What does it mean L1 loss is not differentiable? I was looking through this lecture
https://davidrosenberg.github.io/ml2015/docs/3a.loss-functions.pdf

Slide 3: Absolute or Laplace or L1 loss not differentiable

What does it mean L1 loss not differentiable? I understand that derivative not exist at x=0, but what practical problems can arise from this fact?
What does it mean gives median regression?
Update:
Looks like answer to the second question:
https://stats.stackexchange.com/a/363369/16843
Update 2:
Here is related question: https://stackoverflow.com/questions/41518869/how-does-tensorflow-handle-the-differentials-for-l1-regularization
It doesn't describe how tensorflow implement it, but example shows that at x = 0.0 the gradient = 0.0
 A: $L_1$ loss uses the absolute value of the difference between the predicted and the actual value to measure the loss (or the error) made by the model. The absolute value (or the modulus function), i.e. $f(x) = |x|$ is not differentiable is the way of saying that its derivative is not defined for its whole domain. For modulus function the derivative at $x=0$ is undefined, i.e. we have:
$$
\frac{d|x|}{dx} = \begin{cases}
-1, & x < 0 \\
1, & x > 0
\end{cases}
$$
A: I understand that derivative not exist at x=0, but what practical problems can arise from this fact?
$$ L = |x*a - y|; $$
$$  \frac{\partial L}{\partial a} = \dfrac{x\left(xa-y\right)}{\left|xa-y\right|} $$
When faced with loss equals zero for any sample you train your model with, the gradient calculator will need to divide expression by zero, which will cause error.
This is usually mitigated by, for example, adding a small value to denominator when $L$ is zero.
A: +1 to both Tomasz and Alexey posts.
I would add that a good surrogate for the $L_1$ loss is the Pseudo-Huber loss function: $ L_{\delta }(x) =$ $\delta ^{2}\left({\sqrt {1+(x/\delta )^{2}}}-1\right)$ with $\delta = 1$. It allows us to approximate the $L_1$ rather faithfully away from $x=1$ and within $[-1,1]$ behaves like a quadratic. It has first (and second) derivatives everywhere. 

