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While some order of derivative function of an original function exists in an interval, is it necessary and sufficient to say that the derivative function at that order is continuous in that interval?

Or can you show me a counter example?

I googled, but I don't know how to ask this question smartly and accurately. If it's too elementary or a duplicate. Please give me a pointer and advice. Thank you!

M. Chen
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    If $f'(x)$ exists for all $x\in\Bbb R$, say, this does not imply that the function $f'$ is continuous. However, $f'$ always has (like a continuous function) the intermediate value property. Nevertheless, the functions where the derivative exists and is continuous play such an important role that they have a spcial name, $C^1$ functions, or $C^1(\Bbb R)$ if the domain is $\Bbb R$, for example. – Hagen von Eitzen Jul 19 '17 at 02:06
  • Sorry, what's the question? There are differentiable functions for which the derivative is not continuous. Is that what you are asking? – lulu Jul 19 '17 at 02:11
  • @HagenvonEitzen Oh, thank you, Hagen, you answer is what I am looking for. Can you give me a counter example or a pointer or reference of a more detailed reasoning? I'm reading Thomas Calculus, are there some discussions about this in this book that I didn't find? – M. Chen Jul 19 '17 at 02:32
  • @lulu Lulu, I just want ask is it suffice to say if a derivative of a function is continuous everywhere provided the derivative of the original function exists everywhere in the domain of the original function. – M. Chen Jul 19 '17 at 02:35

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The basic example you're looking for is $f(x) = x^2\sin\left(\frac 1x\right)$ with $f(0) = 0$. This is differentiable for all $x \neq 0$, obviously. To show differentiability at $x=0$, you want to use the definition of the derivative.

More generally, if you want an $n$ times differentiable function ($n \geq 1$) with all (locally) bounded derivatives, but whose $\text{n}$th derivative is discontinuous, let $f(x) = x^{2n}\sin\left(\frac 1x\right)$ with $f(0) = 0$. You can justify this with induction.

Note that if $f:[a,b] \to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then the set of points where $f'(x)$ is continuous is quite "large" in some sense. In particular, it's non-empty. See this amazing answer by Dave L. Renfro.

MathematicsStudent1122
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