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!

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.