Find :
Question:

Find $\frac{d y}{d x}$ :

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

Solution:

The given relationship is $y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$

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

$\Rightarrow \tan y=\frac{3 x-x^{3}}{1-3 x^{2}}$    …(1)

It is known that, $\tan y=\frac{3 \tan \frac{y}{3}-\tan ^{3} \frac{y}{3}}{1-3 \tan ^{2} \frac{y}{3}}$   …(2)

Comparing equations (1) and (2), we obtain

$x=\tan \frac{y}{3}$

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}(x)=\frac{d}{d x}\left(\tan \frac{y}{3}\right)$

$\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{d}{d x}\left(\frac{y}{3}\right)$

$\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{1}{3} \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{3}{\sec ^{2} \frac{y}{2}}=\frac{3}{1+\tan ^{2} \frac{y}{2}}$

$\therefore \frac{d y}{d x}=\frac{3}{1+x^{2}}$