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I know that there exists functions where the derivatives are discontinuous, but most of the examples look sort of artificial. For example: the function $ f(x) = x^2 \sin (1/x)$ when x is non zero and taking value zero when x is zero. Here the original function itself has an undefined point. So I am a bit confused as to whether seemingly ordinary functions can have discontinuous derivatives.

Hence I am looking for any sort of lemmas or theorems which will give hints to as when a function will have a continuous derivative.

N.S.JOHN
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  • Written as $f(x)=\begin{cases}x^2\sin(1/x)&x\ne 0\\0&x=0\end{cases}$, the original function does not have undefined points ... – Hagen von Eitzen Sep 02 '19 at 11:00
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    This maybe relative-Looks like it is a dominant example: :https://math.stackexchange.com/questions/292275/discontinuous-derivative – NoChance Sep 02 '19 at 11:02
  • Sum, differences, products, quotients (where defined), compositions of functions with continuous derivatives are again functions with continuous derivatives. – Hagen von Eitzen Sep 02 '19 at 11:02
  • Note also that if $f$ is differentiable on $(a,b)$ and $f’$ has a limit at $b$, then $f$ is differentiable at $b$ and $f’$ is continuous at $b$. – Aphelli Sep 02 '19 at 12:23
  • @Mindlack sorry can you explain it non rigorously? – N.S.JOHN Sep 02 '19 at 12:28
  • If the (directional) derivative does not exist at a point, it has no limit at this point. Say that $f(x)$ is smooth on $x > 0$ but not differentiable at $0$: then $f’(x)$ has no likit when $x \rightarrow 0$. – Aphelli Sep 02 '19 at 12:43
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    Possible duplicate of [Discontinuous derivative.](https://math.stackexchange.com/questions/292275/discontinuous-derivative) – Xander Henderson Sep 02 '19 at 12:44
  • I picked the dupe target on the basis of [this answer](https://math.stackexchange.com/a/423279/468350), which I think addresses this question. [This answer](https://math.stackexchange.com/a/112133/468350) to [another question](https://math.stackexchange.com/a/112133/468350) may also be relevant. – Xander Henderson Sep 02 '19 at 12:46
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    A derivative always satisfies the intermediate value theorem, whether it is continuous or not. So if it's discontinuous it can't have a simple jump discontinuity, it has to fail to have a limit entirely. – Matt Samuel Sep 02 '19 at 20:10
  • @MattSamuel thankyou for that comment. I was looking for such theorems. – N.S.JOHN Sep 03 '19 at 04:11

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