# Derivatives of Inverse Circular Functions

The derivatives of inverse circular functions are given below:

# Derivative of $\sin^{-1}x$

Let $y=\sin^{-1}x$. Then $\sin y=x$

Differentiating both sides with respect to $y$, we get, $\cos y=\frac{dx}{dx}$ $\sqrt{1-\sin^2 y}=\frac{dx}{dy}$ $\sqrt{1-x^2}=\frac{dx}{dy}$ $\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$ $\text{or, }\frac{d(\sin^{-1}x)}{dx}=\frac{1}{\sqrt{1-x^2}}$

# Derivative of $\cos^{-1}x$

Let $y=\cos^{-1}x$. Then $\cos y=x$

Differentiating both sides with respect to $y$, we get, $-\sin y=\frac{dx}{dy}$ $-\sqrt{1-\cos^2y}=\frac{dx}{dy}$ $-\sqrt{1-x^2}=\frac{dx}{dy}$ $\therefore \frac{dy}{dx}=\frac{-1}{\sqrt{1-x^2}}$ $\text{or, }\frac{d(\cos^{-1}x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$

# Derivative of $\tan^{-1}x$

Let $y=\tan^{-1}x$. Then $\tan y=x$

Differentiating both sides with respect to $y$, we get, $\sec^2 y=\frac{dx}{dy}$ $1+\tan^2 y=\frac{dx}{dy}$ $1+x^2=\frac{dx}{dy}$ $\therefore \frac{dy}{dx}=\frac{1}{1+x^2}$ $\text{or, }\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}$

# Derivative of $\operatorname{cosec}^{-1}x$

Let $y=\operatorname{cosec}^{-1}x$. Then $\operatorname{cosec}y=x$

Differentiating both sides with respect to $y$, we get, $-\operatorname{cosec}y.\cot y=\frac{dx}{dy}$ $-\operatorname{cosec}y(\sqrt{\operatorname{cosec}^2y-1})=\frac{dx}{dy}$ $\frac{dx}{dy}=-x\sqrt{x^2-1}$ $\therefore \frac{dy}{dx}=\frac{-1}{x\sqrt{x^2-1}}$ $\text{or, }\frac{d(\operatorname{cosec}^{-1}x)}{dx}=\frac{-1}{x\sqrt{x^2-1}}$

# Derivative of $\sec^{-1}x$

Let $y=\sec^{-1}x$. Then $\sec y=x$

Differentiating both sides with respect to $y$, we get, $\sec y.\tan y=\frac{dx}{dy}$ $\sec y(\sqrt{\sec^2y-1})=\frac{dx}{dy}$ $\frac{dx}{dy}=x\sqrt{x^2-1}$ $\therefore \frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$ $\text{or, }\frac{d(\sec^{-1}x)}{dx}=\frac{1}{x\sqrt{x^2-1}}$

# Derivative of $\cot^{-1}x$

Let $y=\cot^{-1}x$. Then $\cot y=x$

Differentiating both sides with respect to $y$, we get, $-\operatorname{cosec}^2y=\frac{dx}{dy}$ $-(1+\cot^2y)=\frac{dx}{dy}$ $-(1+x^2)=\frac{dx}{dy}$ $\therefore \frac{dy}{dx}=\frac{-1}{1+x^2}$ $\text{or, }\frac{d(\cot^{-1}x)}{dx}=\frac{-1}{1+x^2}$

## Example: Find the derivative of $\sec^{-1}(\tan x)$.

$\frac{d[\sec^{-1}(\tan x)]}{dx}$$=\frac{1}{\tan x\sqrt{\tan^2x-1}}.\frac{d(\tan x)}{dx}$ $=\frac{1}{\tan x\sqrt{\tan^2x-1}}.\sec^2x$ $=\frac{\sec^2x}{\tan x\sqrt{\tan^2x-1}}$

## Example: Find the derivative of $\tan^{-1}\frac{2x}{1-x^2}$.

Let $y=\tan^{-1}\frac{2x}{1-x^2}$. Put $x=\tan\theta$ $y=\tan^{-1}\frac{2\tan\theta}{1-\tan^2\theta}$ $y=\tan^{-1}(\tan 2\theta)$ $y=2\theta$ $y=2\tan^{-1}x$ $\therefore \frac{dy}{dx}=\frac{2}{1+x^2}$