Absolute value of differentiable function Given differentiable function $f:\mathbb{R}\to\mathbb{R}$ find points where $|f|$ is not differentiable. 
I think it's not differentiable in all zeros of $f$ and differentiable everywhere else. Is it true? Does it suffice to show that left and right derivatives are different in these points?
 A: In order to show a function is not differentiable at a point, it does suffice to show that the left and right derivatives are different at that point.  In general that is not an equivalent condition, but in the case of the absolute value of a differentiable function, that is the only thing that can go wrong.
The problem is not just when $f(a) = 0$, but when $f$ crosses the $x$ axis at nonzero slope at $x=a$.  Here is an exhaustive list of cases, given $a\in\mathbb R$:


*

*$f(a)\neq 0$.  You can use continuity to find a neighborhood of $a$ on which $|f(x)|=f(x)$, or on which $|f(x)|=-f(x)$, depending on the sign of $f(a)$.  

*$f(a) = 0$ and $f'(a)\neq 0$.  You can use the definition of the derivative to show the left and right derivatives of $|f|$ at $a$ are unequal.  

*$f(a) = 0$ and $f'(a) = 0$.  You can use the definition of the derivative to show that $|f|$ is differentiable at $a$ with $|f|'(a) = 0$.

A: You are partially right. To show that $|f|$ is not differentiable in the points where $f$ is zero it suffices to show that the left and right derivatives do not agree. However, this does now show that $|f|$ is differentiable everywhere else.
A way you might do this is to look at the situation in the following way. First define the ''absolute value function''
\begin{equation}
 \text{Abs}: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto |x|.
\end{equation}
The function $\text{Abs}$ is differentiable everywhere but zero.
Given a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ we wish to investigate $|f| = \text{Abs} \circ f: \mathbb{R} \rightarrow \mathbb{R}$. By the chain rule, the derivative of $|f|$ exists for all $x$ such that $\text{Abs}$ is differentiable at $f(x)$, which is whenever $f(x) \neq 0$.
This is however not a complete description, because it might be that there are points $x \in \mathbb{R}$ where $f(x) = 0$ and $f'(x) = 0$, as pointed out by Jonas Meyer, for these points the function $|f|$ is differentiable as well, with derivative $|f|'(x) = 0$.
A: Write $g(x) = |f(x)|$. If $f(x_0) \neq 0$, then $g$ is differentiable at $x_0$, since it is either equal to $f(x)$ or $-f(x)$ in a neighborhood of $x_0$.
So the only points where $g$ may not be differentiable are the zeroes of $f$. I'll give you the answer and then you try and prove it.
Suppose then $f(x_0) =0$. Can you prove that $g$ is differentiable at $x_0$ if and only if $f'(x_0) = 0$?
A: I read the first post where he mentioned the importance of the $f'(x)$. But it is not important as I show you here:
$$\lim \limits_{h\rightarrow 0}\frac{|f(x+h)|-|f(x)|}{h}=\lim \limits_{h\rightarrow 0}\frac{|f(x+h)|-|f(x)|}{h}\frac{|f(x+h)|+|f(x)|}{|f(x+h)|+|f(x)|}=\lim \limits_{h\rightarrow 0}\frac{|f(x+h)|^{2}-|f(x)|^{2}}{h(|f(x+h)|+|f(x)|)}=\lim \limits_{h\rightarrow 0}\frac{f(x+h)^{2}-f(x)^{2}}{h(|f(x+h)|+|f(x)|)}=\lim \limits_{h\rightarrow 0}\frac{(f(x+h)-f(x))(f(x+h)+f(x))}{h(|f(x+h)|+|f(x)|)}=\lim \limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\lim \limits_{h\rightarrow 0}\frac{(f(x+h)+f(x))}{|f(x+h)|+|f(x)|}=\lim \limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\frac{f(x)+f(x)}{|f(x)|+|f(x)|}=f'(x)\frac{2f(x)}{2|f(x)|}=f'(x)\frac{|f(x)|}{f(x)}$$
Finally we have:
$$\frac{d|f(x)|}{dx}=\left\{\begin{array}{r} f'(x) \;\;\;\; \text{if} \; f(x)>0\\
-f'(x) \;\;\;\; \text{if} \; f(x)<0\\
\text{depends} \;\;\; \text{if}\; f(x)=0  \end{array}\right.$$
Even if $f'(x)=0$, it does not matter for the purpose of the derivative of $|f(x)|$. If $f(x)=0$ then you need to analyze its roots and calculate its lateral derivatives on those points. For instance, $g(x)=|\frac{1}{2}x^2-10|$, if you want to calculate its derivative when $g(x)=0$, then you will have $x=\pm\sqrt{20}$ and in those points its derivative will be $\pm\sqrt{20}$ (calculate lateral derivative in those points and see what happens) but in any other point the derivative will be $\pm f'(x)$, setting $f(x)=\frac{1}{2}x^2-10$.
