calculus - Derivative defined at some point but not continuous there?

Suppose $f$ is a continuous function, and $f'$ is its derivative-function. Is it possible that $f'(c)$ exists for some point $c$, but $f'$ is not continuous at $c$?

Answer

Yes. The standard example is $f: \mathbb R\to \mathbb R$ with
$$f(x) = \begin{cases}0 & x=0 \\x^2\sin(\frac{1}{x})&x\neq0\end{cases}$$ Check (using the definition) that the derivative exists at the origin and is equal to $0$. But the derivative is not continuous at $0$. We would need $\lim_{x \rightarrow 0} 2x\sin({\frac{1}{x}}) - \cos(\frac{1}{x}) = 0$, which it is not, because of the oscillation.

In fact, there are examples which are even worse. See for instance Volterra's Function http://en.wikipedia.org/wiki/Volterra%27s_function