Suppose that we have a function $f:[0,1]\to\mathbb{R}$ that is differentiable only in $x=1,1/2, 1/3, 1/4, ..., 1/n$ and in $x=0$. So a new function $f':\mathcal{A} \to\mathbb{R}$ arises, where $\mathcal{A}=\{0\}\cup \left\{\frac{1}{n} \right\}_{n\in\mathbb{N}}$.
Can I say that:
$$\lim_{\mathcal{A}\ni x \to 0}\frac{f'(x)-f'(0)}{x-0}$$
is the second derivative of the function $f$ in $x=0$? Or does the second derivative, to be defined as such, requires the existence of the first derivative in a neighborhood of $0$ of the type $[0, a]$, with $0<a\leq 1$?
The definition of the derivative is $$\lim_{x\rightarrow0}\frac{f(x)-f(0)}{x-0}$$ and by definition of the limit this is $$\forall\epsilon>0\,\exists\delta>0\;|\;\left|x\right|<\delta\;\Rightarrow\left|\frac{f(x)-f(0)}{x-0}-L\right|<\epsilon$$ The definition most definitely requires all $\left|x\right|<\delta$ to work, that is to say, at the very least $f$ needs to exist in a neighbourhood $\left(-\delta,+\delta\right)$.
The right one-sided derivative still needs a neighbourhood of the form $\left[0,\delta\right)$ to work.