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.