We check the differentiability of $f(x) = x\left| x \right| $ at $x=0$. Evaluating derivatives from first principles, we find that the left hand and right hand derivatives both equate to $0$, which in principle proves that the function is differentiable at that point.
Now, let us try to interpret this geometrically.
More generally, if $x_{0}$ is a point in the domain of a function $f$, then $f$ is said to be differentiable at ${x}_{0}$ if the derivative $f ′(x_{0})$ exists. This means that the graph of $f$ has a non-vertical tangent line at the point $(x_{0}, f(x_{0}))$.
Source: Wikipedia
If so, where is the so-called unique tangent at $x=0$?
Source: WolframAlpha
Please find flaws in my interpretation, if any.
Thank you.
Geometrically the derivative at a point can be interpreted for strictly convex or strictly concave graphs easily: the derivative at a point $(c,f(c))$ of a strictly convex (or strictly concave) graph exists if there is a unique straight line $g(x):= ax+b$ that intersect the graph of $f$ at the point $(c,f(c))$ such that $f(x)\ge g(x)$ for all $x$ if $f$ is convex, or $f(x)\le g(x)$ for all $x$ when $f$ is concave. Then we says that $a$ is the derivative of $f$ at $c$, that is, that $f'(c)=a$.
For other functions, not necessarily strictly concave or strictly convex, the geometric interpretation is not so clear because there are infinite straight lines that can cut the graph of a function uniquely at a point as in the function of your picture.
But the geometric definition of the derivative at points of strictly convex or strictly concave graphs define analitically the derivative through a limit, and we can extend this definition to other cases not necessarily strictly concave or strictly convex.
Note that there are strictly convex and strictly concave graphs of functions that doesn't have derivative at some points, because such straight line doesn't exists or because there are more than one straight line that intersect the graph of the function uniquely in some point.