Question

Given:
$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}.$ Then find $\frac{dy}{dx}$.

Answer 1

Given, $y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$

or, $y=3{\tan }^{-1}x$ [ Using direct formula]

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{3}{1+x^2}$.