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

 Page 2 / 3

## 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

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

#### Questions & Answers

how does this work
Brad Reply
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
Cosmos Reply
log tan (x/4+x/2)
Rohan
please answer
Rohan
y=(x^2 + 3x).(eipix)
Claudia
is this a answer
Ismael
A Function F(X)=Sinx+cosx is odd or even?
WIZARD Reply
neither
David
Neither
Lovuyiso
f(x)=1/1+x^2 |=[-3,1]
Yuliana Reply
apa itu?
fauzi
determine the area of the region enclosed by x²+y=1,2x-y+4=0
Gerald Reply
Hi
MP
Hi too
Vic
hello please anyone with calculus PDF should share
Adegoke
Which kind of pdf do you want bro?
Aftab
hi
Abdul
can I get calculus in pdf
Abdul
How to use it to slove fraction
Tricia Reply
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
Jean
Im not in college but this will still help
nothing
how can we scatch a parabola graph
Dever Reply
Ok
Endalkachew
how can I solve differentiation?
Sir Reply
with the help of different formulas and Rules. we use formulas according to given condition or according to questions
CALCULUS
For example any questions...
CALCULUS
v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1
log tan (x/4+x/2)
Rohan
what is the procedures in solving number 1?
Vier Reply
review of funtion role?
Md Reply
for the function f(x)={x^2-7x+104 x<=7 7x+55 x>7' does limx7 f(x) exist?
find dy÷dx (y^2+2 sec)^2=4(x+1)^2
Rana Reply
Integral of e^x/(1+e^2x)tan^-1 (e^x)
naveen Reply
why might we use the shell method instead of slicing
Madni Reply
fg[[(45)]]²+45⅓x²=100
albert Reply

### Read also:

#### Get the best Calculus volume 1 course in your pocket!

Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

 By OpenStax By Bonnie Hurst By OpenStax By OpenStax By JavaChamp Team By David Bourgeois By Janet Forrester By OpenStax By OpenStax By