Question

Find $\frac{d}{dx}{\cot }^{-1}\left(\frac{1+x}{1-x}\right)$, if $x>-1$

• A. $\frac{1}{1+x^2}$
• B. $\frac{1}{1-x^2}$
• C. $-\frac{1}{1+x^2}$
• D. None of these

Answer 1

The correct option is C $-\frac{1}{1+x^2}$

Let, $y={\cot }^{-1}\left(\frac{1+x}{1-x}\right)$
Put $x=\tan \theta$
$y=co{t}^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)$
$y={\cot }^{-1}\left(\tan (\frac{\pi }{4}+\theta )\right)$
$y={\cot }^{-1}\left(\cot (\frac{\pi }{2}-\frac{\pi }{4}\theta )\right)$
$y=\frac{\pi }{4}-\theta$
$y=\frac{\pi }{4}-{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=-\frac{1}{1+x^2}$