2/9
The Chain Rule and Inverse Functions (3.6)
Review inverse functions and Chain Rule. Recall that for inverse functions
d
g( f ( x )) = f !( x ) " g!( f ( x )) .
f f !1 ( x ) = f !1 ( f ( x )) = x . And, the Chain Rule states
dx
Derivative of x .
We have already used the Power Rule to find the derivative of f (x) = x ,
1
1
f !(x) = x "1 2 =
. We can also use the Chain Rule to find its derivative.
2
2 x
d
d
Using the Chain Rule, we can write
( x ) 2 = 2 x ! ( x ) , where
dx
dx
d
d
d
2
2
g ( x ) = x and f ( x ) = x . But
( x ) = (x) = 1. Therefore, 2 x ! ( x ) = 1.
dx
dx
dx
d
1
d
( x) =
Solving for
.
( x ) , we get
dx
dx
2 x
Derivative of ln x .
d ln x
d
Since e ln x = x , we must have
e
= (x) = 1. Using the Chain Rule,
dx
dx
d ln x
d
d
e
= (ln x) ! e ln x = (ln x) ! x , where g ( x ) = e x and f ( x ) = ln x . Therefore,
dx
dx
dx
d
d
1
(ln x) ! x = 1 and
(ln x) = .
dx
dx
x
d
SOLVE 3.6.9. j ( x ) = ln e ax + b . Using the Chain Rule,
ln e ax + b =
dx
ax
d ax
1
1
ae
, where g ( x ) = ln x and f ( x ) = eax + b .
e + b ! ax
= a e ax ! ax
= ax
dx
e +b
e +b e +b
1
2x
FIND the tangent line to y = ln x 2 ! 3 at (2,0) . y ! = 2x " 2
, where
= 2
x #3 x #3
2" 2
4
g ( x ) = ln x and f ( x ) = x 2 ! 3 .
So y !(2) = 2
=
= 4 , and the tangent line at
2 #3 4#3
(2,0) is y = 4 ( x ! 2 ) + 0 = 4x ! 8 .
(
)
(
[
]
]
[ ]
[ ]
(
(
[
)
)
( (
)
))
( )
(
Derivative of a x .
Since ln(a x ) = x ln a , we must have
)
[
]
d
d
ln(a x ) = (x ln a) = ln a . Using the
dx
dx
d
d
1
! ln(a x ) # = (a x ) % , where g ( x ) = ln x and f ( x ) = a x . Therefore,
"
$
dx
dx
ax
1 d x
d x
! (a ) = ln a and
(a ) = (ln a)(a x ) .
x dx
dx
a
d sin x
d sin x
FIND
.
2
2
= cos x ! (ln2) ! 2sin x , where g ( x ) = 2 x and
dx
dx
f ( x ) = sin x .
Chain Rule,
(
)
(
)
Derivative of sin!1 x .
Since sin(sin!1 x) = x , we must have
[
]
[
]
d
d
sin(sin!1 x) = (x) = 1. Using the
dx
dx
d
d
sin(sin!1 x) = (sin!1 x) " cos(sin!1 x) , where g ( x ) = sin x and
dx
dx
d
1
d
f ( x ) = sin !1 x . So
(sin!1 x) =
. But
(sin!1 x) " cos(sin!1 x) = 1 and
dx
dx
cos(sin!1 x)
Chain Rule,
what is cos(sin!1 x) equal to? Recall that sin!1 x represents an angle, say ! , whose sine
is equal to the number x. DRAW. Using the Pythagorean Theorem, we can calculate that
the remaining side is equal to 1! x 2 . Therefore, the cos(sin!1 x) = cos " = 1! x 2 .
d
1
(sin!1 x) =
Finally,
.
dx
1! x 2
d
d
SOLVE 3.6.24.
cos(sin!1 x) = (sin!1 x) " [!sin(sin!1 x)] =
dx
dx
1
!1
" [!sin(sin x)], where g ( x ) = cos x and f ( x ) = sin !1 x . But sin(sin!1 x) = x ,
1! x 2
d
x
cos(sin!1 x) = !
so
.
2
dx
1! x
[
[
]
]
SOLVE 3.6.54. Given that f ( x ) = x 3.
a. Find f !(2) . f !( x ) = 3x 2 , f !(2) = 3" 2 2 = 12 .
b. Find f !1( x ) . f !1( x ) = 3 x = x 1 3 .
"
c. Use your answer from part (b) to find f !1 (8) . Since
( )
( f )"( x) = dxd x
( )
( )
1 !2 3
1 !2 3 1 1 3
"
x , f !1 (8) = (8)
= 8
3
3
3
1 !2 1 " 1 % 1 " 1 % 1
(2) = $ 2 ' = $ ' = .
3
3 # 2 & 3 # 4 & 12
!1
13
=
!2
=
( )
"
d. How could you have used your answer from part (a) to find f !1 (8) ?
(
Since f f
!1
( x )) = x , we can use the Chain Rule to take derivatives of
( )" ( x ) # f " ( f
both sides, f !1
!1
( x )) = 1 or ( f !1 )" ( x ) =
( f )"(8) = f " f 1 (8) = f " 1 8 = f "1(2) = 121 .
(
) ( )
!1
!1
3
(
1
f " f !1 ( x )
)
. So