\(\frac{\mathrm{d}}{\mathrm{d} x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=\)

Let \(y=\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\)
\(\begin{aligned}
& =\cos ^{-1}\left(\frac{x^2-1}{x^2+1}\right)=\cos ^{-1}\left[(-1)\left(\frac{1-x^2}{1+x^2}\right)\right] \\
y= & \pi-\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)
\end{aligned}\)
\(\begin{array}{ll}
& \text { Put } x=\tan \theta \\
\therefore \quad & \theta=\tan ^{-1} x \\
\therefore \quad & y=\pi-\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\
\Rightarrow & y=\pi-\cos ^{-1}(\cos 2 \theta) \\
& \Rightarrow y=\pi-2 \theta \\
& \Rightarrow y=\pi-2 \tan ^{-1} x
\end{array}\)
Differentiating w.r.to \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=0-\frac{2}{1+x^2}=\frac{-2}{1+x^2}\)