Q:

Derivative of $\sqrt{\frac{x-1}{x+1}}$

$$f(x)=\sqrt{\frac{x-1}{x+1}}$$
\begin{align}
f'(x) & ={1\over 2} \left(\frac{x-1}{x+1}\right)^{\frac{-1}{2}} \cdot {x+1-(x-1)\over (x+1)^2} \\[10pt]
& =\frac{1}{2}\left(\frac{x-1}{x+1}\right)^{\frac{-1}{2}}\cdot{2\over (x+1)^2} =\left(\frac{x-1}{x+1}\right)^{\frac{-1}{2}}\cdot{1\over (x+1)^2}
\end{align}
Is it valid to write $ (\frac{x-1}{x+1})^{\frac{-1}{2}}= \sqrt{\frac{x+1}{x-1}}$ ?

A:

Notice, you need to apply some constraints for the variable $x$, 
for all $x>1$,  it will be valid to write $$\left(\frac{x-1}{x+1}\right)^{-1/2}=\sqrt{\frac{x+1}{x-1}}$$

