Is the extended arithmetic derivative continuous? Is the extended arithmetic derivative continuous?  Where we extend the normal arithmetic derivative with the quotient rule, making $x':\mathbb Q\mapsto\mathbb Q$.
$$p'=1,~p\text{ is prime}\\(ab)'=a'b+ab'\\(-a)'=-a'\\\left(\frac ab\right)'=\frac{a'b-ab'}{b^2}$$
From here we can derive that $0=0'=1'$.
I can then see, for example, that the following sequence tends to $0$:
$$\lim_{n\to\infty}\left(\frac1{2^n}\right)'=\lim_{n\to\infty}\frac{-n}{2^n}=0$$

Though I fail to manage the general case of whether the arithmetic derivative is continuous in a neighborhood of zero.
Is it continuous anywhere else?

It is said to be continuous at $x$ for $x\in\mathbb Q$ if
$$x'=\lim_{a\to x,~a\in\mathbb Q}a'$$

Let $n$ be a large integer. Let $p=p(n)$ be a prime in the neighborhood of $\sqrt n2^n$. E.g., by Bertrand, we know there's a prime $p$ between $\sqrt n2^n$ and $2\sqrt n2^n$. Note that $(2^n)/p\to0$ as $n\to\infty$. $$\left({2^n\over p}\right)'={pn2^{n-1}-2^n\over p^2}={n\over2}{2^n\over p}-{2^n\over p^2}\to\infty$$ as $n\to\infty$, a pretty serious violation of continuity at zero. 
