Question

$\frac{d}{dx}\left({\tan }^{-1}\frac{2x}{1-x^2}\right)$

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

Answer 1

The correct option is A $2.\frac{1}{1+x^2}$

Let $y={\tan }^{-1}\frac{2x}{1-x^2}$
put $x=\tan \theta$
$\Rightarrow y={\tan }^{-1}\frac{2\tan \theta }{1-{\tan }^2\theta }={\tan }^{-1}\tan 2\theta =2\theta =2{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=2.\frac{1}{1+x^2}$