Question:
If $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right), x>0$. Find $\frac{d y}{d x}$
Solution:
$y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$
Using, $\sec ^{-1} x=\frac{1}{\cos ^{-1} x}$
$y=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$
Using, $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$
$y=\frac{\pi}{2}$
Now differentiating w.r.t $x$ we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=0$
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