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I'm confused about something I SHOULD know the answer to: Can a non-continuous function be differentiable? According to Desmos, it can. This is intuitive to me, as the y-value of a function does not affect the slope, so it doesn't seem like a vertical jump should affect the derivative.

However, I was also under the impression that, because the definition of derivative involves a nonzero $dx$, it is impossible to take the derivative at any endpoint. I suppose the one-sided limit of $\frac{f(x+h)-f(x)}{h}$ would exist, but the two-sided limit would not.

Also, a question from an old AP exam gives a graph of a derivative (no additional info) and the answer key makes the assumption that the original function is continuous. They seem to be implying that, since the derivative exists, the original function must be continuous.

To confuse me even more, I stumbled across separate terms differentiable and continuously differentiable.

Can someone help clarify? Thank you!

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  • $\begingroup$ Differentiable implies continuous, so the contraposition is non-continuous implies non-differentiable (at a point). $\endgroup$
    – Dole
    Mar 31, 2021 at 2:22
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    $\begingroup$ Desmos uses numerical approxmation methods and has to work around machine rounding error issues with floating point reals. The fact that it says $f'(5)$ is $1$ does not actually prove that f'(5) exists and equals $1$. And on top of that, it is possible that it has a logical error in the algorithm that computes $f'$ from your piecewise defined $f$. $\endgroup$
    – 2'5 9'2
    Mar 31, 2021 at 2:29
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    $\begingroup$ Also, "differentiable": a derivative exists. "continuously differentiable": a derivative exists and it is a continuous function. $\endgroup$
    – 2'5 9'2
    Mar 31, 2021 at 2:34

1 Answer 1

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Yes: Differentiability (at a point) implies continuity at the point.

Differentiability on an interval implies continuity on an interval.

The derivative of a differentiable function is not necessarily a continuous function itself, however, as

$$ f(x) = \begin{cases}x^2 \sin(\frac1x) & x \ne 0 \\ 0 & x = 0 \end{cases} $$ shows (after you do a good deal of work).

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  • $\begingroup$ Side note but perhaps of interest to the OP: derivatives do have the intermediate value property, which means that they can't have jump discontinuities. One might think that this would constrain the set of points where a derivative is discontinuous (e.g. that the set might be countable, or small in some other sense), but in fact the set can be quite large, as this great answer by Dave Renfro shows. $\endgroup$
    – user169852
    Mar 31, 2021 at 2:44
  • $\begingroup$ Excellent point, @Bungo. I'd forgotten that answer. $\endgroup$ Mar 31, 2021 at 10:03

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