MHT CET · Maths · Differentiation
If \(y=\sin ^{-1}\left(\frac{\log x^2}{1+(\log x)^2}\right)\), then \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\mathrm{d} x=1}=\)
- A 2
- B \(\frac{1}{2}\)
- C \(\cdot \frac{2}{3}\)
- D -2
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(y=\sin ^{-1}\left(\frac{\log x^2}{1+(\log x)^2}\right)=\sin ^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)\)
Let \(\log x=\tan \theta\)
\(\begin{array}{ll}
\therefore & y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan 2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta \\
\therefore & y=2 \tan ^{-1}(\log x) \\
\therefore & \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \times \frac{1}{1+(\log x)^2} \times \frac{1}{x} \\
\therefore & \left.\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)\right|_{\mathrm{at} x=1}=2
\end{array}\)
Let \(\log x=\tan \theta\)
\(\begin{array}{ll}
\therefore & y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan 2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta \\
\therefore & y=2 \tan ^{-1}(\log x) \\
\therefore & \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \times \frac{1}{1+(\log x)^2} \times \frac{1}{x} \\
\therefore & \left.\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)\right|_{\mathrm{at} x=1}=2
\end{array}\)
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