Question

If $y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$, then $\frac{d y}{d x}$ 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, $y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$
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{d y}{d x}=\frac{d}{d x}\left(2 \tan ^{-1} x\right) $
$ \frac{d y}{d x}-2 \cdot \frac{1}{1+x^2}-\frac{2}{1+x^2} \left(\because \frac{d}{d x} \tan ^{-1} x-\frac{1}{1+x^2}\right)$