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

0

$y =\sec ^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$

$=\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{\pi}{2}$

$\Rightarrow \frac{d y}{d x}=0$

Hence (4) is correct answer.