Why is $|x|$ not differentiable at $x=0$? The definition of a derivative is the slope of a function tangent to a point. It is also defined as $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$
If we apply this to $f(x)= |x|$, we get that it is $\lim\limits_{h\to 0} \dfrac{|h|}{h}$, which is undefined. However, if we look at the graph of $|x|$, we see that there can exist a tangent line at x=0, with slope 0. So why is the derivative undefined instead of 0?
Let $f(x) = |x|$. For $x &gt; 0$, the gradient is $1$, whereas for $x &lt; 0$, the gradient is $-1$. At $x = 0$, you have infinitely many lines,  which are represented by the subgradient $[-1,1]$.
If you draw the graph of the equation, you will get a v-shaped configuration with the vertex located at x=0, y=0.

If you try to put a tangent at that vertex point, you will see that the line can take on one of innumerable slopes and still be "touching at one point". It's undefined.