# Derivatives of inverse trigonometric functions

Inverse trigonometric functions  are the  inverse of trigonometric functions .

For example if, y = sinx  then the inverse function of y = sinx is , is denoted by: x=sin-1y and is called inverse sin function.

You should note that: $y = \sin ^{-1} x$ doesn’t means $y = \frac{1}{\sin x}$ instead

“y = sin -1 x” is the inverse function of “x = sin y”

Derivatives of Inverse Trigonometric Functions:

The derivatives of inverse sine , inverse cos , inverse tan , inverse csc , inverse sec , inverse cot functions are given below:

Derivative of inverse sin function: $\frac{d}{dx} \sin ^{-1} x = \frac{1}{\sqrt{1-x^2}}$

proof:

If , $f(x) = y = \sin ^{-1} x$ then the function is called inverse sin function.

If, $y = \sin ^{-1} x$ then , $x = \sin y$

now if we differentiate $x = sin y$ with respect to x , using implicit differentiation technique then,

$\frac{d}{dx} x = \frac{d}{dx} \sin y$

$or, 1 = \frac{d}{dy} \sin y \times \frac{d}{dx} y$

$or, \frac{d}{dx} y = \frac{1}{cos y}$

Now using the trigonometric formula, $\cos x = \sqrt{1-(sin x)^2}$

$\frac{d}{dx} y = \frac{1}{\sqrt{1-(sin y)^2}}$

Now as , sin y = x

$or, \frac{d}{dx} y = \frac{1}{\sqrt{1-(sin y)^2}} = \frac{1}{\sqrt{1-x^2}}$

Thus , $\frac{d}{dx} \sin ^{-1} x = \frac{1}{\sqrt{1-x^2}}$

Derivative of inverse cos function: $\frac{d}{dx} \cos ^{-1} x = \frac{-1}{\sqrt{1-x^2}}$

proof:

If , $f(x) = y = \cos ^{-1} x$ then the function is called inverse cos function.

And, If, $y = \cos ^{-1} x$ then we can also rewrite is as: $x = \cos y$

now if we differentiate $x = \cos y$ with respect to x , using implicit differentiation technique then,

$\frac{d}{dx} x = \frac{d}{dx} \cos y$

$or, 1 = \frac{d}{dy} \cos y \times \frac{d}{dx} y$

$or, \frac{d}{dx} y = \frac{1}{- \sin y} = \frac{-1}{ \sin y}$

Now using the trigonometric formula, $\sin x = \sqrt{1-(cos x)^2}$

$\frac{d}{dx} y = \frac{-1}{\sqrt{1-(cos y)^2}}$

Now as , cos y = x

$or, \frac{d}{dx} y = \frac{-1}{\sqrt{1-(cos y)^2}} = \frac{-1}{\sqrt{1-x^2}}$

Thus , $\frac{d}{dx} \cos ^{-1} x = \frac{-1}{\sqrt{1-x^2}}$

Derivative of inverse tan function: $\frac{d}{dx} \tan ^{-1} x = \frac{1}{1+x^2}$

proof:

When , $y = \tan ^{-1} x$ then the function “f” or y is called inverse tan function.
and we can also equally re-write above function as: $x = \tan y$
If we differentiate both L.H.S and R.H.S of the equation $x = \tan y$ with respect
to “y”.
then,

$\frac{d}{dy} x = \frac{d}{dy} \tan y$

$or, \frac{d}{dy} x = \sec ^2 y = 1 + \tan ^2 y = 1+x^2$

Now using the concept of differentials we can re write above equation as:

$\frac{d}{dx} y = \frac{1}{1+x^2}$

Thus , $\frac{d}{dx} \tan ^{-1} x = \frac{1}{1+x^2}$

Derivative of inverse csc , inverse sec & inverse cot functions:

We can use the similar method we used above to find derivative of inverse sin , cos and tan function to find the derivatives of inverse csc , sec and cot function.
After differentiation we get following result:
Derivative of inverse csc function: $\frac{d}{dx} \csc ^{-1} x = \frac{-1}{x \sqrt{x^2-1}}$

Derivative of inverse sec function: $\frac{d}{dx} \sec ^{-1} x = \frac{1}{x \sqrt{x^2-1}}$

Derivative of inverse cot function: $\frac{d}{dx} \csc ^{-1} x = \frac{-1}{1+x^2}$

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