let's say $$A=\{ f:[-1,1]\to \mathbb R | f \text{ is differentiable at } [-1,1]\},$$ $$C=\{f:[-1,1]\to \mathbb R| f \text{ is differentiable and } f' \text{ is continuous at }[-1,1]\}$$ I wanted to prove that $A$ doesn't imply $C$.
I had this counter example in mind: $$f:[-1,1]\to\Bbb R,\quad f(x)= \begin{cases}x^{2}\sin(1/x), & x\neq0 \\ 0, & x=0 .\end{cases}$$ Then $$f':[-1,1]\to\Bbb R,\quad \begin{cases}2x\sin(1/x)+\cos(1/x), & x\neq0 \\ 0, & x=0 .\end{cases}$$
I tried to prove that $f'$ isn't continuous but I'm stuck.
I started with this: Take a random $a\in [-1,1]$.
We're looking for an $\epsilon>0$ such that for a random $\delta>0$ and a fixed $x \in [-1,1]$, so that $|x-a|<\delta$, $|f(x)-f(a)|>\epsilon$.
I tried a lot finding that $\epsilon$, but without success.
Hint:
It should be intuitively clear that, in $\:2x\sin(1/x)+\cos(1/x)$, the first term tends to $0$ when $x$ tends to $0$, whereas the second term has no limit.
So try to define a first sequence $(x_n)$ which tends to $0$ such that the second term is always equal to $1$, and a second sequence such that the second term is always $0$, or $-1$, or whatever you want between $-1$ and $1$.
Explicitly, for instance, $\cos1/x=1=\cos 0\iff 1/x\equiv 0\mod 2\pi$, i.e. $1/x=2n\pi$ for some $n$, or $x=1/2n\pi$.