7 Inverse Functions and L’Hôpital’s Rule 7.1 Inverse Functions 7.3 Exponential and Logarithmic Functions

7.2 Derivatives of Inverse Functions

In this section we will figure out how to differentiate the inverse of a function. To do so, we recall that if $f$ and $g$ are inverses, then $f(g(x))=x$ for all $x$ in the domain of $f$ . Differentiating and simplifying yields:

	$\displaystyle f(g(x))$	$\displaystyle=x$
	$\displaystyle f\,^{\prime}(g(x))g\mkern 1.35mu ^{\prime}(x)$	$\displaystyle=1$
	$\displaystyle g\mkern 1.35mu ^{\prime}(x)$	$\displaystyle=\frac{1}{f\,^{\prime}(g(x))}\quad\text{assuming $f\,^{\prime}(x)% $ is nonzero}$

Note that the derivation above assumes that the function $g$ is differentiable. It is possible to prove that $g$ must be differentiable if $f\,^{\prime}$ is nonzero, but the proof is beyond the scope of this text. However, assuming this fact we have shown the following:

Theorem 7.2.1 Derivatives of Inverse Functions

Let $f$ be differentiable and one-to-one on an open interval $I$ , where $f\,^{\prime}(x)\neq 0$ for all $x$ in $I$ , let $J$ be the range of $f$ on $I$ , let $g$ be the inverse function of $f$ , and let $f(a)=b$ for some $a$ in $I$ . Then $g$ is a differentiable function on $J$ , and in particular,

	$\displaystyle\left(f\mkern 1.35mu ^{-1}\right)^{\prime}(b)$	$\displaystyle=g\mkern 1.35mu ^{\prime}(b)=\frac{1}{f\,^{\prime}(a)}$
	$\displaystyle\left(f\mkern 1.35mu ^{-1}\right)^{\prime}(x)$	$\displaystyle=g\mkern 1.35mu ^{\prime}(x)=\frac{1}{f\,^{\prime}(g(x))}$

The results of Theorem 7.2.1 are not trivial; the notation may seem confusing at first. Careful consideration, along with examples, should earn understanding.

Watch the video:
Derivative of an Inverse Function, Ex 2 from https://youtu.be/RKfGMX0pn2k

In the next example we apply Theorem 7.2.1 to the arcsine function.

Example 7.2.1 Finding the derivative of an inverse trigonometric function

Let $y=\sin^{-1}x$ . Find $y\mkern 1.35mu ^{\prime}$ using Theorem 7.2.1.

SolutionAdopting our previously defined notation, let $g(x)=\sin^{-1}x$ and $f(x)=\sin x$ . Thus $f\,^{\prime}(x)=\cos x$ . Applying Theorem 7.2.1, we have

	$\displaystyle g\mkern 1.35mu ^{\prime}(x)$	$\displaystyle=\frac{1}{f\,^{\prime}(g(x))}$
		$\displaystyle=\frac{1}{\cos(\sin^{-1}x)}.$

^†^†margin: Figure 7.2.1: A right triangle defined by

y=\sin^{-1}(x/1)

with the length of the third leg found using the Pythagorean Theorem. Λ

This last expression is not immediately illuminating. Drawing a figure will help, as shown in Figure 7.2.1. Recall that the sine function can be viewed as taking in an angle and returning a ratio of sides of a right triangle, specifically, the ratio “opposite over hypotenuse.” This means that the arcsine function takes as input a ratio of sides and returns an angle. The equation $y=\sin^{-1}x$ can be rewritten as $y=\sin^{-1}(x/1)$ ; that is, consider a right triangle where the hypotenuse has length 1 and the side opposite of the angle with measure $y$ has length $x$ . This means the final side has length $\sqrt{1-x^{2}}$ , using the Pythagorean Theorem.

Therefore $\cos(\sin^{-1}x)=\cos y=\sqrt{1-x^{2}}/1=\sqrt{1-x^{2}}$ , resulting in

\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sin^{-1}x\bigr{)}=g% \mkern 1.35mu ^{\prime}(x)=\frac{1}{\sqrt{1-x^{2}}}.

Remember that the input $x$ of the arcsine function is a ratio of a side of a right triangle to its hypotenuse; the absolute value of this ratio will be less than 1. Therefore $1-x^{2}$ will be positive.

^†^†margin: Figure 7.2.2: Graphs of

y=\sin x

and

y=\sin^{-1}x

along with corresponding tangent lines. Λ

In order to make $y=\sin x$ one-to-one, we restrict its domain to $[-\pi/2,\pi/2]$ ; on this domain, the range is $[-1,1]$ . Therefore the domain of $y=\sin^{-1}x$ is $[-1,1]$ and the range is $[-\pi/2,\pi/2]$ . When $x=\pm 1$ , note how the derivative of the arcsine function is undefined; this corresponds to the fact that as $x\to\pm 1$ , the tangent lines to arcsine approach vertical lines with undefined slopes.

In Figure 7.2.2 we see $f(x)=\sin x$ and $f\mkern 1.35mu ^{-1}(x)=\sin^{-1}x$ graphed on their respective domains. The line tangent to $\sin x$ at the point $(\pi/3,\sqrt{3}/2)$ has slope $\cos\pi/3=1/2$ . The slope of the corresponding point on $\sin^{-1}x$ , the point $(\sqrt{3}/2,\pi/3)$ , is

\frac{1}{\sqrt{1-(\sqrt{3}/2)^{2}}}=\frac{1}{\sqrt{1-3/4}}=\frac{1}{\sqrt{1/4}% }=\frac{1}{1/2}=2,

verifying Theorem 7.2.1 yet again: at corresponding points, a function and its inverse have reciprocal slopes.

Using similar techniques, we can find the derivatives of all the inverse trigonometric functions after first restricting their domains according to Figure 7.1.5 to allow them to be invertible.

Theorem 7.2.2 Derivatives of Inverse Trigonometric Functions

The inverse trigonometric functions are differentiable on all open sets contained in their domains (as listed in Figure 7.1.5) and their derivatives are as follows:

(a)

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sin^{-1}x% \bigr{)}=\frac{1}{\sqrt{1-x^{2}}}$
(b)

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\sec^{-1}x% \bigr{)}=\frac{1}{\left\lvert x\right\rvert\sqrt{x^{2}-1}}$
(c)

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\tan^{-1}x% \bigr{)}=\frac{1}{1+x^{2}}$
(d)

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\cos^{-1}x% \bigr{)}=-\frac{1}{\sqrt{1-x^{2}}}$
(e)

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\csc^{-1}x% \bigr{)}=-\frac{1}{\left\lvert x\right\rvert\sqrt{x^{2}-1}}$
(f)

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\bigl{(}\cot^{-1}x% \bigr{)}=-\frac{1}{1+x^{2}}$

Note how the last three derivatives are merely the negatives of the first three, respectively. Because of this, the first three are used almost exclusively throughout this text.

Example 7.2.2 Finding derivatives of inverse functions

Find the derivatives of the following functions:

\text{1.}\quad f(x)=\cos^{-1}(x^{2})\qquad\text{2.}\quad g(x)=\frac{\sin^{-1}x% }{\sqrt{1-x^{2}}}\qquad\text{3.}\quad f(x)=\sin^{-1}(\cos x)

Solution

(a)

We use Theorem 7.2.2 and the Chain Rule to find:

$f\,^{\prime}(x)=-\frac{1}{\sqrt{1-(x^{2})^{2}}}(2x)=-\frac{2x}{\sqrt{1-x^{4}}}$
(b)

We use Theorem 7.2.2 and the Quotient Rule to compute:

$\displaystyle g\mkern 1.35mu ^{\prime}(x)$ $\displaystyle=\frac{\left(\frac{1}{\sqrt{1-x^{2}}}\right)\sqrt{1-x^{2}}-(\sin^% {-1}x)\left(\frac{1}{2\sqrt{1-x^{2}}}(-2x)\right)}{\left(\sqrt{1-x^{2}}\right)% ^{2}}$

$\displaystyle=\frac{\sqrt{1-x^{2}}+x\sin^{-1}x}{\left(\sqrt{1-x^{2}}\right)^{3}}$
(c)

We apply Theorem 7.2.2 and the Chain Rule again to compute:

$\displaystyle f\,^{\prime}(x)$ $\displaystyle=\frac{1}{\sqrt{1-\cos^{2}x}}(-\sin x)$

$\displaystyle=\frac{-\sin x}{\sqrt{\sin^{2}x}}$

$\displaystyle=\frac{-\sin x}{\left\lvert\sin x\right\rvert}.$

Theorem 7.2.2 allows us to integrate some functions that we could not integrate before. For example,

\int\frac{\operatorname{d}\!x}{\sqrt{1-x^{2}}}=\sin^{-1}x+C.

Combining these formulas with $u$ -substitution yields the following:

Theorem 7.2.3 Integrals Involving Inverse Trigonometric Functions

Let $a>0$ . Then

(a)

$\displaystyle\int\frac{1}{a^{2}+x^{2}}\operatorname{d}\!x=\frac{1}{a}\tan^{-1}% \left(\frac{x}{a}\right)+C$
(b)

$\displaystyle\int\frac{1}{\sqrt{a^{2}-x^{2}}}\operatorname{d}\!x=\sin^{-1}% \left(\frac{x}{a}\right)+C$
(c)

$\displaystyle\int\frac{1}{x\sqrt{x^{2}-a^{2}}}\operatorname{d}\!x=\frac{1}{a}% \sec^{-1}\left(\frac{\left\lvert x\right\rvert}{a}\right)+C$

We will look at the second part of this theorem. The other parts are similar and are left as exercises.

First we note that the integrand involves the number $a^{2}$ , but does not explicitly involve $a$ . We make the assumption that $a>0$ in order to simplify what follows. We can rewrite the integral as follows:

\int\frac{\operatorname{d}\!x}{\sqrt{a^{2}-x^{2}}}=\int\frac{\operatorname{d}% \!x}{\sqrt{a^{2}(1-(x/a)^{2})}}=\int\frac{\operatorname{d}\!x}{a\sqrt{1-(x/a)^% {2}}}

We next use the substitution $u=x/a$ and $\operatorname{d}\!u=\operatorname{d}\!x/a$ to find:

	$\displaystyle\int\frac{\operatorname{d}\!x}{a\sqrt{1-(x/a)^{2}}}$	$\displaystyle=\int\frac{a}{a\sqrt{1-u^{2}}}\operatorname{d}\!u$
		$\displaystyle=\int\frac{\operatorname{d}\!u}{\sqrt{1-u^{2}}}$
		$\displaystyle=\sin^{-1}u+C$
		$\displaystyle=\sin^{-1}(x/a)+C$

We conclude this section with several examples.

Example 7.2.3 Finding antiderivatives involving inverse functions

Find the following integrals.

\text{1.}\quad\int\frac{\operatorname{d}\!x}{100+x^{2}}\qquad\text{2.}\quad% \int\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\operatorname{d}\!x\qquad\text{3.}\quad% \int\frac{\operatorname{d}\!x}{x^{2}+2x+5}

Solution

(a)

$\displaystyle\int\frac{\operatorname{d}\!x}{100+x^{2}}=\int\frac{\operatorname% {d}\!x}{10^{2}+x^{2}}=\frac{1}{10}\tan^{-1}(x/10)+C$
(b)

We use the substitution $u=\sin^{-1}x$ and $\operatorname{d}\!u=\frac{\operatorname{d}\!x}{\sqrt{1-x^{2}}}$ to find:

$\int\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\operatorname{d}\!x=\int u\operatorname{d% }\!u=\frac{1}{2}u^{2}+C=\frac{1}{2}\left(\sin^{-1}x\right)^{2}+C$
(c)

This does not immediately look like one of the forms in Theorem 7.2.3, but we can complete the square in the denominator to see that

$\int\frac{\operatorname{d}\!x}{x^{2}+2x+5}=\int\frac{\operatorname{d}\!x}{(x^{% 2}+2x+1)+4}=\int\frac{\operatorname{d}\!x}{4+(x+1)^{2}}$

We now use the substitution $u=x+1$ and $\operatorname{d}\!u=\operatorname{d}\!x$ to find:

$\int\frac{\operatorname{d}\!x}{4+(x+1)^{2}}=\int\frac{\operatorname{d}\!u}{4+u% ^{2}}\\ =\frac{1}{2}\tan^{-1}(u/2)+C=\frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C.$

Exercises 7.2

Terms and Concepts

1.

If $(1,10)$ lies on the graph of $y=f(x)$ and $f\,^{\prime}(1)=5$ , what can be said about $y=f^{-1}(x)$ ?
2.

Since $\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\sin^{-1}x+\cos^{-1}x)=0$ , what does this tell us about $\sin^{-1}x+\cos^{-1}x$ ?

Problems

In Exercises 3–8., an invertible function $f(x)$ is given along with a point that lies on its graph. Using Theorem 7.2.1, evaluate $\left(f^{-1}\right)^{\prime}(x)$ at the indicated value.

3.
$f(x)=5x+10$ Point $=(2,20)$ Evaluate $\left(f^{-1}\right)^{\prime}(20)$
4.
$f(x)=x^{2}-2x+4$ , $x\geq 1$ Point $=(3,7)$ Evaluate $\left(f^{-1}\right)^{\prime}(7)$
5.
$f(x)=\sin 2x$ , $-\pi/4\leq x\leq\pi/4$ Point $=(\pi/6,\sqrt{3}/2)$ Evaluate $\left(f^{-1}\right)^{\prime}(\sqrt{3}/2)$
6.
$f(x)=x^{3}-6x^{2}+15x-2$ Point $=(1,8)$ Evaluate $\left(f^{-1}\right)^{\prime}(8)$
7.
$\displaystyle f(x)=\frac{1}{1+x^{2}}$ , $x\geq 0$ Point $=(1,1/2)$ Evaluate $\left(f^{-1}\right)^{\prime}(1/2)$
8.
$f(x)=6e^{3x}$ Point $=(0,6)$ Evaluate $\left(f^{-1}\right)^{\prime}(6)$

In Exercises 9–18., compute the derivative of the given function.

9.

$h(t)=\sin^{-1}(2t)$
10.

$f(t)=\sec^{-1}(2t)$
11.

$g(x)=\tan^{-1}(2x)$
12.

$f(x)=x\sin^{-1}x$
13.

$g(t)=\sin t\cos^{-1}t$
14.

$f(t)=e^{t}\ln t$
15.

$\displaystyle h(x)=\frac{\sin^{-1}x}{\cos^{-1}x}$
16.

$g(x)=\tan^{-1}(\sqrt{x})$
17.

$f(x)=\sec^{-1}(1/x)$
18.

$f(x)=\sin(\sin^{-1}x)$

In Exercises 19–22., compute the derivative of the given function in two ways:

(a)

By simplifying first, then taking the derivative, and
(b)

by using the Chain Rule first then simplifying.

Verify that the two answers are the same.

19.

$f(x)=\sin(\sin^{-1}x)$
20.

$f(x)=\tan^{-1}(\tan x)$
21.

$f(x)=\sin(\cos^{-1}x)$
22.

$f(x)=\sin(\tan^{-1}x)$

In Exercises 23–24., find the equation of the line tangent to the graph of $f$ at the indicated $x$ value.

23.

$f(x)=\sin^{-1}x$ at $x=\frac{\sqrt{2}}{2}$
24.

$f(x)=\cos^{-1}(2x)$ at $x=\frac{\sqrt{3}}{4}$

In Exercises 25–30., compute the indicated integral.

25.

$\displaystyle\int_{1/\sqrt{2}}^{1/2}\frac{2}{\sqrt{1-x^{2}}}\operatorname{d}\!x$
26.

$\displaystyle\int_{0}^{\sqrt{3}}\frac{4}{9+x^{2}}\operatorname{d}\!x$
27.

$\displaystyle\int\frac{\sin^{-1}r}{\sqrt{1-r^{2}}}\operatorname{d}\!r$
28.

$\displaystyle\int\frac{x^{3}}{4+x^{8}}\operatorname{d}\!x$
29.

$\displaystyle\int\frac{e^{t}}{\sqrt{10-e^{2t}}}\operatorname{d}\!t$
30.

$\displaystyle\int\frac{1}{\sqrt{x}(1+x)}\operatorname{d}\!x$

31.

A regulation hockey goal is 6 feet wide. If a player is skating towards the end line on a line perpendicular to the end line and 10 feet from the imaginary line joining the center of one goal to the center of the other, the angle between the player and the goal first increases and then begins to decrease. In order to maximize this angle, how far from the end line should the player be when they shoot the puck?

	$\displaystyle g\mkern 1.35mu ^{\prime}(x)$	$\displaystyle=\frac{\left(\frac{1}{\sqrt{1-x^{2}}}\right)\sqrt{1-x^{2}}-(\sin^% {-1}x)\left(\frac{1}{2\sqrt{1-x^{2}}}(-2x)\right)}{\left(\sqrt{1-x^{2}}\right)% ^{2}}$
		$\displaystyle=\frac{\sqrt{1-x^{2}}+x\sin^{-1}x}{\left(\sqrt{1-x^{2}}\right)^{3}}$

	$\displaystyle f\,^{\prime}(x)$	$\displaystyle=\frac{1}{\sqrt{1-\cos^{2}x}}(-\sin x)$
		$\displaystyle=\frac{-\sin x}{\sqrt{\sin^{2}x}}$
		$\displaystyle=\frac{-\sin x}{\left\lvert\sin x\right\rvert}.$