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 $\frac{dv}{d\theta} =3 \:\therefore \:\:\: \frac{du}{dv} = \frac{du}{d\theta} \times\frac{d\theta}{dv} =\frac{2}{3}$