Does there exist a function $f(x)$ such that $f'(x)$ exists for all $x \in A$ but $f'(x)$ is not continuous on $A$?

Question

I don't think this can be the case, since the derivative is defined as a limit

$$f'(x) = \lim_{a\to x} \frac{f(x)-f(a)}{x-a}$$

And a function that equals its limit is by definition continuous but this feels like a non-rigorous proof so I do not want to depend on it.

If a function does exist such that $\forall \, x \in A \,\, \exists \,y : f'(x) = y $, a good answer would give an example of such a function, or if such a function does not exist, a good answer would give pointers as to how I could prove its non existence

Answer 1

There do exist functions for which the derivative is defined for all $x \in \mathbb{R}$ but not continuous. An example of such a function is

$$f(x) = \begin{cases}x^2 \sin \left(\frac{1}{x}\right) &\text{if }x\neq0\\ 0 &\text{if }x=0 \end{cases}$$ whose derivative is $$f'(x) = \begin{cases}2x \sin \left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) &\text{if }x\neq0\\ 0 &\text{if }x=0 \end{cases}$$

$f'(0)$ exists and is equal to zero, however $\lim_{x\to0} f'(x)$ does not exist due to the term $\cos\left(\frac{1}{x}\right)$ whose limit is undefined as $x\to0$.

Therefore, $f(x)$ is an example of a function that is differentiable everywhere but whose derivative is not continuous.

I found this function in this question, however that question did not pop up in my initial search$^1$, so, I decided to use the "Answer you own question" feature so that future people with the same question may find the answer more easily.

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Probably because it is much more specific.