Derivative of a quotient inside a square root

Question

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...

Answer 1

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}

Answer 2

Hints:

1. $|a|=\sqrt{a^2}$
2. $\sqrt{ab}=\sqrt a \sqrt b$
3. $(a+b)(a-b)=a^2-b^2$.