What is the derivative of a polynomial at $\infty$? Let $f$ be a polynomial defined on the Riemann sphere. I'm struggling to understand in what sense such a map can be said to be "holomorphic" at $\infty$. What is the derivative of $f$ at $\infty$?
I have a chart $z\to\frac1z$ mapping $\infty$ to $0$ and vice versa. So I think I need to work out the derivative of $1/f(\frac 1 z)$ at $z=0$. So:
$$\lim_{z\to 0} \frac {\frac{1}{f(\frac1z)}-\frac1{f(\frac 1 0)}} {z}=\lim_{z\to0}\frac{1}{zf(\frac 1 z)}$$
Expanding the polynomial $f$, we see that if $\deg f>1$, $zf(\frac 1 z)\to \infty$ as $z\to 0$, so the derivative of $f$ at infinity is $0$, but if $f$ is affine of leading coefficient $a$, the derivative will be $\frac 1 a$.
Is this correct? And what is the meaning of the calculation I've just done? In particular, does this result not depend on the choice of chart?
 A: To avoid confusion, it's convenient to use different variables for different coordinate functions.
Let $z$ be the standard (affine) coordinate, and define $w = 1/z$. Your question is whether $f(z)$ is differentiable at $z = \infty$, which is the same thing as whether $f(1/w)$ is differentiable at $w = 0$.
To allay worries about coordinate charts, you should compute a differential rather than the derivative with respect to some coordinate, since that is a genuine operation on scalar fields. That is, you want to find whether 
$$ \mathrm{d} f(z) = f'(z) \mathrm{d} z $$
doesn't have a pole at $z = \infty$. This is awkward since $\mathrm{d}z$ has a double pole at $\infty$; so you need $f'(z)$ to have a double zero at $\infty$.
But again the change of coordinate is our friend:
$$ \mathrm{d}z = -z^2 \mathrm{d} w = -\frac{\mathrm{d} w}{w^2}$$
so you can  rewrite
$$ \mathrm{d} f(z) = - f'(z) z^2 \mathrm{d} w = -f'(1/w) \frac{\mathrm{d}w}{w^2} $$
In any case, the end result is that only the constant polynomials are differentiable at $\infty$.
