Does the derivative of a function need to be continuous if the function is differentiable

Question

I have the following problem:

I know the following:

I. It must be true because if f(7) was not continuous (undefined), then the difference of quotients would not be 4, but undefined

II. It must be true because the difference of quotients is equal to the derivative. The derivative at 7 exists (which is 4), so it must be differentiable.

III. This is the one where I am not certain. I know a derivative at x = 7 must exists, but I am not sure if it is continuous.

How do you know if the derivative of function is continuous at a certain point?

Answer 1

The derivative does not need to be continuous. Here is a standard example of this phenomenon: $f(x) = x^2 \sin(1/x)$. It is easy to see this function can be extended differentiably to take the value $0$ at $0$. However, the derivative blows up at $0$.

It is also true that derivatives always have the intermediate value property, even though they are discontinuous. Both of these are well-explored on this site in other questions.

Answer 2

To have a concrete counterexample to that, as @Alfred Yerger has suggested, one takes $f(x)=(x-7)^{2}\sin(1/(x-7))+4(x-7)$ and $f(7)=0$, then $f'(7)=4$, and $f'(x)=2(x-7)\sin(1/(x-7))-\cos(1/(x-7))+4$ for $x\ne 7$. So there would be a problem for $\lim_{x\rightarrow 7}f'(x)$.