Let $k$ be a field of characteristic zero, $f=f(t), g=g(t) \in k[t]$, $\deg(f)=n, \deg(g)=m$.
Assume that $n \geq 2, m \geq 2$ and $k(f,g)=k(t)$.
(1) Can one tell when $f'=f'(t) \in k[f,g]$?
Trivially, if $k[f,g]=k[t]$ then of course $f' \in k[f,g]$, but we do not know whether $k[f,g]=k[t]$ or not (we do know that $k(f,g)=k(t)$, but I am not sure how this helps).
(2) It would be nice to see an example where $f' \in k[f,g]$ but $g' \notin k[f,g]$.
Any comments are welcome!
Edit: Perhaps the following arguments may help: Write $t=\frac{u}{v}$, where $u,v \in k[f,g]$. Write $f=\sum_{0}^{n} a_it^i$, $g=\sum_{0}^{m} b_jt^j$, $a_i,b_j \in k$. Then $f'=\sum_{1}^{n} ia_it^{i-1}=\sum_{1}^{n} ia_i(\frac{u}{v})^{i-1}=\frac{w}{v^{n-1}}$, where $w \in k[u,v]$. (Then $v^{n-1}f'=w \in k[u,v] \subseteq k[f,g]$). Consider the localization $S^{-1}k[f,g]$, where $S=\{v^i\}_{i \in \mathbb{N}}.$ We have just obtained that $f' \in S^{-1}k[f,g]$. Notice that $t \in S^{-1}k[f,g]$ (since $t=\frac{u}{v}$). Therefore, $S^{-1}k[f,g]=S^{-1}k[t]$ is a UFD (a localization of a UFD is a UFD). If $S$ is primal then $k[f,g]$ is a UFD by Nagata's criterion (Theorem 15.41). However, we do not know if $S$ is primal or not.