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

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