Question

Differentiate the following w.r.t. $x$:
$\sin \left({\tan }^{-1}x\right)$

• A. $\cos \left({\tan }^{-1}x\right)$
• B. $-\cos \left({\tan }^{-1}x\right)$
• C. $\cos \left({\tan }^{-1}x\right)\frac{1}{1+x^2}$
• D. $-\cos \left({\tan }^{-1}x\right)\frac{1}{1+x^2}$

Answer 1

The correct option is C $\cos \left({\tan }^{-1}x\right)\frac{1}{1+x^2}$

Given equation is $\sin \left({\tan }^{-1}x\right)$

$\frac{d}{dx}\left[\sin \left({\tan }^{-1}x\right)\right]=\cos ({\tan }^{-1}x)\left(\frac{d}{dx}{\tan }^{-1}x\right)=\cos \left({\tan }^{-1}x\right)\frac{1}{1+x^2}$

Since $\frac{d}{dx}\left(\sin \right(f\left(x\right))=\cos (f\left(x\right)\left)\frac{d}{dx}\right(f\left(x\right))$

Therefore derivative of $\sin \left({\tan }^{-1}x\right)$ with respect to $x$ is

$\cos \left({\tan }^{-1}x\right)\frac{1}{1+x^2}$