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Of course, differentiability implies continuity, but for a function to be differentiable on a set, say $[a,b]$, then, for the limit to exist, would we not need it to be defined on the set? I hear teachers talk as if it's implied at times, and I have an epsilon from the limit, but I do not have a delta. Could anyone provide a counterexample? Thanks!

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  • $\begingroup$ Whenever differentiability is given in a closed bounded set, I think the standard assumption is one sided differentiability on the end points. $\endgroup$ – DonAntonio Mar 16 '16 at 18:27
  • $\begingroup$ It sounds like you're confused about the definition of "continuously differentiable". Reread it carefully. $\endgroup$ – Nate Eldredge Mar 16 '16 at 18:28
  • $\begingroup$ I'm pretty sure it means that the derivative of the function is continuous. $\endgroup$ – Jon Mar 16 '16 at 18:41
  • $\begingroup$ A differentiable function can have a very discontinuous derivative. Here is a nice discussion: math.stackexchange.com/questions/112067/… $\endgroup$ – Bungo Mar 16 '16 at 18:43
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What exactly does "continuously differentiable" mean? If it means the derivative is continuous, then it does not follow.

Let $$ f(x) = \begin{cases} x^2\sin\dfrac 1 x & \text{if } x\ne 0, \\[8pt] 0 & \text{if } x = 0. \end{cases} $$

Finding $f'(0)$ is done by going back to the definition of "derivative" and finding the limit by squeezing, relying on the fact that the sine of any real number at all is $\le 1$ and $\ge -1$. You get $f'(0)=0$.

But the lack of continuity of the derivative is shown by looking at the wild oscillations of the derivative as $x\to 0$. To see that, you will need to compute the derivative at $x\ne0$.

So this function $f$ is everywhere continuous and everywhere differentiable but its derivative is not everywhere continuous.

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