[Math] Are all functions that have a primitive differentiable

Question

Are all functions that have a primitive differentiable?

For some background, I know that not all functions that are integrable are differentiable. For example:

$$
f =
\begin{cases}
0 & x \neq 0 \\
1 & x = 0
\end{cases}
$$

is integrable over $\mathbb{R}$, and $\int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) \space \forall x \in \mathbb{R}$ does not exist.

But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.

Thanks !

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Edit: here is a very important comment from @HenningMakholm that I thought would be useful for other students encountering this question

Also, (real) functions that have a primitive are not necessarily continuous. For example,

$$
f(x) =
\begin{cases}
0 & \text{for } x = 0 \\
2x \sin(\frac{1}{x}) – \cos(\frac{1}{x}) & \text{otherwise}
\end{cases}
$$

is discontinuous at $x=0$, but nevertheless the derivative of

$$
F(x) =
\begin{cases}
0 & \text{ for } x = 0 \\
x^2 \sin(\frac{1}{x}) & \text{ otherwise}
\end{cases}
$$

Functions that have a primitive do have the intermediate-value property but that is weaker than being continuous.

Answer 1

A primitive for $f(x)=x^{1/3}$ is $F(x)=\frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{x^{1/3}}{x}=\lim\limits_{x\to0}\dfrac{1}{x^{2/3}}=\infty$$