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
$$x'=\lim_{a\to x,~a\in\mathbb Q}a'$$
 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. 
A: A $p$-adic viewpoint gives some nice intuition into the global discontinuity.
First fix a prime $p$ and define the restricted arithmetical derivative as:
$$
(p^n)'^{_p} = n p^{n-1}\\
a'^{_p} = 0
$$
Where $a \in \mathbb{Z}, (a,p) = 1$, and then extending to all of $\mathbb{Q}$ using the Leibniz rule for products and quotients.
The normal arithmetic derivative is then given for $x \in \mathbb{Q}$ by:
$$
x' = \sum_{p \in \mathbb{P}} x'^{_p}
$$
Now consider $x \in \mathbb{Q}_p$, any $x$ can be written as $x=p^n a$, where $n \in \mathbb{Z}, a \in \mathbb{Z}_p$, where $\mathbb{Z}_p$ denotes the $p$-adic integers. By the product rule, $x'^{_p} = (p^n a)'^{_p} = (p^n)'^{_p} a \ + p^n a'^{_p} = n p^{n-1}a = \frac{nx}{p}$.
For any sequence $(x_i)$ converging to $x$, we must have that, for all $i>N$ for some sufficiently large $N$, that $|x_i|_p = |x|_p$, and therefore that $x_i = p^n b$ for the same $n$ as $x$. Then we can show continuity of this derivative. 
$\forall \varepsilon > 0$ if we have $|x_i - x|_p < \delta$ then $|(x_i)'^{_p}-x'^{_p}|_p=\left|\frac{nx_i}{p}-\frac{nx}{p}\right|_p=p \ |n|_p |x_i-x|_p$ $< p \ |n|_p \delta = \varepsilon$, for $\delta = \frac{\varepsilon}{p \ |n|_p}$.
Therefore the restricted arithmetic derivative is continuous over $\mathbb{Q}_p$ for all $p$, and as $\mathbb{Q} \subset \mathbb{Q}_p$, it is continuous under the $p$-adic metric over the rationals.
The trouble comes when we attempt to unrestrict the derivative. Taking the normal arithmetic derivative, we have a sum of restricted derivatives who are each continuous over a different $\mathbb{Q}_p$. A sequence $(x_i)$ that gets closer in one metric, may not get closer with respect to another, thus leading to sequences that converge over the real metric, but will not converge over every $p$-adic metric, thus implying there will always exist infinitely many sequences that are not continuous under the mapping given by the arithmetic derivative, no matter how small of a neighborhood you consider.
