Derivatives of Inverse Functions - Classwork
General Problem: Find the derivative of the inverse function of f ( x ) at x = k.
Method 1: Simply finding the inverse function. This works when it is easy to generate the inverse function.
a) Find the inverse function by interchanging x and y and solving for y
b) Take the derivative of this new y. That will be the derivative of the inverse function.
c) Plug in your given k value
Method 2: Not finding the inverse function because it is too difficult
a) Find the inverse function by interchanging x and y .
dy
implicitly
b) find
dx
dy
c) Solve for
. It will be in terms of y.
dx
d) Replace the value of k for x in your inverse function from step a above and solve for y
dy
e) Plug that value of y into
dx
2
Example: If f ( x ) = x , x ; 0 , find the derivative of f #1 ( x ) at x = 4
METHOD 1
a) y = x 2 , so the inverse is x = y 2
METHOD 2
a) y = x 2 , so the inverse is x = y 2
dy
b) 1 = 2 y
dx
dy 1
c)
=
dx 2 y
therefore y = x (first quadrant)
b) y " =
1
2 x
1
1
c) y "(4) =
=
2 4 4
d) 4 = y 2 4 y = 2(quad I)
e)
So take the
reciprocal of
the slope here!
1
1
dy 1
=
=
=
dx 2 y 2(2) 4
f ( x) = x2
You want the
slope of the
tangent line here!
(2, 4)
f #1 ( x ) = x
(4, 2)
Note: It was necessary to restrict the domain of f ( x ) to x # 0 so that its inverse is a function: i.e. that f ( x ) is
one-to-one (passes the horizontal line test).
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AB Solutions
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Stu Schwartz
Example: Find the derivative of the inverse function of f ( x) = x 3 # 4 x 2 + 7 x # 1 at x = 1.
Method 1 will be too difficult. y = x 3 # 4 x 2 + 7 x # 1 so the inverse is x = y 3 # 4 y 2 + 7 y # 1
dy
dy
1
a) 1 = 3 y 2 # 8 y + 7
4
= 2
dx
dx 3 y # 8 y + 7
(
)
b) Set y 3 # 4 y 2 + 7 y # 1 = 1 . Graphically, you get y= .349.
c)
1
dy
= .219
=
2
dx 3(.349) # 8(.349) + 7
Example: Find the derivative of the inverse function of f ( x) = e x + ln x at x = 3
Method 1 will be too difficult. y = e x + ln x so the inverse is x = e y + ln y
1 0 dy
dy
1
dy
y
a) 1 = /e y + 2 4
=
or
=
y
1
y 1 dx
dx
dx
ye + 1
.
ey +
y
y
b) Set e + ln y = 3. Graphically, you get y = 1.0744.
c)
dy
=
dx
1
1.0744
e
1
+
1.0744
= .259
1
Note: After you graphically intersect, you can easily get the answer by nDeriv(Y1,X,X)
2
Example: Find the derivative of the inverse function of y = e x , x> 0
Inverse Function : x = e y
2
Solution: ln x = y 2
y = ln x = ( ln x )
1
2
#1 1
dy 1
1
= ( ln x ) 2 =
dx 2
x 2 x ln x 12
( )
Sample Problems: Find the derivative of the inverse function of (use method 2 only if method 1 won’t work)
y = x + sin x at x = %
y = x 3 + 1 at x = 9
y = x 3 + 5 x # 1 at x = 5
1
a) dy
b) dy
c) dy
1
1
1
1
=
= DNE
= 2=
= 2
=
dx ( x = %) 1 + cos x
dx ( x = 2) 3 x 12
dx ( x =1) 3 x + 5 8
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AB Solutions
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Stu Schwartz
Derivatives of Inverse Functions - Homework
For the problems below, find the derivative of f #1 for the function f at the specified value of x. No calculators.
1. f ( x ) = x 3 + 2 x # 1 at x = 2
2. f ( x ) = 2 x 5 + x 3 + 1 at x = 4
x3 + 2x #1 = 2 4 x = 1
dy
1
1
= 2
=
dx ( x =1) 3 x + 2 5
2x5 + x3 + 1 = 4 4 x = 1
dy
1
1
=
=
4
2
dx ( x =1) 10 x + 3 x
13
%
%
1
5x5
at x =
2
2
2
%
1
sin x = 4 x =
2
6
1
2
dy
=
=
- %0
dx /. x = 21 cos x
3
6
3. f ( x ) = sin x
5. f ( x ) = x 3 #
#
4
x
x >0
at x = 6
4. f ( x ) = cos2 x
#
%
%
5x5
2
2
at x = 1
cos 2 x = 1 4 x = 0
dy
#1
=
= DNE
dx ( x = 0) 2 sin 2 x
6.
4
=64 x =2
x
dy
1
1
=
=
dx ( x = 2) 3 x 2 + 4 13
x2
x3 #
f ( x) = x # 4
at x = 2
x#4 =24 x =8
dy
1
=
=4
1
dx ( x = 8)
2 x#4
For the problems below, find the derivative of f #1 for the function f at the specified value of x. Use calculators.
3
2
f ( x) = x # 2 x + 5 x # 1
at x = 2
7. dy
1
= 2
= .272
dx ( x = .737) 3 x # 4 x + 5
f ( x) =
x
+ sin 2 x
2
f ( x ) = 3 3x # 5
8.
dy
=
dx ( x =#7.333)
at x = 3
at x = #3
1
=9
1
23
(3x # 5)
f ( x ) = xe cos x
at x = 3
9. dy
1
=
= .818
dx ( x = 4.309) 1
+ 2 sin x cos x
2
10. dy
1
= cos x
= 0.287
dx ( x = 4.335) xe (# sin x ) + e cos x
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AB Solutions
Stu Schwartz