Derivative exists implies derivative continuous

Question

It seems to me that if $f^{'}$ exists in an open interval $I$, then $f^{'}$ is continuous in the interval.

I can't come up with an example that $f^{'}$ exists, but $f^{'}$ not continuous.

Thanks for your help in advance.

Answer 1

The following function: $$ f(x) = \begin{cases} x^2 \sin \frac{1}{x} & x \ne 0 \\ 0 & x = 0 \end{cases} $$ is differentiable everywhere (the derivative at $0$ is $0$, by the squeeze lemma), but the derivative is (wildly) discontinuous at $0$.

This is a great function to keep in your pocket as you look at conjectures in analysis, because it tends to break things. :)

Your conjecture is almost right, in the sense that the derivative of an everywhere-differentiable function is not guaranteed to be continuous, but it does have the intermediate value property. That's known as Darboux's theorem. See this description for some details.