TITLE: Is it true that if the derivative of a function is not continuous then the function is not differentiable?
QUESTION [7 upvotes]: It has been along time since I did real analysis. Someone asked me this question and I could not respond. I am not sure if this is the right place to ask.
He said: 

The function $f: [0, \infty)\mapsto [0, \infty)$ given by $f(x)=\sqrt{x}$ is not differentiable at $0$ because its derivative $f'(x)=1/(2\sqrt{x})$ is not continuous at $0$.

I said (what I could remember):

It is not differentiable at $0$ because $\lim_{x\to 0}(f(x)-f(0))/(x-0)=\infty$.

He said that I know that but what I told is it true?

REPLY [3 votes]: Just to add a thought to what others already said.
As Dr MV pointed out, a function is differentiable at a point if the limit of the difference quotient exists and is finite. That's it. Continuity of the derivative is something more. Therefore your answer was correct.
However, continuity is required when we say that a function $f$ is in $\mathcal{C}^1(I)$ (here $I$ is an interval): $f$ is in $\mathcal{C}^1(I)$ if it is differentiable in $I$, and the derivative is also continuous in $I$.
In this context, the example given by Dr MV is an example of a function which is differentiable everywhere, but does not belong to $\mathcal{C}^1(\mathbb{R})$.