Question

If f(x) = $\sin^{-1}$ $\left(\frac{2x}{1+x^{2}}\right)$, then f' $(\sqrt{3})$ is

• A. $-\frac {1}{2}$
• B. $\frac {1}{2}$
• C. $\frac {1}{\sqrt{3}}$
• D. $-\frac {1}{\sqrt{3}}$

Answer 1

Answer: (B)

$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
Put $x=\tan\, \theta$, where $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$f(x)=\sin ^{-1}\left(\frac{2 \tan \,\theta}{1+\tan ^{2} \theta}\right)$
$\Rightarrow f(x)=\sin ^{-1}(\sin 2 \theta)$
$\Rightarrow f(x)=2 \,\theta=2 \tan ^{-1} x \,\,\left(\because \theta=\tan ^{-1} x\right)$
$f'(x)=\frac{2}{1+x^{2}}$
$\therefore f'(\sqrt{3})=\frac{2}{1+3}=\frac{1}{2}$