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This question was labeled as either prove or disprove. I think that this is impossible, here is my reasoning:

Any function $f(x)$ with asymptotic behavior at some point $x=a$ will hold this asymptotic behavior in the function $F(x)=\int f(x) \, dx$. We have not yet covered any technical terms with integration, only differentiation.

Help greatly appreciated!

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The absolute value function is continuous so has an antiderivative. The antiderivative is differentiable at 0, but its derivative (the absolute value function) is not.

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One of the simplest is $f(x) = \begin{cases} \phantom{+}x^2 & \text{if } x\ge0, \\ -x^2 & \text{if } x<0. \end{cases}$

The derivative is $f'(x) = \begin{cases} \phantom{-} 2x & \text{if } x>0, \\ -2x & \text{if } x<0, \\ \text{what?} & \text{if } x=0. \end{cases}$

At $x=0$ one can compute $\displaystyle f'(x)= \lim_{h\to0} \frac{f(x+h) - f(x)} h = \lim_{h\to0} \frac{\pm h^2 - 0} h =0.$

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    $\begingroup$ The derivative can also be rewritten more recognisably as $f'(x) = 2|x|$. The fact that this isn’t differentiable at 0 is then a commonly given early example of non-differentiability. $\endgroup$ Apr 8, 2020 at 13:59
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Hint:

Consider the function

$$f(x) = \begin{cases} x^3\sin\left(\frac{1}{x}\right), & \text{ if } x \neq 0 \\ 0, & \text{ if } x = 0 \end{cases} \tag{1}\label{eq1A}$$

at $x = 0$.

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  • $\begingroup$ More precisely, something like $x^3 \sin(1/x)$ if $x \neq 0$, and $0$ if $x=0$, right? That function is undefined at $x=0$. $\endgroup$
    – David Lui
    Apr 8, 2020 at 4:17
  • $\begingroup$ Might as well take $x^2\sin(1/x)$ also $\endgroup$
    – pancini
    Apr 8, 2020 at 4:18
  • $\begingroup$ @DavidLui Thanks for the feedback. You're right. I've made that change. $\endgroup$ Apr 8, 2020 at 4:21
  • $\begingroup$ @ElliotG I see that saulspatz has used your suggestion in an answer. I realize now I was considering something else, so your idea is valid. As such, I've deleted my previous $2$ comments. Also, thanks again for the feedback. $\endgroup$ Apr 8, 2020 at 4:33
  • $\begingroup$ It's a still a worthwhile example because you can modify it to get 1) $f$ not continuous, 2) $f$ continuous but not differentiable, 3) $f$ differentiable but $f'$ not continuous, 4) $f$ differentiable and $f'$ continuous but not differentiable, or 5) $f$ and $f'$ both differentiable. $\endgroup$
    – pancini
    Apr 8, 2020 at 4:36
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Consider $$f(x)=\cases{x^2\sin\left(\frac1x\right),&$x\neq0$\\0,&$x=0$}$$

We have $$f'(0)=\lim_{x\to0}\frac{x^2\sin\left(\frac1x\right)-0}{x-0}=\lim_{x\to0}x\sin\left(\frac1x\right)=0$$

For $x\neq0$, we have $$ f'(x)=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)$$ which is not continuous at $x=0$, let alone differentiable.

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    $\begingroup$ isn’t $\lim_{x\to0} x\sin(1/x)=1$ $\endgroup$
    – drfrankie
    Apr 8, 2020 at 4:33
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    $\begingroup$ @drfrankie No, you're thinking of $\frac{\sin x}{x}$. We have $x\to0$ and $|\sin(1/x)|\leq1$ $\endgroup$
    – saulspatz
    Apr 8, 2020 at 4:35
  • $\begingroup$ ah yes that’s what I was thinking of $\endgroup$
    – drfrankie
    Apr 8, 2020 at 4:37
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A general class of examples for our time is splines. Let $f(x)$ be a cubic spline with a knot at $x_0$. Then $f$ is continuous with a derivative at $x_0$, its derivative $f'$ likewise, but $f''$ is not differentiable at $x_0$. See https://en.wikipedia.org/wiki/Spline_(mathematics)

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