Question

Solve : $\frac{d}{dx}\left[Ta{n}^{-1}\right(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\left)\right]$ =

Answer 1

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

Let $x=a\tan \theta$

$\frac{dx}{d\theta }=a{\sec }^2\theta$

$\Rightarrow dx=a{\sec }^2\theta d\theta$

Substituting we get,

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

$=\frac{d}{a{\sec }^2\theta d\theta }\left[{\tan }^{-1}\right(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\left)\right]$ $(\because \tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta })$

$=\frac{1}{a{\sec }^2\theta }\frac{d}{d\theta }{\tan }^{-1}\left(\tan 3\theta \right)$

$=\frac{1}{a{\sec }^2\theta }\frac{d}{d\theta }\left(3\theta \right)$

$=\frac{3}{a{\sec }^2\theta }$