Question

If $y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$, then find $\frac{dy}{dx}$.

Answer 1

Let $x=\tan \theta$

$\Rightarrow y={\tan }^{-1}\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]$

$y={\tan }^{-1}\left(\tan 2\theta \right)$

$y=2\theta$

$y=2{\tan }^{-1}x$

Differentiate w.r.t to $x$

$\frac{dy}{dx}=2\left[\frac{1}{1+x^2}\right]$

$\Rightarrow \frac{dy}{dx}=\frac{2}{1+x}$.