# Derivative of an algebraic power series in positive characteristic

Let $$K$$ be a field. It is easy to see that if the characteristic of $$K$$ is $$0$$ and $$f(T)=\sum_{n\ge0}a_nT^n$$ is a power series algebraic over $$K(T)$$, then $$f'$$ belongs to $$K(T)(f)$$. Indeed let $$P(X)=\sum_{i=0}^lP_i(T)X^i$$ be the minimal polynomial of $$f$$ over $$K(T)$$. By differentiating $$P(f(T))$$, one has $$Q(f(T))+f'(T)R(f(T))=0$$ with $$Q(X)=\sum_{i=0}^lP_i'(T)X^i$$ and $$R(X)=\sum_{i=1}^niP_i(T)X^{i-1}$$. Since $$R(f(T))$$ can not be zero (otherwise $$R$$ would vanish $$f$$ with a smaller degree than $$P$$) one has $$f'(T)=Q(f(T))/R(f(T))$$ But this proof does not work if the characteristic of $$K$$ is positive ($$R(f(T))$$ could be zero). So here is my question. Does $$f'$$ still belong to $$K(T)(f)$$ if the characteristic of $$K$$ is positive?

Let $$p=\operatorname{char}(K)$$.
By minimality of $$P$$, if $$R(f(T))$$ was zero, then $$R(X)=0$$. Thus, $$p$$ divides the degree of each nonzero coefficient of $$P(X)$$, so $$P$$ is not separable.
But the extension $$K((T))/K(T)$$ is separable, see: Why is $K_{\upsilon}|K$ separable for a global field $K$?