Suppose $f:(a,b)\to\mathbb R$, and $f$ differentiable at $(a,b)$. Let $x\in(a,b)$, and $f'$ discontinuous at $x$. Then, $f'(t)$ can't tend to $\pm\infty$ as $t\to x$, right?
@Fargle If $f'(x)$ tends to $\pm$ infinity then $|f'(x)|>|f'(0)|$ for all $x$ with $0<|x|\delta$ for some $\delta$. But this violates intermediate value property.
Hang on, I had a misapprehension here. I was thinking, "Well duh, if f'(t) tends to infinity, then you didn't have differentiability at x in the first place," but that's not the case---say f'(x) just happened to be 3. Oops.
Why the reference to f'(0)? We don't have that (a,b) contains 0 necessarily
@Fargle sorry for that. Rewriting with new info: If $f'$ tends to $\pm$ infinity at $x$ then $|f'(t)|>|f'(x)|$ for $0<|t|<\delta$ for some delta, which violates Darboux Theorem
Oh, I think, to apply that theorem correctly, I should write something like $|f'(t)|>1+|f'(x)|$, and then conclude, right?
@TobiasKildetoft I was thinking of something like $\sqrt[3]{x}$, which has a derivative tending to infinity. But in that case you actually fail to have differentiability at $0$
@MartinSleziak, So, can I conclude from this discussion that, if $f:(a,b)\to\Bbb R$ differentiable at every point in $(a,b)$, and $f'$ discontinuous at $x\in(a,b)$, then $f'$ oscillates from both sides of $x$?
I was supposed to write $\sup_{t\in (x,u)} f'(t) > \inf_{t\in (x,u)} f'(t)$ for every $u>x$?
But in short, I am asking what you mean by saying that $f'$ oscillates.
Of course, it is well-known that a function is continuous at a point iff oscillation at that point is zero. Which is why I mentioned similar expression to the definition of oscillation, with the change that I looked only on one side.
Certainly, if I want oscillation $\omega_f(x)$ to be positive, the positive value, this difference has to be positive on the right or on the left (or both).
This is definition of discontinuity by oscillation I have. So, by Darboux Theorem, I can't have neither jump discontinuity nor removable discontinuity for derivative. And, from discussion before, I can't have $f'$ tending to +- infinity at discontinuity. So, at discontinuity, $f'$ oscillates, right?
@MartinSleziak Wow! you are right :) That is superb text for self study.
If you look at first example - the one with $f(x)=x^2\sin\frac1x$, you can see that you get arbitrarily large positive and negative values of $f'(x)$ close to zero.
So if we use definitions from that book, in this case $f'(x)$ has infinite discontinuity at zero.
You can see this also from the graph of $f'(x)$, which is included in that answer.
I am not sure why Desmos fails so badly on this function, but at the first glance you can see that the red curve is not the graph of this function. Clearly you have $-x^2\le x^2\sin\frac1x \le x^2$ and the graph drawn there is not between these two parabolas.
I see that you do not believe the graphs from Marc McClure's answer. :-)
