Let $f$ be the restriction of a function on the 2-adic numbers (and with the 2-adic topology), to the conventional integers $\Bbb Z$.
Let $f(x)=(3x+2^{\nu_2(x)})/4$ for positive integers
Let $f(x)=(3x+2^{\nu_2(x)})/2$ for negative integers
Let $f(0)=0$
Then change my mind, but I have that $f$ is differentiable everywhere including zero using:
$f'(x)=\lim_{a\to x}\dfrac{f(x)-f(a)}{x-a}$
Including derivative $f'(0)=1$ at zero.
Question
How should $f$ be extended such that it is differentiable over all the periodic 2-adic numbers? Can the same positive / negative rule be used to define the derivative elsewhere or does it need to be enhanced, or is it even impossible?
My thoughts
FWIW if the answer to this falls into the "it can be done but you need to construct the divisor carefully" category, then the following is what I have worked out: One can uniquely factor a periodic 2-adic unit into $2^tu+n$ where $u$ is a periodic 2-adic unit whose binary representation is an infinitely repeating Lyndon word, and $t\in\Bbb Z$ and $n\in\Bbb N:n<2^tu$ the binary representation of $u$.
Then at one extreme, representing the divisor in $f$ when $x$ is a negative integers we have the Lyndon word $\overline1_2\mapsto \frac12$ and at the other extreme representing the divisor in $f$ when $x$ is a positive integer, we have $\overline0_2\mapsto \frac14$. Maybe by a suitable definition of a bijection from the Lyndon words to power of $2$ divisors we can maintain continuity? If so, what's the correct function?