Give an example, with proof, of a function differentiable at a point, whose derivative is non-differentiable that point. 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!
 A: 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. 
A: 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 $\displaystyle f'(x)= \lim_{h\to0} \frac{f(x+h) - f(x)} h = \lim_{h\to0} \frac{\pm h^2 - 0} h =0.$
A: 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 $x = 0$.
A: 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.
A: 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) 
