# Derivatives of Inverse Trigonometric functions

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

The derivatives of inverse trigonometric functions are also algebraic (even though the inverse trigonometric functions are themselves transcendental).

The process of finding the derivatives of a function by using the differentiation from first principles or by delta method. In this section we will find the derivatives of
$sin^{-1}(x), cos^{-1}(x),tan^{-1}(x),sec^{-1}(x),csc^{-1}(x) and cot^{-1}(x)$.

Derivatives of inverse trigonometric functions
1) If f(x) = $sin^{-1}(x) = arcsin(x)$
$x\in (-1,1) and \frac{-\pi}{2} \leq f(x)\leq \frac{\pi}{2}$ then,
f '(x) = $\frac{1}{\sqrt{1-x^{2}}}$

2) If f(x) = $cos^{-1}(x) = arccos(x)$
$x\in (-1,1) and 0 \leq f(x) \leq \pi$ then,
f '(x) = $\frac{-1}{\sqrt{1-x^{2}}}$

3) If f(x) = $tan^{-1}(x) = arctan(x),\frac{-\pi}{2} < f(x) < \frac{\pi}{2}$ then,
f '(x) = $\frac{1}{1+x^{2}}$

4) If f(x) = $cot^{-1}(x) = arccot(x),0 < f(x) < \pi$ then,
f '(x) = $\frac{-1}{1+x^{2}}$

5) If f(x) = $sec^{-1}(x) = arcsec(x)$
$x\in R -[-1,1] , 0 \leq f(x) \leq \pi and f(x)\neq \frac{\pi}{2}$ then,
f '(x) = $\frac{1}{x\sqrt{x^{2}-1}}$

6) If f(x) = $csc^{-1}(x) = arccsc(x)$
$x\in R -[-1,1] , \frac{-\pi}{2} \leq f(x) \leq \frac{\pi}{2} and f(x) \neq 0$ then,
f '(x) = $\frac{-1}{x\sqrt{x^{2}-1}}$

Here we will prove the derivative of inverse trigonometric function using the definition of inverse trigonometric function.

Prove that : If $x \in (-1,1)$,then the differentiation of $sin^{-1}x$ with respect to x is $\frac{1}{\sqrt{1-x^{2}}}$
i.e $\frac{\text{d}}{\text{d}x}(sin^{-1}x) = \frac{1}{\sqrt{1-x^{2}}}$

Proof :
Let y = $sin^{-1}x$
Take sine on both side so we get,
sin(y) = sin($sin^{-1}x$ )
Sin(y) = x
Now differentiate with respect to y,
$\frac{\text{d}}{\text{d}y}(sin(y)) = \frac{\text{d}x}{\text{d}y}$

Cos(y) = $\frac{\text{d}x}{\text{d}y}$ -----(1)

As we know that $sin^{2}\theta + cos^{2}\theta$ = 1

So $sin^{2}y + cos^{2}y$ = 1

$cos^{2}y = 1 - sin^{2}y$
$cos^{2}y = 1 - x^{2}$
$cos(y) = \sqrt{1 - x^{2}}$
So equation (1) becomes,

$\sqrt{1 - x^{2}} = \frac{\text{d}x}{\text{d}y}$

$\frac{\text{d}y}{\text{d}x} = \frac{1}{\frac{\text{d}x}{\text{d}y}}$

$\frac{\text{d}y}{\text{d}x} = \frac{1}{\sqrt{1 - x^{2}}}$

$\frac{\text{d}}{\text{d}x}(sin^{-1}x) = \frac{1}{\sqrt{1-x^{2}}}$

Prove that : If $x \in (-1,1)$,then the differentiation of $cos^{-1}x$ with respect to x is $-\frac{1}{\sqrt{1-x^{2}}}$
i.e $\frac{\text{d}}{\text{d}x}(cos^{-1}x) = -\frac{1}{\sqrt{1-x^{2}}}$

Proof :
Let y = $cos^{-1}x$
x = cos(y)
Now differentiate with respect to x and use a chain rule ,
1 = $\frac{\text{d}}{\text{d}x}(cos(y))$

1= -sin(y). $\frac{\text{d}y}{\text{d}x}$

$\frac{\text{d}y}{\text{d}x} = \frac{1}{-sin(y)}$ ---(1)

As we know that $sin^{2}\theta + cos^{2}\theta$ = 1

So $sin^{2}y + cos^{2}y$ = 1

$sin^{2}y = 1 - cos^{2}y$
$sin^{2}y = 1 - x^{2}$
$sin(y) = \sqrt{1 - x^{2}}$
$sin(y) = \sqrt{1 - x^{2}}$
So equation (1) becomes,

$\frac{\text{d}y}{\text{d}x} = \frac{1}{\bf-\sqrt{1 - x^{2}}}$

$\frac{\text{d}}{\text{d}x}(cos^{-1}x) = \bf-\frac{1}{\sqrt{1-x^{2}}}$

Note : Similarly you can prove the other inverse trigonometric functions.

## Examples on Derivatives of Inverse Trigonometric Functions

Find the derivative of the following function

Example 1 : f(x) = arcsin(2x)
Solution : f(x) = arcsin(2x)
Here we will apply the chain rule as well as derivative of inverse trigonometric function.

$\frac{\text{d}}{\text{d}x}(arcsin(2x))=\frac{1}{\sqrt{1-(2x)^{2}}}.\frac{\text{d}}{\text{d}x}(2x)$

= $\frac{1}{\sqrt{1-4x^{2}}}.2$

$\frac{\text{d}}{\text{d}x}(arcsin(2x))=\frac{\text{d}}{\text{d}x}sin^{-1}(2x)=\frac{2}{\sqrt{1-4x^{2}}}$

Example 2 : f(x) = arccos(x-1)
Solution : f(x) = arccos(x-1)
Here we will apply the chain rule as well as derivative of inverse trigonometric function.

$\frac{\text{d}}{\text{d}x}(arccos(x-1))=\frac{\bf-1}{\sqrt{1-(x-1)^{2}}}.\frac{\text{d}}{\text{d}x}(x-1)$

= $\frac{\bf-1}{\sqrt{1-(x-1)^{2}}}.1$

$\frac{\text{d}}{\text{d}x}(arccos(x-1))=\frac{\text{d}}{\text{d}x}cos^{-1}(x-1)=\frac{\bf-1}{\sqrt{1-(x-1)^{2}}}$

Example 3 : f(x) = $x^{2}.arctan(x)$
Solution : f(x) = $x^{2}.arctan(x)$
Here we will apply the product rule as well as derivative of inverse trigonometric function.

$\frac{\text{d}}{\text{d}x}(x^{2}.arctan(x))=arctan(x).\frac{\text{d}}{\text{d}x}x^{2} + x^{2}.\frac{\text{d}}{\text{d}x}(arctan(x))$

= $arctan(x).2x + x^{2}.\frac{1}{1 + x^{2}}$

$\frac{\text{d}}{\text{d}x}(x^{2}.arctan(x))= 2x.arctan(x) + \frac{x^{2}}{1 + x^{2}}$