# 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: $\frac{d}{dx} \csc ^{-1} x = \frac{-1}{1+x^2}$

Related posts:

1. Derivatives of Trigonometric functions. As you know, The functions SINE x(sin x) , CO-SECANT...
2. Second and higher derivatives. Think of a function y=f(x) , and let y=f(x) be...
3. The Trigonometric Functions. The trigonometric functions are sin , cos , tan ,...
4. Derivative of simple algebraic or polynomial functions. The derivative and calculations on finding derivative of simple algebraic functions...
5. The Power Rule. Power Rule is one of the Techniques of Differentiation. The...