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This is the problem I'm working on.

If $f$ is differentiable on $(a,b)$ and $c \in (a,b)$, then $f′$ is continuous at $c$.

I believe it to be true, but I'm having a hard time proving it. I don't see how a function's derivative can both be defined at a point inside an open interval and be discontinuous at that same point.

GhostyOcean
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    It's not true. The standard counterexample is $f(x)=x^2\sin(1/x)$ for $x\neq0$ and $f(0)=0$. Proving discontinuity of $f'$ is frequently given as an exercise. – Vercassivelaunos Nov 04 '20 at 17:06
  • For your example, wouldn't $f'$ not be differentiable on $(-1,1)$ because it's not differentiable at $0$? – GhostyOcean Nov 04 '20 at 17:32
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    Wait, I just calculated $\lim \limits_{x \to 0} \frac{f(x)-f(0)}{x-0} = 0$, so the derivative at $0$ is $0$, but the limit to the derivative osscilates wildly and doesn't converge to $0$. – GhostyOcean Nov 04 '20 at 17:42
  • Which means $f'$ is not continuous at $0$, even though it is defined everywhere on $(-1,1)$. $f'$ is not differentiable on $(-1,1)$ because it is not differentiable at $0$ because it is not even continuous at $0$. – Arturo Magidin Nov 04 '20 at 17:46

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