Question

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

• A. $\frac{-3}{1+x^2}$
• B. $\frac{3}{1+2x^2}$
• C. $\frac{3}{1-x^2}$
• D. $\frac{3}{1+x^2}$

Answer 1

The correct option is D $\frac{3}{1+x^2}$

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

Let $x=\tan \theta$

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

$\therefore y={\tan }^{-1}\left(\tan 3\theta \right)$

$\therefore y=3\theta$

$\therefore y=3{\tan }^{-1}x$

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

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