If $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$, then $\frac{d y}{d x}=$

If $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$, then $\frac{d y}{d x}=$
  1. $\frac{-1}{1+x^2}$
  2. $\frac{-2}{1+x^2}$
  3. $\frac{2}{1-x^2}$
  4. $\frac{1}{1+x^2}$

Solution

$\begin{aligned} & y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)=\sec ^{-1}\left(\frac{x^2+1}{x^2-1}\right)=\sec ^{-1}\left(-\frac{1+x^2}{1-x^2}\right) \\ & \Rightarrow y=\sec ^{-1}\left(-\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)=\sec ^{-1}\left(\frac{-1}{\cos 2 \theta}\right)=\sec ^{-1}(-\sec 2 \theta) \\ & \Rightarrow y=\pi-2 \theta=\pi-2 \tan ^{-1} x \\ & \Rightarrow \frac{d y}{d x}=0-\frac{2}{1+x^2}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 1)

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