Question

The derivative of ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)w.r.t.{\cot }^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)$ is

• A. $\frac{3}{2}$
• B. 1
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

Ley $u={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ and $v={\cot }^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)$
Put $x=\tan \theta$
$u={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$ and $v={\cot }^{-1}\left(\frac{1-3{\tan }^2\theta }{3\tan \theta -{\tan }^3\theta }\right)$
$u={\cos }^{-1}[\cos 2\theta ]$ and $v={\cot }^{-1}[\cot 3\theta ]$
$u=2\theta$ and $v=3\theta$
$\frac{du}{d\theta }=2$ and $\frac{dv}{d\theta }=3\therefore \frac{du}{dv}=\frac{du}{d\theta }\times \frac{d\theta }{dv}=\frac{2}{3}$