If $(f ∘ f)$ is differentiable, is $f$ also differentiable? Question:

If $(f \circ f)$ is differentiable on $\mathbb R$, then $f$ is differentiable on $\mathbb R$.

Is this statement true or false and why?
I have had a look at this question and really can't get my head around it.
I have thought that it is False, because if we let $f(x) = 2$, $(f\circ f)$ can't be defined as $(f(f(2))$ doesn't exist.  So the statement would be false as we can't define $(f\circ f)$ so it can't be differentiable on $\mathbb R$.
Is this way of looking at it right or not?
 A: There is a very simple counterexample. Consider $$
f(x)=\begin{cases}2 & x\geq 0\\1 & x<0\end{cases}
$$ which is not differentiable (not even continuous) at $0$ but $f\circ f\equiv 2$ is constant, so differentiable on $\mathbb R$.
A: We can do even better than Mundron Schmidt's counterexample.  For example, the Dirichlet function: $$f(x) = \begin{cases}1 & \text{if } x \in \mathbb Q \\ 0 & \text{if } x \notin \mathbb Q\end{cases}$$ is discontinuous (and thus non-differentiable) everywhere on $\mathbb R$, but $f \circ f$ is constant.

A slight modification will also provide a counterexample such that $(f \circ f)' \ne 0$.  For example, the function: $$g(x) = \begin{cases}\phantom{+}x & \text{if } x \in \mathbb Q \\ -x & \text{if } x \notin \mathbb Q\end{cases}$$ is discontinuous almost everywhere on $\mathbb R$ (except at the origin), while $g \circ g$ is the identity function (and thus $(g \circ g)' = 1$).  Or, for a counterexample that's truly discontinuous everywhere, we could instead pick, say: $$h(x) = \begin{cases}-x & \text{if } x \in \mathbb Q \\ 1/x & \text{if } x \notin \mathbb Q\end{cases}$$ which also satisfies $(h \circ h)(x) = x$.
A: Your assumption that the statement is false is correct, but your argument is wrong. With $f(x)=2$ you certainly can form $f\circ f$:
$$(f\circ f)(x) = f(f(x)) = f(2) = 2$$
Indeed, whenever the range if $f$ is a subset of its domain, $f\circ f$ can be formed.
Also, besides your wrong argument, $f(x)=2$ isn't a valid counterexample, as it is a differentiable function, with $f'(x)=0$. Any valid counterexample to the claim involves a non-differentiable function.
A list of valid counterexamples has already been given by Mundron Schmidt and Ilmari Karonen.
A: Let
$$f(x)=\begin{cases}1&\text{if } x=2 \\ 0 & \text{otherwise}\end{cases}$$
then
$$f(f(x))=0$$
is differentiable, although $f$ is not at $x=2$.
For $f$ being the Dirichlet's function:
$$f(x)=\begin{cases}1&\text{if } x\in\mathbb Q \\ 0 & \text{otherwise}\end{cases}$$
you have
$$f(f(x))=1$$
differentiable, although $f$ is not differentiable, and even not continuous anywhere.
