Question

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

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

Answer 1

Answer: (B)

$\frac{3}{(1+x^2)}$

given

$y={\tan }^{-1}\frac{3x-x^3}{1-3x^3},\frac{-1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

let $x=\tan \theta$ $\frac{-1}{\sqrt{3}}<\tan \theta <\frac{1}{\sqrt{3}}$ $\therefore dx={\sec }^2\theta$

$\frac{d\theta }{dx}=\frac{1}{{\sec }^2\theta }=\frac{1}{1+x^2}$ $\therefore \frac{-\pi }{3}<\theta <\frac{\pi }{3}$

$\therefore -\pi <\theta <\pi$

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

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

$y=3\theta$

$\therefore \frac{dy}{dx}=\frac{d\left(3\theta \right)}{dx}$

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