calculus - Taking derivative of an absolute function

Let $$f(x) = \left|\frac{μ}{μ-1}-\frac{1}{x}\right|^\frac{1}{μ-1}.$$ If I want to get the derivative of this absolute value function, how would I go about doing this?

I tried using the fact that $\sqrt{x^2} = |x|$, and applied the chain rule to obtain an answer as the following:
$$f'(x) = \frac{\left(\left(\sqrt{\dfrac{μ}{μ-1}-\dfrac{1}{x}}\right)^2\right)^\frac{-μ+2}{μ-1}\left(\dfrac{μ}{μ-1}-\dfrac{1}{x}\right)}{x^2(μ-1)\left|\dfrac{μ}{μ-1}-\dfrac{1}{x}\right|},$$ where $μ$ is just a constant such as $μ = 2,3,\cdots, 100$.

The derivative I have works for $μ = 2$, but does not work for other values.Is my algebra off or is there something else that I should note? Am I on the right path to getting $f'(x)$?

Answer

For $\frac{\mu}{\mu-1}-\frac{1}{x}>0\iff \frac{\mu}{\mu-1}>\frac1x\iff x>1-\frac1\mu$, $$f'(x)=\frac d{dx}\left(\left(\frac{\mu}{\mu-1}-\frac{1}{x}\right)^\frac1{\mu-1}\right)=\frac1{\mu-1}\left(\frac{\mu}{\mu-1}-\frac{1}{x}\right)^{\frac1{\mu-1}-1}\cdot\frac1{x^2}$$

For $x<1-\frac1\mu$, $$f'(x)=\frac d{dx}\left(\left(-\frac{\mu}{\mu-1}+\frac{1}{x}\right)^\frac1{\mu-1}\right)=\frac1{\mu-1}\left(-\frac{\mu}{\mu-1}+\frac{1}{x}\right)^{\frac1{\mu-1}-1}\cdot\left(-\frac1{x^2}\right)$$

For $x=1-\frac1\mu$, plug this value into each of the above derivatives. If they are the same, that is the derivative at this $x$. If they are different, then the derivative does not exist.