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Let $q$ be a power of a prime $p$ and $w$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_w$ the completion of an algebraic closure of $K_w$, the completion of $\mathbb F_q(T)$ for the normalized valuation $v$ ($v(w)=1$). Denote by $K^{\text{sep}}_w$ the separable closure of $\mathbb F_q(T)$ in $\mathbb C_w$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K^{\text{sep}}_w$ (one still denotes it by $'$). My question: for every $\alpha\in K^{\text{sep}}_w$, does one have $v(\alpha')=v(\alpha)-1$ or $v(\alpha')\ge v(\alpha)$ if $p\mid v(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.

Let $q$ be a power of a prime $p$ and $w$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_w$ the completion of an algebraic closure of $K_w$, the completion of $\mathbb F_q(T)$ for the normalized valuation $v$ ($v(w)=1$). Denote by $K^{\text{sep}}_w$ the separable closure of $K_w$ in $\mathbb C_w$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K^{\text{sep}}_w$ (one still denotes it by $'$). My question: for every $\alpha\in K^{\text{sep}}_w$, does one have $v(\alpha')=v(\alpha)-1$ or $v(\alpha')\ge v(\alpha)$ if $p\mid v(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.

Let $q$ be a power of a prime $p$ and $P$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_P$ the completion of an algebraic closure of $\mathbb F_q(T)$ for the normalized valuation $v_P$ ($v_P(P)=1$) Denote by $K_P$ the separable closure of $\mathbb F_q(T)$ in $\mathbb C_P$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K_P$ (one still denotes it by $'$). My question: for every $\alpha\in K_P$ algebraic over $\mathbb F_q(T)$, does one have $v_P(\alpha')=v_P(\alpha)-1$ or $v_P(\alpha')\ge v_P(\alpha)$ if $p\mid v_P(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.

Let $q$ be a power of a prime $p$ and $w$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_w$ the completion of an algebraic closure of $K_w$, the completion of $\mathbb F_q(T)$ for the normalized valuation $v$ ($v(w)=1$). Denote by $K^{\text{sep}}_w$ the separable closure of $\mathbb F_q(T)$ in $\mathbb C_w$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K^{\text{sep}}_w$ (one still denotes it by $'$). My question: for every $\alpha\in K^{\text{sep}}_w$, does one have $v(\alpha')=v(\alpha)-1$ or $v(\alpha')\ge v(\alpha)$ if $p\mid v(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.

Let $q$ be a power of a prime $p$ and $P$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_P$ the completion of an algebraic closure of $\mathbb F_q(T)$ for the normalized valuation $v_P$ ($v_P(P)=1$) Denote by $K_P$ the separable closure of $\mathbb F_q(T)$ in $\mathbb C_P$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K_P$ (one still denotes it by $'$). My question: for every $\alpha\in K_P$, does one have $v_P(\alpha')=v_P(\alpha)-1$ or $v_P(\alpha')\ge v_P(\alpha)$ if $p\mid v_P(\alpha)$. Obviously, it is true for $\alpha\in\mathbb F_q(T)$.

Let $q$ be a power of a prime $p$ and $P$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_P$ the completion of an algebraic closure of $\mathbb F_q(T)$ for the normalized valuation $v_P$ ($v_P(P)=1$) Denote by $K_P$ the separable closure of $\mathbb F_q(T)$ in $\mathbb C_P$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K_P$ (one still denotes it by $'$). My question: for every $\alpha\in K_P$ algebraic over $\mathbb F_q(T)$, does one have $v_P(\alpha')=v_P(\alpha)-1$ or $v_P(\alpha')\ge v_P(\alpha)$ if $p\mid v_P(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.