Q62 / 180 · KCET 2020 · Mathematics
If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, then $f^{\prime}(\sqrt{3})$ is
- A$-\frac{1}{2}$
- Correct answerB$\frac{1}{2}$
- C$\frac{1}{\sqrt{3}}$
- D$-\frac{1}{\sqrt{3}}$
Solution
We have, $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
On putting $x=\tan \theta$
$\Rightarrow \quad \theta=\tan ^{-1} x$
Then we get, $f(x)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$\begin{aligned}
&=\sin ^{-1}(\sin 2 \theta) \\
&=2 \theta=2 \tan ^{-1} x...(i)
\end{aligned}$
On differentiating Eq. (i) both sides w.r.t. $x$, we get
$\begin{aligned}
f^{\prime}(x) &=\frac{2}{1+x^{2}} \\
\therefore \quad f^{\prime}(\sqrt{3}) &=\frac{2}{1+(\sqrt{3})^{2}}=\frac{2}{1+3}=\frac{2}{4}=\frac{1}{2}
\end{aligned}$
Hence, the correct answer is option: 'B'.