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?

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 $\infty$.
