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If a function $f$ defined on $[a,b]$ and differentiable everywhere, does it mean that its derivative $f'(x)$ is continuous everywhere on $[a,b]$?

My understanding is 'yes'. Because if there were 'a gap' in $f'(x)$, we could integrate back to $f(x)$ and show that where the gap is, there would be a sharp "angle" where $f(x)$ is not differentiable.

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Not necessarily. The standard example is $$ f(x)=\begin{cases} x^2\sin\dfrac1x & x\ne0,\\0 & x=0. \end{cases} $$ $f'(0)=0$, but $\lim_{x\to0}f'(x)$ does not exist.

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  • $\begingroup$ Can there be other examples? I found $x^2\cos{\frac1x}$, but can't think of any other. $\endgroup$
    – Kay K.
    May 16, 2017 at 15:14
  • $\begingroup$ You can try $(x-x_0)^a\sin((x-x_0)^b)$, $(x-x_0)^a\cos((x-x_0)^b)$ for certain values of $a$ and $b$ and any $x_0$. Then you can combine them. $\endgroup$ May 16, 2017 at 16:13
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    $\begingroup$ Thanks, but would there be some other form...? $\endgroup$
    – Kay K.
    May 16, 2017 at 16:20
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    $\begingroup$ I think $a(x-x_0)^{m+1}\sin^p\left(\frac1{(x-x_0)^m}\right) + b(x-x_1)^{n+1}\cos^q\left(\frac1{(x-x_1)^n}\right)$ work all as you said, but still wondering if there's other form. $\endgroup$
    – Kay K.
    May 16, 2017 at 16:30

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