3

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 $c$?

mosceo
  • 2,035
  • http://math.stackexchange.com/questions/602568/if-a-function-is-continuous-and-differentiable-everywhere-is-the-derivative-cont/602571#602571 – hrkrshnn Feb 25 '14 at 06:38

1 Answers1

2

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

user127.0.0.1
  • 7,267
Euler....IS_ALIVE
  • 4,707
  • The derivative at the origin is zero. But how come I can't get it by applying the derivative-function to $0$, in which case i will be dividing by zero? In other words, the derivative-function says that it's undefined at zero. – mosceo Feb 25 '14 at 06:58
  • @Graduate: the given example is not completely correct. It should be e.g. $f(x)=\begin{cases}0&x=0\x^2\sin(1/x)&x\neq0\end{cases}$ – user127.0.0.1 Feb 25 '14 at 07:04