$p$- - adic derivatives of the Iwasawa logarithm

Question

The Iwasawa logarithm is the unique group homomorphism: $$ \log_p:\mathbf{C}_p ^\times \to \mathbf{C}_p $$ such that $\log_p(p)=0$ and if $|x-1|<1$ $$ \log_p(x) = \sum_{n=0}^\infty (-1)^{n+1}\frac{(x-1)^n}{n} $$ i.e. the classic power-series. This is a locally analytic function and so I think it is possible to define the derivative of $\log_p$, I think that should be computed in this way: $$ \frac{d}{dx}\log_p(x) = \frac{1}{x} $$ if $|x-1|<1$ . In the other cases $\log_p(x) = \log_p(p^{-v_p(x)} \cdot \mu_x \cdot x)$with $|p^{-v_p(x)} \cdot \mu_x \cdot x-1|<1$ for a certain root of 1 $\mu_x$. Then in this situation $$ \frac{d}{dx}\log_p(x) = \frac{p^{v_p(x)}}{\mu_x \cdot x} $$ Is this argument right?

Answer 1

Maybe this is the way to look at things:

You choose the root of $p$ and the root of unity to make $p^\alpha$ and $\mu$ satisfy $p^\alpha\mu x\in1+\mathfrak M$, where $\mathfrak M$ is the set of things with $|z|<1$. Then since $\log p^\alpha=0$ and $\log\mu=0$, you extend by linearity, $\log x=\log(\frac x{p^\alpha\mu})$, where now what’s in the parentheses is in the principal units $1+\mathfrak M$, and calculable by some formula that we can ignore for now, except for the fact that when $x\in1+\mathfrak M$, the derivative is $1/x$.

Now apply high-school Calculus: the derivative that we want is the reciprocal of what’s in the parentheses, multiplied by the derivative of what’s in the parentheses, that is, $$ \frac{p^\alpha\mu}x\cdot\frac1{p^\alpha\mu}=1/x\,, $$ just the result we’ve all been hoping for.