How do I differentiate $f(x)=\log(x)$ in $\mathbb{Q}_2$? How do I differentiate these functions in $\mathbb{Q}_2$?


$f(x)=x^2$

$f(x)=1/x$

$f(x)=\log(x)$

I've been struggling with this for a couple of days now.

Please bear in mind I'm working without a teacher here so no laughing...

$\displaystyle f(x)=x^2: f'(x)=\lim_{\lvert a\rvert_2\to0}\frac{x^2-(x-a)^2}{a}=\lim_{\lvert a\rvert_2\to0}\frac{a^2+2ax}{a}=2x$

$f(x)=1/x:\displaystyle f'(x)=\lim_{\lvert a\rvert_2\to0}\frac{\frac{1}{x}-\frac{1}{x-a}}{a}=\lim_{\lvert a\rvert_2\to0}\frac{-1}{x(x-a)}=-x^{-2}$

But I think we would need to define $\displaystyle f'(x)=0$ for $x=0$?

As for $f(x)=\log(x)$ I have little idea where to start. Other than knowing that $\log$ is defined in $\mathbb{Q}_2$
Let's follow Wikipedia and define the 2-adic (Iwasawa) logarithm on $\Bbb{Q}_2^*$ as follows. Recall that the group $\Bbb{Q}_2^*$ is a direct product
$$
\Bbb{Q}_2^*=\langle 2\rangle\times(1+2\Bbb{Z}_2).
$$
The usual Taylor series of $\log(1+x)$ familiar from calculus
$$
x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}n
$$
converges w.r.t. the 2-adic topology whenever $x\in2\Bbb{Z}_2$. This is because for all $n$ we have $\nu_2(x^n)=n\nu_2(x)\ge n$ but we easily see that $\nu_2(n)\le n/2$. Therefore the general term $x^n/n\to0$ which is sufficient for convergence in a complete non-archimedean case. We can thus use this series and define
$$
\log(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}n
$$
whenever $x\in 2\Bbb{Z}_2$. The familiar power series identities then show that the rule $$\log(xy)=\log x+\log y\qquad(*)$$ holds whenever $x,y\in1+2\Bbb{Z}_2$.
We can extend the definition of the logarithm to all of $\Bbb{Q}_2$ (actually even further, see WP) in such a way that $(*)$ holds by declaring that $\log(2)=0$, and, as a consequence of $(*)$ that
$$
\log(2^n(1+x))=\log(1+x)=x-\frac{x^2}2+\cdots
$$ 
for all integers $n$ and all $x\in 2\Bbb{Z}_2$.

The usual business of termwise differentiation of a power series carries over to the 2-adics. Therefore when $x,a\in1+2\Bbb{Z}_2$ we get
$$
\lim_{x\to a}\frac{\log x-\log a}{x-a}=1-(a-1)+(a-1)^2-(a-1)^3+\cdots=\frac1a
$$
because the geometric series converges as $a-1\in2\Bbb{Z}_2$.

As we took care to have $(*)$, the usual calculation of the real limit goes thru, and the result $\log'(x)=1/x$ holds everywhere.
Hint

$$f(x)=\log(x)\implies f'(x)=\lim_{ a\to 0}\frac{\log(a+x)-\log(x)}{a}=\lim_{ a\to 0}\frac{\log\left(1+\frac a x\right)}{a}$$ Now, using equivalents $$\log\left(1+\frac a x\right)\sim  \frac a x$$