sum of reciprocals of derivative of polynomial at its roots If $P(x)$ is a polynomial of degree $n > 1$ with only simple roots $a_1,\ldots,a_n$, is it true that $\frac 1{P'(a_1)} + \cdots + \frac 1{P'(a_n)} = 0$, and, if so, what is the proof?  I can see this directly for $n = 2,3,4$ with some brute force for $4$.

Consider the meromorphic differential form $\omega (z)= \frac {1}{P(z)}dz$.
Its residue at infinity is $Res(\omega;\infty)=0$  and its residue at $a_j$ is $Res(\omega;a_j)= \frac {1}{P'(a_j)}$ (a standard basic result following from the definition)  
Now if you remember that the sum of the residues of a  meromorphic forms on the extended plane $\hat { \mathbb C}$ [= Riemann sphere $\mathbb P^1(\mathbb C)$] is zero, you get as required $$\Sigma_{P\in {\hat {\mathbb    C}}} Res(\omega; P)=0+ \Sigma Res(\omega; a_j)=0+ \Sigma \frac {1}{P'(a_j)}=0$$
Edit: an optional exercise
Just as a little exercise in residues of differential forms, let's analyze what happens in degree 1 if $P(x)=c\cdot (x-a)$ and $\omega (z)=\frac {1}{c\cdot (z-a)} dz$ (this is the case, correctly excluded by Joe, where the formula $\Sigma \frac {1}{P'(a_j)}=0$ is obviously false).
We still have  $Res(\omega;a)= \frac {1}{P'(a)}=\frac {1}{c}$.
At infinity we write $z=\frac {1}{t}$ and $\omega(z)dz= \frac {1}{c\cdot (z-a)}\cdot dz=\frac {t}{ c(1-at)} \cdot\frac{-1}{t^2} dt$
Hence $Res(\omega:\infty)=\frac{-1}{c} $ by the "standard basic result " evoked above.
The sum of the residues of $\omega$ is again zero, as it must: 
$\frac {1}{c}+(-\frac {1}{c})=0$
