 2 +12%&
2x 2 +1
(2x ) ( 3x 2 " 5)
3
1
( 3x 2 " 5) + (24x ) (2x 2 +1) 2 ( 3x 2 " 5)
(27x
2x 2 +1
2
+ 7)
3
2) Find the equation of the tangent line to the curve
(2x +1)
y=
2
x 2 +1
1
1
"
2 #1&
2
2
2 2 2x +1 2 " 2x +1
x
+1
x
+1
( ) ( ) ( ) ( ) ( ) %$ 2 (' ( ) 2 (2x )
y! =
(
x 2 +1
1
=
=
=
2
4 ( x 2 +1) 2 ( 2x +1) " ( 2x +1) ( x 2 +1)
(x
( x 2 +1)
"
1
2
"
1
2
( x)
+1)
2
(2x +1) )*4 ( x 2 +1) " (2x +1) ( x )+,
(x
2
+1)
(2x +1) )*4x 2 + 4 " 2x 2 " x+,
(x
=
)
2
3
2
+1) 2
(2x +1) (2x 2 " x + 4 )
3
( x 2 +1) 2
Slope at (0,1) :
f !(0) =
2
(2 (0) +1) (2 (0) " (0) + 4 )
(( 0 )
=
2
)
+1
3
2
(1) ( 4 )
1
=4
Equation of the tangent at (0,1) :
y !1
=4
x!0
y !1 = 4x
0 = 4x ! y +1
3) Find the derivatives of the following:
a) F(x) = [ h(x)]
2
3
F !(x) = 3 [ h(x)] ( h!(x))
b) G(x) = h ( x 3 )
G!(x) = h! ( x 3 ) ( 3x 2 )
at the point (0,1) .