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There have been MANY variations of this question answered elsewhere, but I could find no example of this particular question, and it's important:

Say you have a piecewise function defined by different functions on different domain intervals. At the point where the two intervals meet:

-> The function value is well defined

-> The derivative of that function is the same from the left and the right, therefore it is continuous

-> But the function itself is not continuous at that point - it has a break.

Do we say the derivative exists at that point?

For a concrete example:

y = 0 if x < 0

y = cos x if x >= 0

In this example, at the point x=0, the derivative of y = cos x is - sin x = -sin 0 = 0. And the derivative of y = 0 is always 0. So the derivative, y', is continuous across x = 0. It is the same slope from the right or the left side.

BUT, at x = 0, the graph itself is at 1, but jumps to 0 on the left - it is defined at 0 but discontinuous.

So when a function is discontinuous (but defined) at a point, yet the derivative IS continuous at that point (same from both directions), do we say the derivative at that point exists? Note in this case it is not a multiple valued function - the point at 0 is well defined - it equals 1.

This question can be generalized: if at a given point, the Nth derivative of a function is not continuous, do we consider the (N+i)th derivative at that point for all i to also be undefined?

Thanks so much for your thoughts.

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    $\begingroup$ Welcome to MSE! <> In a word, no. You can check that the definition, a limit of difference quotients, does not exist at that point (and are welcome to post your own answer if you like). $\endgroup$ Oct 10, 2023 at 22:34
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    $\begingroup$ If your definition of the derivative is $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$, note that the numerator is $f(x+h)-f(x)$, not $f(x+h)-\lim_{t \to x}f(t)$. So continuity is necessary. $\endgroup$
    – Brian Tung
    Oct 10, 2023 at 23:08

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Remember that a differentiable function is always continuous (this is well known and you can check it in many books), and differentiable is the same as the derivative exists at any point. Therefore, if a function is not continuous at one point, then the derivative does not exist at that point. Simple as that. Hope this helps.

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  • $\begingroup$ Markdown isn't quite LaTeX, so we use asterisks to do italics; for example, *differentiable* yields differentiable. (That term is probably preferable to "derivable.") $\endgroup$
    – Brian Tung
    Oct 10, 2023 at 23:06
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – CrSb0001
    Oct 10, 2023 at 23:11

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