# 3.7 Derivatives of inverse functions  (Page 2/3)

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## Extending the power rule to rational exponents

The power rule may be extended to rational exponents. That is, if $n$ is a positive integer, then

$\frac{d}{dx}\left({x}^{1\text{/}n}\right)=\frac{1}{n}{x}^{\left(1\text{/}n\right)-1}.$

Also, if $n$ is a positive integer and $m$ is an arbitrary integer, then

$\frac{d}{dx}\left({x}^{m\text{/}n}\right)=\frac{m}{n}{x}^{\left(m\text{/}n\right)-1}.$

## Proof

The function $g\left(x\right)={x}^{1\text{/}n}$ is the inverse of the function $f\left(x\right)={x}^{n}.$ 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,

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

Finally,

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

To differentiate ${x}^{m\text{/}n}$ we must rewrite it as ${\left({x}^{1\text{/}n}\right)}^{m}$ and apply the chain rule. Thus,

$\frac{d}{dx}\left({x}^{m\text{/}n}\right)=\frac{d}{dx}\left({\left({x}^{1\text{/}n}\right)}^{m}\right)=m{\left({x}^{1\text{/}n}\right)}^{m-1}·\frac{1}{n}{x}^{\left(1\text{/}n\right)-1}=\frac{m}{n}{x}^{\left(m\text{/}n\right)-1}.$

## Applying the power rule to a rational power

Find the equation of the line tangent to the graph of $y={x}^{2\text{/}3}$ at $x=8.$

First find $\frac{dy}{dx}$ and evaluate it at $x=8.$ Since

$\frac{dy}{dx}=\frac{2}{3}{x}^{-1\text{/}3}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}|\begin{array}{l}\\ {}_{x=8}\end{array}=\frac{1}{3}$

the slope of the tangent line to the graph at $x=8$ is $\frac{1}{3}.$

Substituting $x=8$ into the original function, we obtain $y=4.$ Thus, the tangent line passes through the point $\left(8,4\right).$ Substituting into the point-slope formula for a line, we obtain the tangent line

$y=\frac{1}{3}x+\frac{4}{3}.$

Find the derivative of $s\left(t\right)=\sqrt{2t+1}.$

${s}^{\prime }\left(t\right)={\left(2t+1\right)}^{\text{−}1\text{/}2}$

## Derivatives of inverse trigonometric functions

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 $g\left(x\right)={\text{sin}}^{-1}x.$

Since for $x$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right],f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ is the inverse of $g\left(x\right)={\text{sin}}^{-1}x,$ begin by finding ${f}^{\prime }\left(x\right).$ Since

${f}^{\prime }\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(g\left(x\right)\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left({\text{sin}}^{-1}x\right)=\sqrt{1-{x}^{2}},$

we see that

${g}^{\prime }\left(x\right)=\frac{d}{dx}\left({\text{sin}}^{-1}x\right)=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)}=\frac{1}{\sqrt{1-{x}^{2}}}.$

## Applying the chain rule to the inverse sine function

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

Applying the chain rule to $h\left(x\right)={\text{sin}}^{-1}\left(g\left(x\right)\right),$ we have

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

Now let $g\left(x\right)=2{x}^{3},$ so ${g}^{\prime }\left(x\right)=6x.$ Substituting into the previous result, we obtain

$\begin{array}{cc}\hfill {h}^{\prime }\left(x\right)& =\frac{1}{\sqrt{1-4{x}^{6}}}·6x\hfill \\ & =\frac{6x}{\sqrt{1-4{x}^{6}}}.\hfill \end{array}$

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

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

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

$\phantom{\rule{0.4em}{0ex}}\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{-1}x=\frac{1}{\sqrt{1-{\left(x\right)}^{2}}}$
$\phantom{\rule{0.29em}{0ex}}\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{-1}x=\frac{-1}{\sqrt{1-{\left(x\right)}^{2}}}$
$\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{-1}x=\frac{1}{1+{\left(x\right)}^{2}}$
$\phantom{\rule{0.23em}{0ex}}\frac{d}{dx}{\text{cot}}^{-1}x=\frac{-1}{1+{\left(x\right)}^{2}}$
$\phantom{\rule{1.5em}{0ex}}\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{-1}x=\frac{1}{|x|\sqrt{{\left(x\right)}^{2}-1}}$
$\phantom{\rule{1.75em}{0ex}}\frac{d}{dx}\phantom{\rule{0.1em}{0ex}}{\text{csc}}^{-1}x=\frac{-1}{|x|\sqrt{{\left(x\right)}^{2}-1}}$

## Applying differentiation formulas to an inverse tangent function

Find the derivative of $f\left(x\right)={\text{tan}}^{-1}\left({x}^{2}\right).$

Let $g\left(x\right)={x}^{2},$ so ${g}^{\prime }\left(x\right)=2x.$ Substituting into [link] , we obtain

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

Simplifying, we have

${f}^{\prime }\left(x\right)=\frac{2x}{1+{x}^{4}}.$

## Applying differentiation formulas to an inverse sine function

Find the derivative of $h\left(x\right)={x}^{2}{\text{sin}}^{-1}x.$

By applying the product rule, we have

${h}^{\prime }\left(x\right)=2x\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{-1}x+\frac{1}{\sqrt{1-{x}^{2}}}·{x}^{2}.$

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