When a field extension $E\subset F$ has degree $n$, can I find the degree of the extension $E(x)\subset F(x)?$ This is not a problem I've found stated anywhere, so I'm not sure how much generality I should assume. I will try to ask my question in such a way that answers on different levels of generality could be possible. I'm also not sure that this question isn't trivial.
Let $E\subset F$ be a fields (both can be finite if needed), and let $n$ be the degree of the field extension ($n$ can be finite if needed). Can we find the degree of the extension $E(x)\subset F(x)$ of rational function fields? 
Say $E=\mathbb F_2$ and $F=\mathbb F_4$. Then $(F:E)=2.$ I can take $\{1,\xi\}$ to be an $E$-basis of $F$. Now let $f\in F(x),$ $$f(x)=\frac{a_kx^k+\cdots +a_0 } {b_lx^l+\cdots+b_0 }$$ for $a_0,\ldots a_k,b_0,\ldots,b_l\in F$
I can write $a_0,\ldots a_k,b_0,\ldots,b_l$ in the basis: $$\begin{eqnarray}&a_i&=p_i\xi+q_i\\&b_j&=r_j\xi+s_j\end{eqnarray}$$
But all I get is $$f(x)=\frac{p_k\xi x^k+\cdots+p_0+q_kx^k+\cdots+q_0} { r_k\xi x^k+\cdots+r_0+s_kx^k+\cdots+s_0},$$
and I have no idea what to do with this. On the other hand, my intuition is that the degree of the extension of rational function fields should only depend on the degree of the extension $E\subset F,$ even regardless of any finiteness conditions.
 A: My proof is concerned with commutative algebra. Probably there is also field-theoretic one. I prove that: if $E \subseteq F$ is a finite field extension, then $[F(x) : E(x)] = [F \colon E]$.
This follows from: if $E \subseteq F$ is an algebraic field extension, then $F(x) \simeq E(x) \otimes_E F$ as $E(x)$-vector space.
The ring $F[x]$ is integral over $E[x]$. Consider the multiplicative subset $S = E[x] \setminus \{ 0 \}$ and the ring $A = S^{-1} F[x]$. Obviously $A$ is integral over the field $E(x)$, hence $A$ is a field (Atiyah-Macdonald 5.7); but $A$ is contained in $F(x)$ and contains $F[x]$, hence $A = F(x)$. Therefore
$$
F(x) = S^{-1} F[x] = E(x) \otimes_{E[x]} F[x] = E(x) \otimes_E F.
$$
A: Suppose $F = E(\alpha)$, where $\alpha$ has minimal polynomial $f(t)$.
Then
$$F(x) = E(\alpha)(x) = E(\alpha, x) = E(x)(\alpha)$$
and $\alpha$ still has minimal polynomial $f(t)$. If you're not sure on that point, the roots of $f(t)$ in $\overline{F}$ are in $\overline{F}$. The roots of $f(t)$ in $\overline{F(x)}$ are still in $\overline{F}$. (you may want to prove a theorem that says this reasoning makes sense, and implies what we want it to imply!)
Every finite example is either this one, or can be formed by iterating this one finitely many times.
