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So I know that it's possible for a differentiable function to have a discontinuous derivative. For example, https://calculus.subwiki.org/wiki/Derivative_of_differentiable_function_need_not_be_continuous

But I don't know how to visualize it in general cases.

Just say for example a differentiable function f where f '(x) = 0 for all x $\neq$ 0, but f '(x) = 1 for x = 0. What does f look like? I can't seem to wrap my head around it.

mark
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  • derivatives satisfy the intermediate value property so your example is not a derivative – Conrad Sep 30 '20 at 22:19
  • Take $f(x)=|x|$, then $f'(x)=1$ for $x>0$ and $f'(x)=-1$ for $x<0$ (unless you want $f$ to be differentiable at $x=0$). – Tuvasbien Sep 30 '20 at 22:19
  • @Tuvasbien : Of course, OP states that $f$ is required to be differentiable – MPW Sep 30 '20 at 22:29

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While indeed differentiable functions need not to have continuous derivatives

(see $f(x)=x^2\sin 1/x, x \ne 0, f(0)=0$),

the derivatives satisfy the intermediate value property by Darboux Theorem, so your example is not the derivative of an everywhere differentiable function.

Conrad
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  • Ahh, I was not aware of this theorem... thank you! – mark Sep 30 '20 at 22:29
  • happy to be of help - the structure of (non-continuous) derivatives is quite tricky and one of those "deep" real analysis problems that give rise to weird examples/counterexamples (they have dense points of continuity for example so not all Darboux functions can be derivatives either as there some of those everywhere discontinuous) – Conrad Sep 30 '20 at 22:33
  • ok so just so I'm clear on what's going on - the derivative I gave as an example in my original post - is that just not a possible derivative for any kind of function (differentiable or not)? Even though f '(x) exists for all x? – mark Sep 30 '20 at 22:40
  • What does it mean derivative of a function that is not differentiable? – Conrad Sep 30 '20 at 23:39
  • Hmm, good point. So I guess the bottom line is that it isn't enough for a derivative f '(x) to simply be defined at every point to say that f is differentiable. Is that right? – mark Sep 30 '20 at 23:47
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    yes - $f'$ needs to be a very special kind of function if it is not continuous; in other words if you give a function $g$ that is not continuos like in your example, it is highly unlikely that you will find an $f$ for which $f'=g$ everywhere - obviously you can find $f$ with $f'=g$ outside of $0$ but that is trivial since $g$ is continuous there; no value you give $g$ at $0$ except the trivial case $g(0)=0$ (making $g$ everywhere continuous) would give you an $f$ with $f'=g$ everywhere – Conrad Oct 01 '20 at 00:11