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Can a derivative function have a jump discontinuity at a point and the function still be differentiable at that point? I'm pretty sure it cannot, since it means that the secant slopes do not tend to the same number, but I am not sure how to prove it from the definitions. I have read that the derivative function can have an essential discontinuity at a point and the function still be differentiable at that point. i.e.

$$ f(x) = \left\{ \begin{array}{ll} x^2sin\left(\frac{1}{x}\right) & \quad x \ne 0 \\ 0 & \quad x = 0 \end{array} \right. $$.

Are there any other kinds of discontinuity or are these only the two cases. Many thanks.

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    $\begingroup$ The derivative need not be continuous, but it must satisfy the IVP. $\endgroup$ – Randall Jan 11 at 15:34
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It follows from Darboux's theorem, that if $f$ is a differentiable function and if $f'$ is discontinuous at $a\in D_f$, then the only possible cause of that discontinuity is that the limit $\lim_{x\to a}f'(x)$ doesn't exist.

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  • $\begingroup$ but can f'(a) still exist then? I know intuitively it can't but I am not too sure how to show it. Can the derivative function look like $$ f'(x) = \left\{ \begin{array}{ll} -1 & \quad x < 0 \\ 0 & \quad x=0\\ 1 & \quad x > 0 \end{array} \right. $$ $\endgroup$ – matt Jan 11 at 16:04
  • $\begingroup$ No, by Darboux's theorem. $\endgroup$ – José Carlos Santos Jan 11 at 16:08

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