# 3.7 Derivatives of inverse functions

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

${\left({f}^{-1}\right)}^{\prime }\left(a\right)=\frac{1}{{f}^{\prime }\left({f}^{-1}\left(a\right)\right)}.$

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

$1={f}^{\prime }\left({f}^{-1}\left(x\right)\right)\left({f}^{-1}{\right)}^{\prime }\left(x\right)\right).$

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

${\left({f}^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{{f}^{\prime }\left({f}^{-1}\left(x\right)\right)}.$

We summarize this result in the following theorem.

## Inverse function theorem

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

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

Alternatively, if $y=g\left(x\right)$ is the inverse of $f\left(x\right),$ then

$g\left(x\right)=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)}.$

## Applying the inverse function theorem

Use the inverse function theorem to find the derivative of $g\left(x\right)=\frac{x+2}{x}.$ Compare the resulting derivative to that obtained by differentiating the function directly.

The inverse of $g\left(x\right)=\frac{x+2}{x}$ is $f\left(x\right)=\frac{2}{x-1}.$ Since ${g}^{\prime }\left(x\right)=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)},$ begin by finding ${f}^{\prime }\left(x\right).$ Thus,

${f}^{\prime }\left(x\right)=\frac{-2}{{\left(x-1\right)}^{2}}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(g\left(x\right)\right)=\frac{-2}{{\left(g\left(x\right)-1\right)}^{2}}=\frac{-2}{{\left(\frac{x+2}{x}-1\right)}^{2}}=-\frac{{x}^{2}}{2}.$

Finally,

${g}^{\prime }\left(x\right)=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)}=-\frac{2}{{x}^{2}}.$

We can verify that this is the correct derivative by applying the quotient rule to $g\left(x\right)$ to obtain

${g}^{\prime }\left(x\right)=-\frac{2}{{x}^{2}}.$

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

${g}^{\prime }\left(x\right)=-\frac{1}{{\left(x+2\right)}^{2}}$

## Applying the inverse function theorem

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

The function $g\left(x\right)=\sqrt[3]{x}$ is the inverse of the function $f\left(x\right)={x}^{3}.$ Since ${g}^{\prime }\left(x\right)=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)},$ begin by finding ${f}^{\prime }\left(x\right).$ Thus,

${f}^{\prime }\left(x\right)=3{x}^{3}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(g\left(x\right)\right)=3{\left(\sqrt[3]{x}\right)}^{2}=3{x}^{2\text{/}3}.$

Finally,

${g}^{\prime }\left(x\right)=\frac{1}{3{x}^{2\text{/}3}}=\frac{1}{3}{x}^{-2\text{/}3}.$

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

$g\left(x\right)=\frac{1}{5}{x}^{\text{−}4\text{/}5}$

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

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