Question

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

• A. $2.\frac{1}{1+x^2}$
• B. $4.\frac{1}{1-x^2}$
• C. $4.\frac{1}{1+x^2}$
• D. $2.\frac{1}{1-x^2}$

Answer 1

Answer: (A)

$2.\frac{1}{1+x^2}$

Let $y={\sin }^{-1}\frac{2x}{1+x^2}$

put $x=\tan \theta$

$\Rightarrow y={\sin }^{-1}\frac{2\tan \theta }{1+{\tan }^2\theta }={\sin }^{-1}\sin 2\theta =2\theta =2{\tan }^{-1}x$

$\therefore \frac{dy}{dx}=2.\frac{1}{1+x^2}$