Question

If $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

Answer: (B)

Given, $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
Putting $\theta ={\tan }^{-1}x$ i.e., $x=\tan \theta$
Then, $y={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)={\sin }^{-1}(\sin 2\theta )=2\theta =2{\tan }^{-1}x$
$\left(\because \sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{d}{dx}\left(2{\tan }^{-1}x\right)$
$\frac{dy}{dx}-2\cdot \frac{1}{1+x^2}-\frac{2}{1+x^2}\left(\because \frac{d}{dx}{\tan }^{-1}x-\frac{1}{1+x^2}\right)$