sum of reciprocals of derivative of polynomial at its roots If $P(x)$ is a polynomial of degree $n &gt; 1$ with only simple roots $a_1,\ldots,a_n$, is it true that $\frac 1{P&#39;(a_1)} + \cdots + \frac 1{P&#39;(a_n)} = 0$, and, if so, what is the proof?  I can see this directly for $n = 2,3,4$ with some brute force for $4$.
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 

$$0= \sum_{i=1}^n \frac{1}{P'(a_i)}$$
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