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 $4$.
 A: 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: 
$\frac {1}{c}+(-\frac {1}{c})=0$
A: In fact, we have 
$$\sum_{j=1}^n\frac{f(a_j)}{p'(a_j)}=f[x_1,\ldots,x_n],$$
where $f=1$ and $f[x_1,\ldots,x_n]$ is the $n$-th divided difference of $f$. 
Here is a proof of the used identity.
A: It's the residue theorem, applied to the function $g(z):={1\over f(z)}$
and a large circle $\gamma_R:=\partial D_R$:
As $p:={\rm deg}(f)\ge2$ one has
$$\left|\int_{\gamma_R}{1\over f(z)}\ dz\right|  \leq\  C\ {1\over  R^p}\ 2\pi R \to 0\qquad (R\to\infty)\ .$$
This implies that the sum of the residues of ${1\over f}$ at the poles $a_k$ of ${1\over f}$ is zero. Since all zeros of $f$ are supposed to be simple we can write $f$ near a zero $a_k$ as $f(z)=(z-a_k)g(z)$ with $g$ analytic near $a_k$ and $g(a_k)=f'(a_k)\ne0$. It follows that the residue of ${1\over f}$ at $a_k$ is $${1\over g(a_k)}={1\over f'(a_k)}\ .$$
Altogether we see that $$\sum_{k=1}^n {1\over f'(a_k)}\ =\ 0\ .$$
A: The partial fraction decomposition yields
$$\frac{1}{P(x)}= \sum_{i=1}^n \frac{c_i}{x-a_i}  $$
Multiplying by $x-a_j$ yields 
$$\frac{x-a_j}{P(x)}= \sum_{i=1}^n \frac{c_i(x-a_j)}{x-a_i}  (*)$$
from where $c_j =\frac{1}{P'(a_j)}$.  [ED: $P(x)=(x-a_j)Q(x)$, sub it back up and note that $P'(a_j)=Q(a_j)$]
Hence
$$1=\sum_{i=1}^n \frac{1}{P'(a_i)} \frac{(x-a_1)\cdots(x-a_n)}{x-a_i}$$
The coefficient of $x^{n-1}$ is 
$$0= \sum_{i=1}^n \frac{1}{P'(a_i)}$$
A: Here is a link to a youtube video which I have made proving this result using complex analysis and the residue theorem:
https://www.youtube.com/watch?v=m_LWdPl6npw
