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If $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{-2}{1+x^2}$
$\begin{aligned} & y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)=\sec ^{-1}\left(\frac{x^2+1}{x^2-1}\right)=\sec ^{-1}\left(-\frac{1+x^2}{1-x^2}\right) \\ & \Rightarrow y=\sec ^{-1}\left(-\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)=\sec ^{-1}\left(\frac{-1}{\cos 2 \theta}\right)=\sec ^{-1}(-\sec 2 \theta) \\ & \Rightarrow y=\pi-2 \theta=\pi-2 \tan ^{-1} x \\ & \Rightarrow \frac{d y}{d x}=0-\frac{2}{1+x^2}\end{aligned}$
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