Question

Find $\frac{dy}{dx}$ of $y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Answer 1

$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$
Put $x=\tan \theta$
$\Rightarrow y={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$
$={\tan }^{-1}\tan \left(3\theta \right)=3\theta =3{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=\frac{3}{1+x^2}$