KCET · Maths · Application of Derivatives
If \(f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\), then \(f^{\prime}(\sqrt{3})\) is
- A \(-\frac{1}{2}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{\sqrt{3}}\)
- D \(-\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
f^{\prime}(x) &=\frac{2}{1+x^{2}} \\
\therefore \quad f^{\prime}(\sqrt{3}) &=\frac{2}{1+(\sqrt{3})^{2}}=\frac{2}{1+3}=\frac{2}{4}=\frac{1}{2}
\end{aligned}\)
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
\(\begin{aligned}
f^{\prime}(x) &=\frac{2}{1+x^{2}} \\
\therefore \quad f^{\prime}(\sqrt{3}) &=\frac{2}{1+(\sqrt{3})^{2}}=\frac{2}{1+3}=\frac{2}{4}=\frac{1}{2}
\end{aligned}\)
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