# Find $$\frac {dy}{dx}$$ in the following: $y = tan^{-1} \left(\frac{3x - x^3}{1 - 3x^2}\right), - \; { \frac{1}{\sqrt3} } < x < \;{\frac{1}{\sqrt3} }\\$

$\begin{array}{1 1} \large \frac{3}{1+x^2} \\ \large \frac{3}{1-x^2} \\ \large \frac{3x-x^3}{1-3x^2} \\ -\large \frac{3x-x^3}{1-3x^2} \end{array}$

Toolbox:
• $\tan 3\theta = \large \frac{3 \tan \theta - (\tan \theta)^3}{1 - 3(\tan \theta)^2}$
• $\; \large \frac{d(tan^{-1}x)}{dx} $$= \large\frac{1}{1+x^2} Given y = tan^{-1} \left(\large \frac{3x - x^3}{1 - 3x^2}\right),$$ - \; { \frac{1}{\sqrt3} } < x < \;{\frac{1}{\sqrt3} }\\$
Substitute $\tan^{-1} x= \theta \rightarrow x = \tan \theta$
Therefore $y = tan^{-1} \left(\large \frac{3 \tan \theta - (\tan \theta)^3}{1 - 3(\tan \theta)^2}\right)$
We know that $\tan 3\theta = \large \frac{3 \tan \theta - (\tan \theta)^3}{1 - 3(\tan \theta)^2}$
Therefore, $y = tan^{-1} \tan 3 \theta = 3\theta = 3\tan^{-1} x$
We know that $\; \large \frac{d(tan^{-1}x)}{dx} $$= \large\frac{1}{1+x^2} Differentiaing both sides: \Rightarrow dy = 3 \large \frac{1}{1+x^2}$$\; dx$
$\Rightarrow \large \frac{dy}{dx} = \frac{3}{1+x^2}$