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)

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