## Section2.7Derivatives of Inverse Functions

Recall that a function $$y=f(x)$$ is said to be one-to-one if it passes the horizontal line test; that is, for two different $$x$$ values $$x_1$$ and $$x_2\text{,}$$ we do not have $$f\mathopen{}\left(x_1\right)\mathclose{}=f\mathopen{}\left(x_2\right)\mathclose{}\text{.}$$ In some cases the domain of $$f$$ must be restricted so that it is one-to-one. For instance, consider $$f(x)=x^2\text{.}$$ Clearly, $$f(-1)= f(1)\text{,}$$ so $$f$$ is not one-to-one on its regular domain, but by restricting $$f$$ to $$(0,\infty)\text{,}$$ $$f$$ is one-to-one.

Now recall that one-to-one functions have inverses. That is, if $$f$$ is one-to-one, it has an inverse function, denoted by $$f^{-1}\text{,}$$ such that if $$f(a)=b\text{,}$$ then $$f^{-1}(b) = a\text{.}$$ The domain of $$f^{-1}$$ is the range of $$f\text{,}$$ and vice-versa. For ease of notation, we set $$g=f^{-1}$$ and treat $$g$$ as a function of $$x\text{.}$$

Since $$f(a)=b$$ implies $$g(b)=a\text{,}$$ when we compose $$f$$ and $$g$$ we get a nice result:

\begin{equation*} f\big(g(b)\big) = f(a) = b\text{.} \end{equation*}

In general, $$f\big(g(x)\big) = x$$ and $$g\big(f(x)\big) = x\text{.}$$ This gives us a convenient way to check if two functions are inverses of each other: compose them and if the result is $$x$$ (on the appropriate domains), then they are inverses.

For a refresher on how to determine the inverse of a given function, watch the following video.

When the point $$(a,b)$$ lies on the graph of $$f\text{,}$$ the point $$(b,a)$$ lies on the graph of $$g\text{.}$$ This leads us to discover that the graph of $$g$$ is the reflection of $$f$$ across the line $$y=x\text{.}$$ In Figure 2.7.4 we see a function graphed along with its inverse. See how the point $$(1,1.5)$$ lies on one graph, whereas $$(1.5,1)$$ lies on the other. Because of this relationship, whatever we know about $$f$$ can quickly be transferred into knowledge about $$g\text{.}$$

For example, consider Figure 2.7.5 where the tangent line to $$f$$ at the point $$(1,1.5)$$ is drawn. That line has slope $$3\text{.}$$ Through reflection across $$y=x\text{,}$$ we can see that the tangent line to $$g$$ at the point $$(1.5,1)$$ has slope $$1/3\text{.}$$ Their slopes are reciprocals. This should make sense since reflecting a line (such as a tangent line) across the line $$y=x$$ switches the $$x$$ and $$y$$ values. Also consider the point $$(0,0.5)$$ on the graph of $$f\text{,}$$ where the tangent line is horizontal. At the point $$(0.5,0)$$ on $$g\text{,}$$ the tangent line is vertical.

More generally, consider the tangent line to $$f$$ at the point $$(a,b)\text{.}$$ That line has slope $$\fp(a)\text{.}$$ Through reflection across $$y=x\text{,}$$ we can extend our above observation to say that the tangent line to $$g$$ at the point $$(b,a)$$ should have slope $$1/\fp(a)\text{.}$$ This then tells us that $$g'(b)=1/\fp(a)\text{.}$$

The information from these two graphs is summarized in Table 2.7.6 below:

We have discovered a relationship between $$\fp$$ and $$g'$$ in a mostly graphical way. We can realize this relationship analytically as well. Let $$y = g(x)\text{,}$$ where again $$g = f^{-1}\text{.}$$ We want to find $$y'\text{.}$$ Since $$y = g(x)\text{,}$$ we know that $$f(y) = x\text{.}$$ Using the The Chain Rule and Implicit Differentiation, take the derivative of both sides of this last equality.

\begin{align*} \lzoo{x}{f(y)} \amp = \lzoo{x}{x}\\ \fp(y)\cdot y' \amp = 1\\ y' \amp = \frac{1}{\fp(y)}\\ y' \amp = \frac{1}{\fp(g(x))}\text{.} \end{align*}

This leads us to the following theorem.

The results of Theorem 2.7.7 are not trivial; the notation may seem confusing at first. Careful consideration, along with examples, should earn understanding.

In the next example we apply Theorem 2.7.7 to the arcsine function.

A word of caution is required here. The function $$\sin(x)$$ is clearly not one-to-one. How can we say that $$\arcsin(x)$$ is the inverse of $$\sin(x)\text{?}$$ To make sense of this, we employ a technique known as restriction of domain: instead of considering the entire domain of the sine function, we consider a portion of it, on which the function is one-to-one, as explained in the following video.

###### Example2.7.10.Finding the derivative of an inverse trigonometric function.

Let $$y = \arcsin(x) = \sin^{-1}(x)\text{.}$$ Find $$y'$$ using Theorem 2.7.7.

Solution

Adopting our previously defined notation, let $$g(x) = \arcsin(x)$$ and $$f(x) = \sin(x)\text{.}$$ Thus $$\fp(x) = \cos(x)\text{.}$$ Applying the theorem, we have

\begin{align*} g'(x) \amp = \frac{1}{\fp(g(x))}\\ \amp = \frac{1}{\cos(\arcsin(x))}\text{.} \end{align*}

This last expression is not immediately illuminating. Drawing a figure will help, as shown in Figure 2.7.11. 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 $$y=\arcsin(x)$$ can be rewritten as $$y=\arcsin(x/1)\text{;}$$ that is, consider a right triangle where the hypotenuse has length $$1$$ and the side opposite of the angle with measure $$y$$ has length $$x\text{.}$$ This means the final side has length $$\sqrt{1-x^2}\text{,}$$ using the Pythagorean Theorem.

Therefore

\begin{align*} \cos\mathopen{}\left(\sin^{-1}(x)\right)\mathclose{} \amp= \cos(y)\\ \amp = \frac{\sqrt{1-x^2}}{1}\\ \amp = \sqrt{1-x^2}\text{,} \end{align*}

resulting in

\begin{equation*} \lzoo{x}{\arcsin(x)} = \frac{1}{\sqrt{1-x^2}}\text{.} \end{equation*}

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 never be greater than $$1\text{.}$$ Therefore the inside of the square root will never be negative.

In order to make $$y=\sin(x)$$ one-to-one, we restrict its domain to $$[-\pi/2,\pi/2]\text{;}$$ on this domain, the range is $$[-1,1]\text{.}$$ Therefore the domain of $$y=\arcsin(x)$$ is $$[-1,1]$$ and the range is $$[-\pi/2,\pi/2]\text{.}$$ When $$x=\pm 1\text{,}$$ note how the derivative of the arcsine function is undefined; this corresponds to the fact that as $$x\to \pm1\text{,}$$ the tangent lines to arcsine approach vertical lines with undefined slopes.

In Figure 2.7.12 we see $$f(x) = \sin(x)$$ and $$f^{-1}(x) = \sin^{-1}(x)$$ graphed on their respective domains. The line tangent to $$\sin(x)$$ at the point $$\left(\pi/3, \sqrt{3}/2\right)$$ has slope $$\cos(\pi)/3 = 1/2\text{.}$$ The slope of the corresponding point on $$\sin^{-1}(x)\text{,}$$ the point $$\left(\sqrt{3}/2,\pi/3\right)\text{,}$$ is

\begin{align*} \frac{1}{\sqrt{1-\left(\sqrt{3}/2\right)^2}} \amp = \frac{1}{\sqrt{1-3/4}}\\ \amp = \frac{1}{\sqrt{1/4}}\\ \amp = \frac{1}{1/2}=2\text{,} \end{align*}

verifying yet again that 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. In Table 2.7.13 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible.

Note how each derivative is the negative of the derivative of its “co” function. Because of this, derivatives of $$\sin^{-1}(x)\text{,}$$ $$\tan^{-1}(x)\text{,}$$ and $$\sec^{-1}(x)$$ are used almost exclusively throughout this text.

In Section 2.3, we stated without proof or explanation that $$\lzoo{x}{\ln(x)}=\frac{1}{x}\text{.}$$ We can justify that now using Theorem 2.7.7, as shown in the example.

###### Example2.7.16.Finding the derivative of $$y=\ln(x)$$.

Use Theorem 2.7.7 to compute $$\lzoo{x}{\ln(x)}\text{.}$$

Solution

View $$y= \ln(x)$$ as the inverse of $$y = e^x\text{.}$$ Therefore, using our standard notation, let $$f(x) = e^x$$ and $$g(x) = \ln(x)\text{.}$$ We wish to find $$g'(x)\text{.}$$ Theorem 2.7.7 gives:

\begin{align*} g'(x) \amp = \frac{1}{\fp(g(x))}\\ \amp = \frac{1}{e^{\ln(x) }}\\ \amp = \frac{1}{x}\text{.} \end{align*}

In this chapter we have defined the derivative, given rules to facilitate its computation, and given the derivatives of a number of standard functions. We restate the most important of these in the following theorem, intended to be a reference for further work.

### ExercisesExercises

###### 1.

• True

• False

Every function has an inverse.

###### 2.

In your own words explain what it means for a function to be “one-to-one.”

###### 3.

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

###### 4.

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

###### Problems

Verify that the given functions are inverses.

###### 5.

$$f(x) = 2x+6$$ and $$g(x) = \frac{1}{2}x-3$$

###### 6.

$$f(x) = x^2+6x+11\text{,}$$ $$x\geq 3$$ and $$g(x) = \sqrt{x-2}-3\text{,}$$ $$x\geq 2$$

###### 7.

$$f(x) = \frac{3}{x-5}\text{,}$$ $$x\neq 5$$ and $$g(x) = \frac{3+5x}{x}\text{,}$$ $$x\neq 0$$

###### 8.

$$f(x) = \frac{x+1}{x-1}\text{,}$$ $$x\neq 1$$ and $$g(x) = f(x)$$

An invertible function $$f(x)$$ is given along with a point that lies on its graph. Using Theorem 2.7.7, evaluate $$\left(f^{-1}\right)'(x)$$ at the indicated value.

###### 9.

The point $${\left(2,20\right)}$$ is on the graph of $$f(x) = {5x+10}\text{.}$$ Find $$\left(f^{-1}\right)'({20})\text{.}$$

###### 10.

The point $${\left(3,7\right)}$$ is on the graph of $$f(x) = {x^{2}-2x+4}, x\geq 1\text{.}$$ Find $$\left(f^{-1}\right)'({7})\text{.}$$

###### 11.

The point $$\left({\frac{\pi }{6}},{\frac{\sqrt{3}}{2}}\right)$$ is on the graph of $$f(x) = {\sin\!\left(2x\right)}, {\frac{-\pi }{4}}\leq x\leq {\frac{\pi }{4}}\text{.}$$ Find $$\left(f^{-1}\right)'\left({\frac{\sqrt{3}}{2}}\right)\text{.}$$

###### 12.

The point $${\left(1,8\right)}$$ is on the graph of $$f(x) = {x^{3}-6x^{2}+15x-2}\text{.}$$ Find $$\left(f^{-1}\right)'({8})\text{.}$$

###### 13.

The point $${\left(1,{\textstyle\frac{1}{2}}\right)}$$ is on the graph of $$f(x) = {\frac{1}{1+x^{2}}}, x\geq 0\text{.}$$ Find $$\left(f^{-1}\right)'\left({{\textstyle\frac{1}{2}}}\right)\text{.}$$

###### 14.

The point $${\left(0,6\right)}$$ is on the graph of $$f(x) = {6e^{3x}}\text{.}$$ Find $$\left(f^{-1}\right)'({6})\text{.}$$

Compute the derivative of the given function.

###### 15.

$$h(t) = {\sin^{-1}\!\left(2t\right)}$$

###### 16.

$$f(t) = {\sec^{-1}\!\left(2t\right)}$$

###### 17.

$$g(x) = {\tan^{-1}\!\left(2x\right)}$$

###### 18.

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

###### 19.

$$g(t) = {\sin\!\left(t\right)\cos^{-1}\!\left(t\right)}$$

###### 20.

$$f(t) = \ln(t) e^t$$

###### 21.

$$h(x) = {\frac{\sin^{-1}\!\left(x\right)}{\cos^{-1}\!\left(x\right)}}$$

###### 22.

$$g(x) = {\tan\!\left(\sqrt{x}\right)}$$

###### 23.

$$f(x) = {\sec\!\left(\frac{1}{x}\right)}$$

###### 24.

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

Compute the derivative of the given function in two ways:

1. By simplifying first, then taking the derivative, and

2. by using the Chain Rule first then simplifying.

Verify that the two answers are the same.

###### 25.

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

###### 26.

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

###### 27.

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

Find the equation of the line tangent to the graph of $$f$$ at the indicated $$x$$ value.

###### 28.

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

###### 29.

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

###### 30.

Find $$\lz{y}{x}\text{,}$$ where $${x^{2}y-y^{2}x}=1\text{.}$$

###### 31.

Find the equation of the line tangent to the graph of $${x^{2}+y^{2}+xy}={7}$$ at the point $$(1,2)\text{.}$$

###### 32.

Let $$f(x) = {x^{3}+x}\text{.}$$ Find $$\lim\limits_{s\to 0} \frac{f(x+s)-f(x)}{s}\text{.}$$