Domain/Points of differentiability How would I find the domain or points of a function where it is differentiable? Let's say I had the function $f: \mathbb R \rightarrow \mathbb R$ defined by $f(x) = |x|$. I know that a function $f: \mathbb R \rightarrow \mathbb R$ differentiable at $x_0 \in \mathbb R$ means that for all $\epsilon > 0$, there exists a $\delta > 0$ such that for some $L \in \mathbb R$, $|\frac{f(x) - f(x_0)}{x - x_0} - L| < \epsilon$ for all $x \in \mathbb R$ with $|x - x_0| < \delta$. However, I'm not sure how to find all possible values of $x_0$ from this alone, so am I supposed to be doing this a different way? I'm confused.
All responses are much appreciated.
 A: For $x>0$ the function is just $f(x)=x$ and you know that this is differentiable.
For $x<0$ the function is just $f(x)=-x$ and this is differentiable.
So the only question is differentiability at $x=0$. Here consider $\lim_{h \to 0} \frac {f(0+h)-f(0)} h=\lim_{h \to 0} \frac {|h|} h$. Since the limit here through positive values of $h$ is $1$ and the limit through negative values of $h$ is $-1$, the limit does not exist, so $f$ is not differentiable at $0$.
A: This is a very broad topic!
Let's give some elements however.
Base differentiable real functions
As differentiable functions have nice properties, we first try to define differentiable functions. For example, one can prove that constant, polynomials, $e^x$, $\log x$, entire functions are differentiable.
Building differentiable functions from differentiable functions
After that, you can prove that, sum, product, ratio, composition of differentiable functions are differentiable. $\int f$ is differentiable if $f$ is continuous...
Using the facts above allows you to quickly prove that a lot of functions are differentiable! Like $x \mapsto \log \left( \frac{x^{24} + 347 x^2 +1}{x^2 + e^x} \right) + \sin \left(\int_0^x \vert t \vert \ dt\right)$...
The example of $\vert f \vert$
Coming back to your precise example, if $f$ is differentiable, one can prove that $\vert f \vert$ is also differentiable at $a$ if $f(a) \neq 0$. $f$ will be differentiable at $a$ if $f(a)=0$ if and only if $f^\prime(a) = 0$. Nice exercise!
The "interesting" cases!
After all those "standard cases", there are some cases where you need to come back to the basic definition and look at what happens in details.
Like a continuous function that is nowhere differentibale. Look here for details.
