Derivative of a quotient inside a square root From my solutions page, I saw that
$f(x)= \sqrt{\frac{x+1}{x-1}}$
$ f^{\prime}(x)=\frac{-1}{\mid x-1\mid\sqrt{x^{2}-1}} $
However, I got that:
$ f^{\prime}(x)=\frac{-1}{\sqrt{x+1} \sqrt{(x-1)^{3}}} $
which is the same answer from this youtube video:
https://www.youtube.com/watch?v=AegzQ_dip8k&t=14247s
Are they both correct? If they are, how does
$ f^{\prime}(x)=\frac{-1}{\sqrt{x+1} \sqrt{(x-1)^{3}}} $ simplify to $ f^{\prime}(x)=\frac{-1}{\mid x-1\mid\sqrt{x^{2}-1}} $? Any and all hints are appreciated so I can understand this solution, because all the justifications I have heard from friends do not make sense...

You have\begin{align}-\frac1{\sqrt{x+1}\sqrt{(x-1)^3}}&=-\frac1{\sqrt{x+1}\sqrt{(x-1)^2}\sqrt{x-1}}\\&=-\frac1{\sqrt{(x+1)(x-1)}|x-1|}\\&=-\frac1{\sqrt{x^2-1}|x-1|}.\end{align}
