Question

Find the differential coefficient of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ with respect to ${\tan }^{-1}x$

Answer 1

Let $y={\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta }$
where $x=\tan \theta$
$\therefore \theta ={\tan }^{-1}x$
$y={\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta }={\sin }^{-1}\sin 2\theta =2\theta$
$y=2\times {\tan }^{-1}x$
differentiate with respect to $x$
$\frac{dy}{dx}=2\times \frac{d}{dx}{\tan }^{-1}x$
$=\frac{2}{1}\times \frac{1}{1+x^2}$
$\frac{dy}{dx}=\frac{2}{1+x^2}...\left(1\right)$
let $t={\tan }^{-1}x$
differentiate with respect to $x$
$\frac{dt}{dx}=\frac{d}{dx}{\tan }^{-1}x$
$=\frac{1}{1+x^2}...\left(2\right)$
$\frac{dy}{dx}=\frac{dy}{dx}\times \frac{dx}{dt}=\frac{2}{1+x^2}\times \frac{1+x^2}{1}=2$