On Wikipedia it is stated that the function $$ f:\mathbb{Q}_p\to \mathbb{Q}_p $$ with $f(x)=(1/|x|_p)^2$ if $x\neq 0$ and $f(0)=0$ is differentiable and its derivative is the zero-function.
How can I see that? More generally, how can I derivate a function of $p$-adic numbers explicitely?
By definition (yes, it's the same as in the general real case):
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{\frac1{|x+h|_p^2}-\frac1{|x|_p^2}}h=$$
$$=\lim_{h\to 0}\frac{p^{2\nu_p(x+h)}-p^{2\nu_p(x)}}h=0$$
Since when $\;h\;$ is very close to zero, clearly both $\;x\;$ and $\;x+h\;$ are divided by the same power of the prime $\;p\;$ (which is what the $\;p-$adic valuation $\;\nu_p(x)\;$ gives) ...