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?



Derivative of an algebraic power series in positive characteristic
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 $K$?](https://mathoverflow.net/questions/287873/why-is-k-upsilonk-separable-for-a-global-field-k)


So this situation cannot occur, and your argument goes through.


