Is the derivative of a differentiable function in a closed domain $J$, always continuous on the same domain $J$.
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It is false. Counterexamples are easy to find on books. P.S: to make sense of a derivative, $J$ must be an interval. – Giuseppe Negro Jun 01 '17 at 17:28
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Hence the fact that "differentiable" and "continuously differentiable" are both part of the vocabulary, instead of just one of the two. – Clement C. Jun 01 '17 at 17:31
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Darboux's theorem, mentioned by Jorge Fernandez Hidalgo, says that, even if f' is not continuous, it still satisfies the "intermediate value theorem": between x= a and x= b, f'(x) takes on all values between f'(a) and f'(b). – user247327 Jun 01 '17 at 17:31
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If you're interested on how discontinuous the function can be take a look at [How discontinuous can a derivative be?](https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be) and [Discontinuous derivative.](https://math.stackexchange.com/questions/292275/discontinuous-derivative) – kingW3 Jun 01 '17 at 17:40
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This statement is false, as seen by the following example:
$f(x) = x^2\sin\left(\frac 1x\right)$, $x \neq 0$ and defining $f(0) = 0$.
Proving that it is differentiable is easy when $x \neq 0$, and at $0$ the derivative is $0$ by applying the limit definition of the derivative.
The derivative can be seen to be discontinuous at $0$.
[EDITS: there were some edit issues, sorry if I accidentally reverted someone's edit]
Tob Ernack
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That's right the derivative isn't continuous on 0 , I'm grateful for your answer – Put Me Jun 01 '17 at 17:32
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No, consider the function $f(x)=x^2\sin(1/x)$ if $x\neq 0$ and $f(0)=0$.
What is true however is that it does not have jump discontinuities, look up Darboux's theorem.
Asinomás
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No.
Consider $$f(x)=\begin{cases}x^2\sin \frac1{x}&\text{if }x\ne 0\\0&\text{if }x=0\end{cases}$$ Then $$f'(x)=\begin{cases} 2x\sin\frac 1x-\cos \frac 1x&\text{if }x\ne 0\\0&\text{if }x=0\end{cases}$$
Note that $\lim_{x\to 0}f'(x)$ does not exist.
Hagen von Eitzen
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