TITLE: Implicit equation between rational function and its derivative
QUESTION [6 upvotes]: Let $f$ be a complex rational function in one variable. How does one find a complex polynomial $P$ in two variables such that $P(f,f^\prime)=0$?
EDIT: as the answers explained there is a standard procedure using resultant to obtain $P$. see also https://en.wikipedia.org/wiki/Resultant#Algebraic_geometry . 
However this process is computationally very heavy. I was wondering if in the case when $g=f^\prime$ we may know $P$ more explicitly, if we had some other, more efficient way to compute it.

REPLY [3 votes]: The magic word is "implicitization". Consider the curve in $\mathbb{P}^2$ given by $C = (f(t), f^\prime(t)).$ (as pointed out by Robert Bryant in the comments). Now, we would like to find the equation satisfied by the curve. This can be done for any pair of rational functions (not just $f, f^\prime$). Namely, we write down the pair of equations:
$$x = \frac{p(t)}{q(t)} \Leftrightarrow x q(t) = p(t)$$
$$y = \frac{r(t)}{s(t)} \Leftrightarrow y s(t) = r(t).$$
We can now eliminate $t$ from these equations, but thinking of both of them as being defined over $\mathbb{C}[x, y],$ and setting the Sylvester matrix determinant resultant to zero.

REPLY [3 votes]: As Igor noted, this is all quite standard, and I'm really not sure it belongs on MO, since I'm sure it would certainly receive a quick answer on MathSE. Anyway, in order to compute an equation for the image of the implicit curve $\bigl(f(t),g(t)\bigr)$, where $f(t)=f_1(t)/f_2(t)$ and $g(t)=g_1(t)/g_2(t)$ are rational functions, compute the $t$-resultant
$$
\text{Resultant}_t\bigl( f_2(t)X-f_1(t), g_2(t)Y-g_1(t) \bigr).
$$
More generally, this sort of question in answered by elimination theory, and as a practical matter, one uses Grobner bases to compute such quantities.