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I've started self-studying for AP Calculus BC, and picked up a nifty textbook called 500 AP Calculus Questions. As the name may suggest, it contains 500 questions on a variety of different subjects. Question 53 asks: if f'(x) exists at every value in the domain of f(x), then is f(x) continuous, and is f'(x) be continuous?

Evidently, the first one is true. For a function to be differentiable at all values in its domain, it must be continuous within its domain as well. My question regards the second part of the question: would the derivative of a function be continuous?

I said yes, because I could not find any true counterexamples. This counterexample came to mind, but something is bugging me about it. Let f(x) be a function, such that:

$f(x) = \begin{cases} 2x + 6 &\text{if }x \ge 0 \\ -5x + 6 &\text{if }x < 0 \end{cases}$

Alright, so the derivative of this function is not continuous, since there will be a sudden jump from 2 to -5 when passing across x = 0. However, what exactly happens at x = 0? Because two lines intersect at that point, with completely different slopes, so it seems to me as though this function is not differentiable at x = 0, and thus, that its not differentiable at all points in its domain in the first place. So it's not really a counterexample.

My question is, could anyone help me come up with some counterexamples to this fact? Functions that are continuous and differentiable at all value in their domain, but have non-continuous derivatives?

LogicBeDamned
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  • There is a counterexample, but the discontinuity cannot be a jump/hole because the derivative of a differentiable function must still satisfy an intermediate value property. As a result, constructing the counterexamples must involve a bit more care/nuance. – Brian Moehring Jan 31 '24 at 20:06
  • As answered, this is a pretty well-known topic. It is also a natural question to ask for the bright and curious ones, so I guess there's probably a duplicate here. – X-Rui Jan 31 '24 at 20:06

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There is a famous counterexample, but it does not usually show up in AP calculus courses. Take $$f(x)=\begin{cases} x^2\sin(1/x), &x\ne 0\\0, &x=0\end{cases}.$$ Then $f’(0)=0$ (compute by the definition of the derivative), but $\lim\limits_{x\to 0} f’(x)$ does not exist (compute the derivative by the product rule and chain rule).

Ted Shifrin
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