Question

Find $dy/dx$ if $y={\tan }^{-1}{\left(\frac{3x-x^3}{1-3x^2}\right)}^{\prime }\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{3}}$.

Answer 1

put $x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$

$y={\tan }^{-1}\left[\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right]$

$y={\tan }^{-1}\left[\tan 3\theta \right]$

$y=3\theta =3\left({\tan }^{-1}x\right)$

Diff. on both sides.

$\frac{dy}{dx}=3.\frac{d\left({\tan }^{-1}x\right)}{dx}=\frac{3}{1+x^2}\left[\right({\tan }^{-1}x{)}^1=\frac{1}{1+x^2}]$