In a lecture, I saw this exercice: Denote by $k$ the field $\mathbb F_q(T)$ of characteristic $p$. Let $H$ be an irreductible polynomial of $\mathbb F_q[T]$. Denote by $v_H$ the $H$ be the valuation of $k$ associated to $H$. 1) Let $u\in k$. Determine $v_H(u')$. 2) Show that the derivative extends in a unique way to $k_H$ the completion of $k$ for the valuation $v_H$ 3) Prove that 1) still holds for $u\in k_H$.
Here is what I did. 1) Write $u=H^r\frac ab$ with $r\in\mathbb Z$, $a,b\in\mathbb F_q[T]$ with $H\not\mid a$ and $H\not\mid b$. Then $u'=rH'H^{r-1}\frac ab+H^r\left(\frac{ab'-ab'}{b^2}\right)$. Since $a'b-ab'\in\mathbb F_q[T]$, $v_H(a'b-ab')\ge0$. So $$v_H(u')\left\{\begin{array}{cl} =r-1&\text{if }p\not\mid r\\ \ge r&\text{if }p\mid r \end{array}\right.$$ 2)Since the derivative is linear and is a bounded operator on $k$ for the $v_H$-adic topology ($v_H(u')\ge v_H(u)-1$ from the previous question) and since $k_H$ is complete, the derivative extends to $k_H$ in a unique way. 3) Let $u\in k_H$ with valuation $r\in\mathbb Z$. There exists a sequence $(u_n)_{n\in\mathbb N}$ of $k$ that converges to $u$. There exists $n_0\in\mathbb N$ and $r\in\mathbb Z$ such that for every $n\ge n_0$, $v_H(u_n)=r$ and then $v_H(u'_n)= r-1$ if $p\not\mid r$. Otherwise $v_H(u_n)\ge r$. By continuity of the derivative on $k_H$ and by question 2, same holds for $u$.
Do you think what I did is correct? Can the case $p\mid r$ be improved? Thanks in advance for any answer