Chapter 4.8: Derivatives of Inverse Functions
Learning Objectives
- Calculate the derivative of an inverse function.
- Recognize the derivatives of the standard inverse trigonometric functions.
In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents.
The Derivative of an Inverse Function
We begin by considering a function and its inverse. If is both invertible and differentiable, it seems reasonable that the inverse of
is also differentiable. (Figure) shows the relationship between a function
and its inverse
. Look at the point
on the graph of
having a tangent line with a slope of
. This point corresponds to a point
on the graph of
having a tangent line with a slope of
. Thus, if
is differentiable at
, then it must be the case that


We may also derive the formula for the derivative of the inverse by first recalling that . Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

Solving for , we obtain

We summarize this result in the following theorem.
Inverse Function Theorem
Let be a function that is both invertible and differentiable. Let
be the inverse of
. For all
satisfying
,

Alternatively, if is the inverse of
, then

Applying the Inverse Function Theorem
Use the Inverse Function Theorem to find the derivative of . Compare the resulting derivative to that obtained by differentiating the function directly.
Solution
The inverse of is
. Since
, begin by finding
. Thus,


Finally,

We can verify that this is the correct derivative by applying the quotient rule to to obtain

Use the inverse function theorem to find the derivative of Compare the result obtained by differentiating
directly.
Show Solution
Hint
Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of .
Solution
The function is the inverse of the function
. Since
, begin by finding
. Thus,

![Rendered by QuickLaTeX.com f^{\prime}(g(x))=3(\sqrt[3]{x})^2=3x^{2/3}](https://ecampusontario.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-361de2cd30623f5e272cfbefdd560b0a_l3.png)
Finally,

Find the derivative of by applying the inverse function theorem.
Solution
Hint
Use the fact that is the inverse of
.
From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form , where
is a positive integer. This extension will ultimately allow us to differentiate
, where
is any rational number.
Extending the Power Rule to Rational Exponents
The power rule may be extended to rational exponents. That is, if is a positive integer, then

Also, if is a positive integer and
is an arbitrary integer, then

Proof
The function is the inverse of the function
. Since
, begin by finding
. Thus,


Finally,

To differentiate we must rewrite it as
and apply the chain rule. Thus,


Applying the Power Rule to a Rational Power
Find the equation of the line tangent to the graph of at
.
Solution
First find and evaluate it at
. Since


the slope of the tangent line to the graph at is
.
Substituting into the original function, we obtain
. Thus, the tangent line passes through the point
. Substituting into the point-slope formula for a line and solving for
, we obtain the tangent line

Find the derivative of .
Solution
Hint
Rewrite as and use the chain rule.
Derivatives of Inverse Trigonometric Functions
We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.
Derivative of the Inverse Sine Function
Use the inverse function theorem to find the derivative of .
Solution
Since for in the interval
is the inverse of
, begin by finding
. Since


we see that

Analysis
To see that , consider the following argument. Set
. In this case,
where
. We begin by considering the case where
. Since
is an acute angle, we may construct a right triangle having acute angle
, a hypotenuse of length 1, and the side opposite angle
having length
. From the Pythagorean theorem, the side adjacent to angle
has length
. This triangle is shown in (Figure). Using the triangle, we see that
.








Now if or
or
, and since in either case
and
, we have

Consequently, in all cases, .
Applying the Chain Rule to the Inverse Sine Function
Apply the chain rule to the formula derived in (Figure) to find the derivative of and use this result to find the derivative of
.
Solution
Applying the chain rule to , we have

Now let , so
. Substituting into the previous result, we obtain

Use the inverse function theorem to find the derivative of .
Solution
Hint
The inverse of is
. Use (Figure) as a guide.
The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.
Derivatives of Inverse Trigonometric Functions

Applying Differentiation Formulas to an Inverse Tangent Function
Find the derivative of .
Applying Differentiation Formulas to an Inverse Sine Function
Find the derivative of .
Solution
By applying the product rule, we have

Applying the Inverse Tangent Function
The position of a particle at time is given by
for
. Find the velocity of the particle at time
.
Solution
Begin by differentiating in order to find
. Thus,

Simplifying, we have

Thus, .
Find the equation of the line tangent to the graph of at
.
Solution
Hint
gives the slope of the tangent line.
Key Concepts
- The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative.
- We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.
Key Equations
- Inverse function theorem
whenever
and
is differentiable.
- Power rule with rational exponents
.
- Derivative of inverse sine function
- Derivative of inverse cosine function
- Derivative of inverse tangent function
- Derivative of inverse cotangent function
- Derivative of inverse secant function
- Derivative of inverse cosecant function
For the following exercises, use the graph of to
- sketch the graph of
, and
- use part a. to estimate
.


Solution
a.
b.


Solution
a.
b.
For the following exercises, use the functions to find
-
at
and
-
.
- Then use part b. to find
at
.
5.
6.
Solution
a. 6
b.
c.
7.
8.
Solution
a. 1
b.
c. 1
For each of the following functions, find .
9.
10.
Solution
11.
12.
Solution
13.
14.
Solution
1
For each of the given functions ,
- find the slope of the tangent line to its inverse function
at the indicated point
, and
- find the equation of the tangent line to the graph of
at the indicated point.
15.
16.
Solution
a. 4
b.
17.
18.
Solution
a.
b.
19.
For the following exercises, find for the given function.
20.
Solution
21.
22.
Solution
23.
24.
Solution
25.
26.
Solution
27.
28.
Solution
29.
For the following exercises, use the given values to find .
30.
Solution
-1
31.
32.
Solution
33.
34.
Solution
35.
36. [T] The position of a moving hockey puck after seconds is
where
is in meters.
- Find the velocity of the hockey puck at any time
.
- Find the acceleration of the puck at any time
.
- Evaluate a. and b. for
, and 6 seconds.
- What conclusion can be drawn from the results in c.?
Solution
a.
b.
c.
d. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.
37. [T] A building that is 225 feet tall casts a shadow of various lengths as the day goes by. An angle of elevation
is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation
when
feet.
38. [T] A pole stands 75 feet tall. An angle is formed when wires of various lengths of
feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle
when a wire of length 90 feet is attached.
Show Solution
-0.0168 radians per foot
39. [T] A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by , where
is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart.
40. [T] A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle (in radians) given by
,
where is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.
- Find
.
- Evaluate
for
, and 20.
- Interpret the results in b.
- Evaluate
for
, and 40
- Interpret the results in d. At what distance
should the person sit to maximize his or her viewing angle?
Show Solution
a.
b.
c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening.
d.
e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should sit for maximizing the viewing angle is 20 feet.
Use the preceding example as a guide.