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. We will use this relationship to find derivatives of inverse trigonometric functions.

The Derivative of an Inverse Function

We begin by considering a function and its inverse. If f(x) is both invertible and differentiable, it seems reasonable that the inverse of f(x) is also differentiable. Let us look at the graphs of a function f(x) and its inverse f^{-1}(x) on Figure 1 below. Consider the point (a,f^{-1}(a)) on the graph of f^{-1}(x) having a tangent line with a slope of \ds (f^{-1})^{\prime}(a). As we discussed in the previous section, the graphs of f and f^{-1} are symmetric with respect to the line y=x. Therefore, the tangent line to the curve y=f^{-1}(x) at the point (a,f^{-1}(a)) must be symmetric to the tangent line to the curve y=f(x) at the (symmetric) point (f^{-1}(a),a). Note that the product of slopes of the lines that are symmetric with respect to the line y=x is 1. Indeed, if a line has equation y=mx+b, symmetric one would have equation x=my+b (switching x and y), which is equivalent to \displaystyle y=\frac1m x-\frac bm, provided m\ne0. Thus, if f^{-1}(x) is differentiable at a, then it must be the case that

\ds (f^{-1})^{\prime}(a)=\frac{1}{f^{\prime}(f^{-1}(a))}.
This graph shows a function f(x) and its inverse f-1(x). These functions are symmetric about the line y = x. The tangent line of the function f(x) at the point (f-1(a), a) and the tangent line of the function f-1(x) at (a, f-1(a)) are also symmetric about the line y = x. Specifically, if the slope of one were p/q, then the slope of the other would be q/p. Lastly, their derivatives are also symmetric about the line y = x.
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 \ds x=f(f^{-1}(x)) for every x in the domain of f. Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

1=f^{\prime}(f^{-1}(x))\cdot(f^{-1})^{\prime}(x).

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

\ds (f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}.

We summarize the above in the following theorem.

Inverse Function Theorem

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

\ds\frac{dy}{dx}=\frac{d}{dx}(f^{-1}(x))=(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}.

In other words, if we let y=g(x) be the inverse of f(x), then

\ds g'(x)=\frac{1}{f^{\prime}(g(x))} whenever f'(g(x))\ne0.

Derivatives of Inverse Trigonometric Functions

We now use the inverse function theorem to find derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text.

Derivative of the Inverse Sine Function

Use the inverse function theorem to find the derivative of \sin^{-1} (x).

Solution

Since f(x)= \sin(x) is differentiable and invertible when restricted to the interval \left[-\dfrac{\pi}2,\dfrac{\pi}2\right], as per the above theorem, for g(x)=\sin^{-1}(x), we have that \ds g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))} whenever f'(g(x))\ne0. Because f^{\prime}(y)=(\sin(y))'= \cos(y), we need to compute f^{\prime}(g(x))= \cos (\sin^{-1} (x)).

Let \sin^{-1}(x)=\theta. Then \ds\theta\in\left[-\frac{\pi}2,\frac{\pi}2\right], and we want to find \cos(\theta). Since \cos(\theta)\ge0 on \ds\left[-\frac{\pi}2,\frac{\pi}2\right], it follows from the Pythagorean trigonometric identity \sin^2(\theta)+\cos^2(\theta)=1 that \ds\cos(\theta)=\sqrt{1-\sin^2(\theta)}=\sqrt{1-x^2}.

The domain of \sin^{-1}(x) is [-1,1] and \sqrt{1-x^2}\ne0 for every x\in(-1,1). Therefore,

\ds\frac{d}{dx}\big(\sin^{-1} (x)\big)=g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}=\frac{1}{\sqrt{1-x^2}}, x\in (-1,1).

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

Answer

\ds g^{\prime}(x)=\frac{1}{1+x^2}

The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. The corresponding formulas are provided in the following theorem.

Derivatives of Inverse Trigonometric Functions

\begin{array}{lllll}\ds\frac{d}{dx}\big(\ds\sin^{-1} (x)\big)=\large \frac{1}{\sqrt{1-x^2}}& \big(x\in(-1,1)\big) & & \ds\frac{d}{dx}\big(\cos^{-1} (x)\big)=\large \frac{-1}{\sqrt{1-x^2}}& \big(x\in(-1,1)\big)\\[4mm] \ds\frac{d}{dx}\big(\tan^{-1} (x)\big)=\large \frac{1}{1+x^2} & \big(x\in\mathbb{R}\big) & & \ds\frac{d}{dx}\big(\cot^{-1} (x)\big)=\large \frac{-1}{1+x^2} & \big(x\in\mathbb{R}\big) \\[4mm] \ds\frac{d}{dx}\big(\sec^{-1} (x)\big)=\large \frac{1}{|x|\sqrt{x^2-1}} & \big(|x|>1\big) & & \ds\frac{d}{dx}\big(\csc^{-1} (x)\big)=\large \frac{-1}{|x|\sqrt{x^2-1}} & \big(|x|>1\big) \end{array}

Applying Formulas for the Derivatives of Inverse Trigonometric Functions

Find the derivatives of the following functions.

  1. \ds f(x)=\cos^{-1}\big(x\sqrt[3]x\big)
  2. \ds f(x)=x^2\sec^{-1}\left(2^x\right)

 

Solution

  1. We apply the chain rule with outside function g(y)=\cos^{-1}(y) and inside function \ds h(x)=x\sqrt[3]x=x^{1+\frac13}=x^{\frac43} to obtain \ds \Big(\cos^{-1}\big(x\sqrt[3]x\big)\Big)'=\big(g(h(x))\big)'=g'(h(x))\cdot h'(x)=-\frac{1}{\sqrt{1-\left(x^{\frac43}\right)^2}}\cdot\frac43x^{\frac13}\ds=-\frac{4x^{\frac13}}{3\sqrt{1-x^{\frac83}}}
  2. Here we need to apply the product rule to the multiples x^2 and \sec^{-1}(2^x), while differentiating \sec^{-1}(2^x) requires using the chain rule. We have \Big(x^2\sec^{-1}\left(2^x\right)\Big)'=(x^2)'\cdot\sec^{-1}(2^x)+x^2\cdot\big(\sec^{-1}(2^x)\big)' \ds=2x\cdot\sec^{-1}(2^x)+x^2\cdot\frac1{|2^x|\sqrt{(2^x)^2-1}}\cdot2^x\ln(2)=2x\cdot\sec^{-1}(2^x)+\frac{x^2\ln(2)}{\sqrt{2^{2x}-1}}.

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
    \ds(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))} whenever f^{\prime}(f^{-1}(x))\ne 0 and f(x) is differentiable.
  • Derivative of inverse sine function
    \ds\frac{d}{dx}\big(\sin^{-1} (x)\big)=\frac{1}{\sqrt{1-x^2}}
  • Derivative of inverse cosine function
    \ds\frac{d}{dx}\big(\cos^{-1} (x)\big)=\frac{-1}{\sqrt{1-x^2}}
  • Derivative of inverse tangent function
    \ds\frac{d}{dx}\big(\tan^{-1} (x)\big)=\frac{1}{1+x^2}
  • Derivative of inverse cotangent function
    \ds\frac{d}{dx}\big(\cot^{-1} (x)\big)=\frac{-1}{1+x^2}
  • Derivative of inverse secant function
    \ds\frac{d}{dx}\big(\sec^{-1} (x)\big)=\frac{1}{|x|\sqrt{x^2-1}}
  • Derivative of inverse cosecant function
    \ds\frac{d}{dx}\big(\csc^{-1} (x)\big)=\frac{-1}{|x|\sqrt{x^2-1}}.

Exercises

For the following exercises, find \ds\frac{dy}{dx}.

1.  y= \sin^{-1}(x^2)

Answer

\ds \frac{2x}{\sqrt{1-x^4}}

2.  y= \cos^{-1}\big(\sqrt{x}\big)

3.  \ds y= \sec^{-1}\left(\frac{1}{x}\right)

Answer

\ds\frac{-1}{\sqrt{1-x^2}}

4.  y=\sqrt{\csc^{-1} (x)}

5.  \ds y=\big(1 + \tan^{-1} (x)\big)^3

Answer

\ds \frac{3\big(1 + \tan^{-1} x\big)^2}{1+x^2}

6.  y= \cos^{-1}(2x) \cdot \sin^{-1}(2x)

7.  \ds y=\frac{1}{\tan^{-1}(x)}

Answer

\ds \frac{-1}{(1+x^2)(\tan^{-1} (x))^2}

8.  y= \sec^{-1}(-x)

9.  y= \cot^{-1} \big(\sqrt{4-x^2}\big)

Answer

\ds \frac{x}{(5-x^2)\sqrt{4-x^2}}

10.  y=x \cdot \csc^{-1} (x)

For the following exercises, find the slope of the tangent line to the given curve at the given point.

11.  y= \pi x+2\cos^{-1}(1-x) at P(1,2\pi).

Answer

\ds \pi+2

12.  y= \ln(\cot^{-1}(e^x)) at the point corresponding to x=0.

13*. y^2=\sec^{-1}(x+y) at P(1,0).

(Hint: Use implicit differentiation.)

Answer

\ds -1

14*.  There is a theorem that if a function f is differentiable on an open interval I and f'(x)=0 on I, then f is constant on I. Using this result, prove that \ds\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}2, x\in\mathbb{R}.

15. [T] A pole stands 75 feet tall. An angle \theta is formed when wires of various lengths x (measured in feet) are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change \ds\frac{d\theta}{dx} of the angle with respect to the wire length when a wire of length 90 feet is attached.

A flagpole is shown with height 75 ft. A triangle is made with the flagpole height as the opposite side from the angle θ. The hypotenuse has length x.

Answer

\dfrac{-1}{18\sqrt{11}} radians per foot

16. [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. After launch, let x be the height of the rocket and \theta be the angle of elevation of the camera. Find the rate of change \ds\frac{d\theta}{dx} of the angle of elevation with respect to the rocket’s height when the camera and the rocket are 5000 feet apart.

A rocket is shown with in the air with the distance from its nose to the ground being x. A triangle is made with the rocket height as the opposite side from the angle θ. The adjacent side has length 2000.

17*. A local movie theatre has a 30-foot-high screen that is 10 feet above a person’s eye level when seated. Suppose that a person is sitting at a distance of x feet from the movie screen and has a viewing angle of \theta radians, see the figure below.

A person is shown with a right triangle coming from their eye (the right angle being on the opposite side from the eye), with height 10 and base x. There is a line drawn from the eye to the top of the screen, which makes an angle θ with the triangle’s hypotenuse. The screen has a height of 30.

  1. Find \ds\frac{d\theta}{dx}.
  2. Use optimization methods learned in Calculus 1 to determine at what distance x the person should sit to maximize his or her viewing angle.

Answer

a. \ds\frac{d\theta}{dx}=\frac{10}{100+x^2}-\frac{40}{1600+x^2}

b. The optimal distance x for maximizing the viewing angle is 20 feet.

 

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Calculus: Volume 2 (Second University of Manitoba Edition) by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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