Question

Differentiate w.r.t 'x', y=$\sec \left({\tan }^{-1}x\right)$

Answer 1

Let $y=\sec \left({\tan }^{-1}x\right)$

$\frac{dy}{dx}=\frac{d}{dx}\left(\sec \left({\tan }^{-1}x\right)\right)$

$=\sec \left({\tan }^{-1}x\right)\tan \left({\tan }^{-1}x\right)\frac{d\left({\tan }^{-1}x\right)}{dx}$

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

$=x\left(\sqrt{1+x^2}\right)\frac{1}{1+x^2}$

$\therefore \frac{dy}{dx}=\left(\frac{x}{1+x^2}\right)\sec \left({\tan }^{-1}x\right)=\frac{x}{\sqrt{1+x^2}}$