What is $\frac{dy}{dx}$ for $y = \tan^{-1}\left(\frac{3x - x^{3}}{1 - 3x^{2}}\right)$, $\lvert x \rvert < \frac{1}{\sqrt{3}}$?

Explanation

1. 01

   Put $x = \tan \theta$: $\frac{3\tan \theta - \tan^{3} \theta}{1 - 3\tan^{2} \theta} = \tan 3\theta$, so the function is $3\theta = 3\tan^{-1} x$.

2. 02

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