How to prove $f$ regular implies ${df/dt}$ regular In this question here How do I prove the differentials of a smooth curve is 1 dimensional?, I asked how to prove that the module of Kahler differentials of a smooth curve $C$, denoted by ${\Omega_C}$, was $1$-dimensional. My attempt in the post didn't seem to work, but a comment was left hinting that one way to do this would be to use the fact that ${k(C)}$ (function field of $C$) had transcendence degree $1$ (from this, it isn't too difficult to see that ${\Omega_C}$ was $1$-dimensional, since if ${\{x\}}$ is a transcendence basis for ${k(C)/k}$ then every ${df}$ can be written in the form ${f'dx}$ for some ${f' \in k(C)}$).
However, I now want to prove the notion of regularity of a differential is well defined. That is, suppose $f$ is regular and let ${t}$ be a local parameter of ${C}$ at a point $P$. Then ${df = (df/dt)dt}$ for some ${df/dt \in k(C)}$ (since ${\Omega_C}$ $1$ dimensional). I want to show that ${df/dt}$ is also regular at $P$. Any ideas?
 A: Question: "However, I now want to prove the notion of regularity of a differential is well defined. That is, suppose $f$ is regular and let ${t}$ be a local parameter of ${C}$ at a point $P$. Then ${df = (df/dt)dt}$ for some ${df/dt \in k(C)}$ (since ${\Omega_C}$ $1$ dimensional). I want to show that ${df/dt}$ is also regular at $P$. Any ideas?!"
Answer: If $d:\mathcal{O}_C \rightarrow \Omega^1_{C/k}$ is the universal derivation, you may look at the stalk of $d$ at the generic point $\eta$ to get a map
$$d_{\eta}:\mathcal{O}_{C,\eta}\cong K(C) \rightarrow \Omega^1_{C/k,\eta} \cong \Omega^1_{K(C)/k}$$
at the level of function fields. You are speaking of a section $s$ that is "regular at $x\in C$". This is best formulated in terms of the stalk at $x$.
If $x\in U:=Spec(A) \subseteq C$ is an open affine subset containing $x$ where $\Omega^1_{C/k}$ trivialize, you get the following. There is an isomorphism
$$(\Omega^1_{C/k})_U \cong \mathcal{O}_U e$$
and there is a map
$$d: \mathcal{O}_U \rightarrow \mathcal{O}_Ue$$
which on local sections is the following: For $V\subseteq U$ an open set and $s,t\in\mathcal{O}_U(V)$ it follows
$$d(s)=D^*(s)e, d(st)=D^*(st)e=sD^*(t)e+D^*(s)te$$
where $D:A \rightarrow A$ is a derivation and $D^*$ is the induced map of sections.
When you pass to the stalk at $x$ you get a canonical map
$$d_x: A_{\mathfrak{p}_x} \rightarrow  \Omega^1_{A_{\mathfrak{p}_x}/k}\cong   A_{\mathfrak{p}_x}e$$
An element $s \in \mathcal{O}_{C,x} \cong A_{\mathfrak{p}_x}$ is by definition on the form $s=u/v$ with $u,v\in A, v\notin \mathfrak{p}_x$, and you get the formula
$$d_x(s)= D'(s)e:=\frac{D(u)v-uD(v)}{v^2}e \in \mathcal{O}_{U,x}e$$
and since $v^2 \notin \mathfrak{p}_x$ it follows $D'(s)$ is "regular at $x$".
Since your curve is smooth it is integral hence there is an inclusion
$i_x:A_{\mathfrak{p}_x} \subseteq K(C)$ hence under this inclusion you may view your regular section $s$ as an element in $K(C)$. By definition a section  $s\in K(C)$ is regular at $x$ iff $s\in Im(i_x)$ for some point $x\in C$. This means $s=u/v$ with $v(x) \neq 0 \in \kappa(x)$ where $\kappa(x)$ is the residue field at $x$.
Note: Whenever you have a derivation $D: A\rightarrow A$ and a prime ideal $\mathfrak{p} \subseteq A$ you get an induced derivation
$D^*: A_{\mathfrak{p}} \rightarrow  A_{\mathfrak{p}}$ defined by the formula
$$D^*(u/v):=\frac{D(u)v-uD(v)}{v^2},$$
and since $\mathfrak{p}$ is a prime ideal it follows $v^2 \notin \mathfrak{p}$. Hence the derivation $D^*$ is "well defined".
