What is the derivative of $|f(x)|^n$ Well, it's time for a trivial question but I really need a clarification about!
Let's say I have to evaluate
$$\frac{\text{d}}{\text{d}x}\ |f(x)|^n$$
Shall I have to reason by treating the function as
$$|f(x)|^n = \left(\sqrt{f(x)^2}\right)^n$$
and then it would be quite easy, or shall I try to apply, somehow, the same rule for the derivative of $|x|$ like?
$$D\ |f(x)|^n = n\ f(x)^{n-1}\frac{|f(x)|}{f(x)}f'(x) = n\ f(x)^{n-2}|f(x)|f'(x)$$
Thank you for the clarification!
 A: $\DeclareMathOperator{\abs}{abs}$If $n$ is an even integer, you may as well write
$$
|f(x)|^{n} = f(x)^{n}
$$
and differentiate using the chain rule. 
Otherwise, if $x \neq 0$, the function $\abs(x) = |x|$ is differentiable at $x$, and
$$
\abs'(x) = \frac{x}{|x|}.
$$
This (plus the chain rule) guarantees that if $f$ is differentiable at $x$ and $f(x) \neq 0$, then your formula is correct:
$$
\frac{d}{dx} |f(x)|^{n} = n\, f(x)^{n-2} |f(x)|\, f'(x).
$$
If $f(x) = 0$ (or if $f$ is not differentiable at $x$), you need to examine the specific form of $f$ to determine whether or not $\abs^{n}(f)(x) = |f(x)|^{n}$ is differentiable at $x$.
A: The biggest issue is that the function $y \mapsto |y| = \sqrt{y^2}$ is of course not differentiable when $y = 0$. Given this, we can only take the derivative of $|f(x)|^n$ on all points where $f(x) \ne 0$.
But as long as we are at $x$ where $f(x) \ne 0$, either of your two proposed methods will work. We have on the one hand
\begin{align*}
\frac{d}{dx} |f(x)|^n
&= \frac{d}{dx} [f(x)^2]^{n/2} \\
&= \frac{n}{2} [f(x)^2]^{\left(\tfrac{n}{2} - 1\right)} \left[2f(x)f'(x)\right]. \\
&= n \cdot f(x) \cdot f'(x) \cdot |f(x)|^{n-2}
\end{align*}
and on the other hand
\begin{align*}
\frac{d}{dx} |f(x)|^n
&= n |f(x)|^{n-1} \left[\frac{d}{dx} |f(x)|\right] \\
&= n |f(x)|^{n-1} \frac{f(x)}{|f(x)|} f'(x) \\
&= n \cdot f(x) \cdot f'(x) \cdot |f(x)|^{n-2}.
\end{align*}
where we used $\frac{d}{dy} |y| = \frac{y}{|y|}$.
You can now see that these two expressions are the same.
Note 1: In your calculations, I saw a couple of small typos or errors.
Note 2:
We have to be careful dealing with the expression $[f(x)^2]^{n/2}$. Note this is not the same as $f(x)^n$. For a potentially negative $a$, we do not have that $(a^b)^c = a^{bc}$.
