3.7 Derivatives of Inverse Functions (2024)

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.

We begin by considering a function and its inverse. If [latex]f(x)[/latex] is both invertible and differentiable, it seems reasonable that the inverse of [latex]f(x)[/latex] is also differentiable. (Figure) shows the relationship between a function [latex]f(x)[/latex] and its inverse [latex]f^{-1}(x)[/latex]. Look at the point [latex](a,f^{-1}(a))[/latex] on the graph of [latex]f^{-1}(x)[/latex] having a tangent line with a slope of [latex](f^{-1})^{\prime}(a)=\frac{p}{q}[/latex]. This point corresponds to a point [latex](f^{-1}(a),a)[/latex] on the graph of [latex]f(x)[/latex] having a tangent line with a slope of [latex]f^{\prime}(f^{-1}(a))=\frac{q}{p}[/latex]. Thus, if [latex]f^{-1}(x)[/latex] is differentiable at [latex]a[/latex], then it must be the case that

[latex](f^{-1})^{\prime}(a)=\frac{1}{f^{\prime}(f^{-1}(a))}[/latex].

3.7 Derivatives of Inverse Functions (1)

Figure 1. The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions.

We may also derive the formula for the derivative of the inverse by first recalling that [latex]x=f(f^{-1}(x))[/latex]. Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

[latex]1=f^{\prime}(f^{-1}(x))(f^{-1})^{\prime}(x))[/latex].

Solving for [latex](f^{-1})^{\prime}(x)[/latex], we obtain

[latex](f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}[/latex].

We summarize this result in the following theorem.

Inverse Function Theorem

Let [latex]f(x)[/latex] be a function that is both invertible and differentiable. Let [latex]y=f^{-1}(x)[/latex] be the inverse of [latex]f(x)[/latex]. For all [latex]x[/latex] satisfying [latex]f^{\prime}(f^{-1}(x))\ne 0[/latex],

[latex]\frac{dy}{dx}=\frac{d}{dx}(f^{-1}(x))=(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}[/latex].

Alternatively, if [latex]y=g(x)[/latex] is the inverse of [latex]f(x)[/latex], then

[latex]g(x)=\frac{1}{f^{\prime}(g(x))}[/latex].

Applying the Inverse Function Theorem

Use the Inverse Function Theorem to find the derivative of [latex]g(x)=\frac{x+2}{x}[/latex]. Compare the resulting derivative to that obtained by differentiating the function directly.

Show Solution

Use the inverse function theorem to find the derivative of [latex]g(x)=\frac{1}{x+2}.[/latex] Compare the result obtained by differentiating [latex]g(x)[/latex] directly.

Show Solution

Hint

Use the preceding example as a guide.

Applying the Inverse Function Theorem

Use the inverse function theorem to find the derivative of [latex]g(x)=\sqrt[3]{x}[/latex].

Show Solution

Find the derivative of [latex]g(x)=\sqrt[5]{x}[/latex] by applying the inverse function theorem.

Show Solution

Hint

Use the fact that [latex]g(x)[/latex] is the inverse of [latex]f(x)=x^5[/latex].

From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form [latex]\frac{1}{n}[/latex], where [latex]n[/latex] is a positive integer. This extension will ultimately allow us to differentiate [latex]x^q[/latex], where [latex]q[/latex] is any rational number.

Extending the Power Rule to Rational Exponents

The power rule may be extended to rational exponents. That is, if [latex]n[/latex] is a positive integer, then

[latex]\frac{d}{dx}(x^{1/n})=\frac{1}{n}x^{(1/n)-1}[/latex].

Also, if [latex]n[/latex] is a positive integer and [latex]m[/latex] is an arbitrary integer, then

[latex]\frac{d}{dx}(x^{m/n})=\frac{m}{n}x^{(m/n)-1}[/latex].

Proof

The function [latex]g(x)=x^{1/n}[/latex] is the inverse of the function [latex]f(x)=x^n[/latex]. Since [latex]g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}[/latex], begin by finding [latex]f^{\prime}(x)[/latex]. Thus,

[latex]f^{\prime}(x)=nx^{n-1}[/latex] and [latex]f^{\prime}(g(x))=n(x^{1/n})^{n-1}=nx^{(n-1)/n}[/latex].

Finally,

[latex]g^{\prime}(x)=\frac{1}{nx^{(n-1)/n}}=\frac{1}{n}x^{(1-n)/n}=\frac{1}{n}x^{(1/n)-1}[/latex].

To differentiate [latex]x^{m/n}[/latex] we must rewrite it as [latex](x^{1/n})^m[/latex] and apply the chain rule. Thus,

[latex]\frac{d}{dx}(x^{m/n})=\frac{d}{dx}((x^{1/n})^m)=m(x^{1/n})^{m-1} \cdot \frac{1}{n}x^{(1/n)-1}=\frac{m}{n}x^{(m/n)-1}[/latex]. [latex]_\blacksquare[/latex]

Applying the Power Rule to a Rational Power

Find the equation of the line tangent to the graph of [latex]y=x^{2/3}[/latex] at [latex]x=8[/latex].

Show Solution

Find the derivative of [latex]s(t)=\sqrt{2t+1}[/latex].

Show Solution

Hint

Rewrite as [latex]s(t)=(2t+1)^{1/2}[/latex] and use the chain rule.

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 [latex]g(x)=\sin^{-1} x[/latex].

Show Solution

Analysis

To see that [latex]\cos (\sin^{-1} x)=\sqrt{1-x^2}[/latex], consider the following argument. Set [latex]\sin^{-1} x=\theta[/latex]. In this case, [latex]\sin \theta =x[/latex] where [latex]-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}[/latex]. We begin by considering the case where [latex]0<\theta <\frac{\pi}{2}[/latex]. Since [latex]\theta[/latex] is an acute angle, we may construct a right triangle having acute angle [latex]\theta[/latex], a hypotenuse of length 1, and the side opposite angle [latex]\theta [/latex] having length [latex]x[/latex]. From the Pythagorean theorem, the side adjacent to angle [latex]\theta[/latex] has length [latex]\sqrt{1-x^2}[/latex]. This triangle is shown in (Figure). Using the triangle, we see that [latex]\cos (\sin^{-1} x)= \cos \theta =\sqrt{1-x^2}[/latex].

3.7 Derivatives of Inverse Functions (2)

Figure 2. Using a right triangle having acute angle [latex]\theta[/latex], a hypotenuse of length 1, and the side opposite angle [latex]\theta[/latex] having length [latex]x[/latex], we can see that [latex]\cos (\sin^{-1} x)= \cos \theta =\sqrt{1-x^2}[/latex].

In the case where [latex]-\frac{\pi}{2}<\theta <0[/latex], we make the observation that [latex]0<-\theta<\frac{\pi}{2}[/latex] and hence [latex]\cos (\sin^{-1} x)= \cos \theta = \cos (−\theta )=\sqrt{1-x^2}[/latex].

Now if [latex]\theta =\frac{\pi}{2}[/latex] or [latex]\theta =-\frac{\pi}{2}, \, x=1[/latex] or [latex]x=-1[/latex], and since in either case [latex]\cos \theta =0[/latex] and [latex]\sqrt{1-x^2}=0[/latex], we have

[latex]\cos (\sin^{-1} x)= \cos \theta =\sqrt{1-x^2}[/latex].

Consequently, in all cases, [latex]\cos (\sin^{-1} x)=\sqrt{1-x^2}[/latex].

Applying the Chain Rule to the Inverse Sine Function

Apply the chain rule to the formula derived in (Figure) to find the derivative of [latex]h(x)=\sin^{-1} (g(x))[/latex] and use this result to find the derivative of [latex]h(x)=\sin^{-1}(2x^3)[/latex].

Show Solution

Use the inverse function theorem to find the derivative of [latex]g(x)=\tan^{-1} x[/latex].

Show Solution

Hint

The inverse of [latex]g(x)[/latex] is [latex]f(x)= \tan x[/latex]. 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

[latex]\begin{array}{lllll}\frac{d}{dx}(\sin^{-1} x)=\large \frac{1}{\sqrt{1-x^2}} & & & & \frac{d}{dx}(\cos^{-1} x)=\large \frac{-1}{\sqrt{1-x^2}} \\ \frac{d}{dx}(\tan^{-1} x)=\large \frac{1}{1+x^2} & & & & \frac{d}{dx}(\cot^{-1} x)=\large \frac{-1}{1+x^2} \\ \frac{d}{dx}(\sec^{-1} x)=\large \frac{1}{|x|\sqrt{x^2-1}} & & & & \frac{d}{dx}(\csc^{-1} x)=\large \frac{-1}{|x|\sqrt{x^2-1}} \end{array}[/latex]

Applying Differentiation Formulas to an Inverse Tangent Function

Find the derivative of [latex]f(x)=\tan^{-1} (x^2)[/latex].

Show Solution

Applying Differentiation Formulas to an Inverse Sine Function

Find the derivative of [latex]h(x)=x^2 \sin^{-1} x[/latex].

Show Solution

Find the derivative of [latex]h(x)= \cos^{-1} (3x-1)[/latex].

Show Solution

Hint

Use (Figure). with [latex]g(x)=3x-1[/latex]

Applying the Inverse Tangent Function

The position of a particle at time [latex]t[/latex] is given by [latex]s(t)= \tan^{-1}(\frac{1}{t})[/latex] for [latex]t\ge \frac{1}{2}[/latex]. Find the velocity of the particle at time [latex]t=1[/latex].

Show Solution

Find the equation of the line tangent to the graph of [latex]f(x)= \sin^{-1} x[/latex] at [latex]x=0[/latex].

Show Solution

Hint

[latex]f^{\prime}(0)[/latex] gives the slope of the tangent line.

For the following exercises, use the graph of [latex]y=f(x)[/latex] to

  1. sketch the graph of [latex]y=f^{-1}(x)[/latex], and
  2. use part a. to estimate [latex](f^{-1})^{\prime}(1)[/latex].

1.
3.7 Derivatives of Inverse Functions (3)

2.
3.7 Derivatives of Inverse Functions (4)

Show Solution

3.
3.7 Derivatives of Inverse Functions (5)

4.
3.7 Derivatives of Inverse Functions (6)

Show Solution

For the following exercises, use the functions [latex]y=f(x)[/latex] to find

  1. [latex]\frac{df}{dx}[/latex] at [latex]x=a[/latex] and
  2. [latex]x=f^{-1}(y)[/latex].
  3. Then use part b. to find [latex]\frac{df^{-1}}{dy}[/latex] at [latex]y=f(a)[/latex].

5.[latex]f(x)=6x-1, \, x=-2[/latex]

6.[latex]f(x)=2x^3-3, \, x=1[/latex]

Show Solution

7.[latex]f(x)=9-x^2, \, 0\le x\le 3, \, x=2[/latex]

8.[latex]f(x)= \sin x, \, x=0[/latex]

Show Solution

For each of the following functions, find [latex](f^{-1})^{\prime}(a)[/latex].

9.[latex]f(x)=x^2+3x+2, \, x\ge -1, \, a=2[/latex]

10.[latex]f(x)=x^3+2x+3, \, a=0[/latex]

Show Solution

11.[latex]f(x)=x+\sqrt{x}, \, a=2[/latex]

12.[latex]f(x)=x-\frac{2}{x}, \, x<0, \, a=1[/latex]

Show Solution

13.[latex]f(x)=x + \sin x, \, a=0[/latex]

14.[latex]f(x)= \tan x+3x^2, \, a=0[/latex]

Show Solution

For each of the given functions [latex]y=f(x)[/latex],

  1. find the slope of the tangent line to its inverse function [latex]f^{-1}[/latex] at the indicated point [latex]P[/latex], and
  2. find the equation of the tangent line to the graph of [latex]f^{-1}[/latex] at the indicated point.

15.[latex]f(x)=\frac{4}{1+x^2}, \, P(2,1)[/latex]

16.[latex]f(x)=\sqrt{x-4}, \, P(2,8)[/latex]

Show Solution

17.[latex]f(x)=(x^3+1)^4, \, P(16,1)[/latex]

18.[latex]f(x)=−x^3-x+2, \, P(-8,2)[/latex]

Show Solution

19.[latex]f(x)=x^5+3x^3-4x-8, \, P(-8,1)[/latex]

For the following exercises, find [latex]\frac{dy}{dx}[/latex] for the given function.

20.[latex]y= \sin^{-1}(x^2)[/latex]

Show Solution

21.[latex]y= \cos^{-1}(\sqrt{x})[/latex]

22.[latex]y= \sec^{-1}(\frac{1}{x})[/latex]

Show Solution

23.[latex]y=\sqrt{\csc^{-1} x}[/latex]

24.[latex]y=(1 + \tan^{-1} x)^3[/latex]

Show Solution

25.[latex]y= \cos^{-1}(2x) \cdot \sin^{-1}(2x)[/latex]

26.[latex]y=\frac{1}{\tan^{-1}(x)}[/latex]

Show Solution

27.[latex]y= \sec^{-1}(−x)[/latex]

28.[latex]y= \cot^{-1} \sqrt{4-x^2}[/latex]

Show Solution

29.[latex]y=x \cdot \csc^{-1} x[/latex]

For the following exercises, use the given values to find [latex](f^{-1})^{\prime}(a)[/latex].

30.[latex]f(\pi)=0, \, f^{\prime}(\pi)=-1, \, a=0[/latex]

Show Solution

31.[latex]f(6)=2, \, f^{\prime}(6)=\frac{1}{3}, \, a=2[/latex]

32.[latex]f(\frac{1}{3})=-8, \, f^{\prime}(\frac{1}{3})=2, \, a=-8[/latex]

Show Solution

33.[latex]f(\sqrt{3})=\frac{1}{2}, \, f^{\prime}(\sqrt{3})=\frac{2}{3}, \, a=\frac{1}{2}[/latex]

34.[latex]f(1)=-3, \, f^{\prime}(1)=10, \, a=-3[/latex]

Show Solution

35.[latex]f(1)=0, \, f^{\prime}(1)=-2, \, a=0[/latex]

36. [T] The position of a moving hockey puck after [latex]t[/latex] seconds is [latex]s(t)= \tan^{-1} t[/latex] where [latex]s[/latex] is in meters.

  1. Find the velocity of the hockey puck at any time [latex]t[/latex].
  2. Find the acceleration of the puck at any time [latex]t[/latex].
  3. Evaluate a. and b. for [latex]t=2,4[/latex], and 6 seconds.
  4. What conclusion can be drawn from the results in c.?

Show Solution

37. [T] A building that is 225 feet tall casts a shadow of various lengths [latex]x[/latex] as the day goes by. An angle of elevation [latex]\theta[/latex] 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 [latex]\frac{d\theta}{dx}[/latex] when [latex]x=272[/latex] feet.

3.7 Derivatives of Inverse Functions (7)

38. [T] A pole stands 75 feet tall. An angle [latex]\theta[/latex] is formed when wires of various lengths of [latex]x[/latex] 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 [latex]\frac{d\theta}{dx}[/latex] when a wire of length 90 feet is attached.

3.7 Derivatives of Inverse Functions (8)

Show Solution

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 [latex]\theta = \tan^{-1}(\frac{x}{2000})[/latex], where [latex]x[/latex] 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.

3.7 Derivatives of Inverse Functions (9)

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 [latex]\theta[/latex] (in radians) given by [latex]\theta = \cot^{-1}(\frac{x}{40})- \cot^{-1}(\frac{x}{10})[/latex],

where [latex]x[/latex] is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.

3.7 Derivatives of Inverse Functions (10)

  1. Find [latex]\frac{d\theta}{dx}[/latex].
  2. Evaluate [latex]\frac{d\theta}{dx}[/latex] for [latex]x=5,10,15[/latex], and 20.
  3. Interpret the results in b.
  4. Evaluate [latex]\frac{d\theta}{dx}[/latex] for [latex]x=25,30,35[/latex], and 40
  5. Interpret the results in d. At what distance [latex]x[/latex] should the person sit to maximize his or her viewing angle?

Show Solution

3.7 Derivatives of Inverse Functions (2024)

References

Top Articles
Latest Posts
Article information

Author: Greg O'Connell

Last Updated:

Views: 5954

Rating: 4.1 / 5 (42 voted)

Reviews: 81% of readers found this page helpful

Author information

Name: Greg O'Connell

Birthday: 1992-01-10

Address: Suite 517 2436 Jefferey Pass, Shanitaside, UT 27519

Phone: +2614651609714

Job: Education Developer

Hobby: Cooking, Gambling, Pottery, Shooting, Baseball, Singing, Snowboarding

Introduction: My name is Greg O'Connell, I am a delightful, colorful, talented, kind, lively, modern, tender person who loves writing and wants to share my knowledge and understanding with you.