Question

$\frac{d}{dx}\left[{\cos }^2{\cot }^{-1}\left\{\sqrt{\frac{1-x}{1+x}}\right\}\right]$ is equal to

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

Answer 1

The correct option is A $\frac{-1}{2}$

$y=\frac{d}{dx}\left[{\cos }^2{\cot }^{-1}\left\{\sqrt{\frac{1-x}{1+x}}\right\}\right]$
Put $x=\cos \theta$
$\Rightarrow y=\frac{d}{dx}\left[{\cos }^2{\cot }^{-1}\left\{\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right\}\right]$

$\Rightarrow y=\frac{d}{dx}\left[{\cos }^2{\cot }^{-1}\left\{\sqrt{\frac{2{\sin }^2\left(\theta /2\right)}{2{\cos }^2\left(\theta /2\right)}}\right\}\right]$

$\Rightarrow y=\frac{d}{dx}\left[{\cos }^2{\cot }^{-1}\left\{\tan \frac{\theta }{2}\right\}\right]$

$\Rightarrow y=\frac{d}{dx}\left[{\cos }^2{\cot }^{-1}\left\{\cot (\frac{\pi }{2}-\frac{\theta }{2})\right\}\right]$
$\Rightarrow y={\cos }^2(\frac{\pi }{2}-\frac{\theta }{2})$
$\Rightarrow y={\sin }^2\frac{\theta }{2}=\frac{1-\cos \theta }{2}=\frac{1-x}{2}$

Hence, the value of the differentiation is
$\frac{dy}{dx}=-\frac{1}{2}$