I have been trying to solve this proof:
My observations were:
The polynomial could be rewritten in factored form: (x-c1)(x-c2)....(x-cn)
The RHS could be rewritten as the sum of each c-term
Taking the common denominator of the RHS would lead p'(c1)*....p'(cn)*the factored form of the polynomial
But I wasn't able to proceed any further than these observations.
Any help appreciated.
Let's consider some polynomials $p, q$ such that $c$ is not a zero of either of them, and. Then by partial fractions we know:
$\begin{equation*} \frac{p(x)}{q(x) (x - c)} = \frac{A}{x - c} + \text{ other junk} \end{equation*}$
The remaining stuff has no denominators divisible by $x - c$. We are after $A$. A nice trick that gets it almost for free is to multiply by $x - c$ and have $x \to c$ in the result. You get:
$\begin{align*} \lim_{x \to c} \frac{p(x)}{q(x)} &= A + \lim_{x \to c} (\text{other junk}) (x - c) \\ \frac{p(c)}{q(c)} &= A \end{align*}$
For your particular case, it is clear that the sum of fractions of this form gives a fraction with denominator $p(x)$. You can write:
$\begin{align*} p(x) &= a \prod_{1 \le i \le n} (x - c_i) \\ \frac{1}{p(x)} &= \sum_{1 \le i \le n} \frac{A_i}{x - c_i} \end{align*}$
where you'd compute using the above trick (remember $p(c_i) = 0$):
$\begin{align*} A_i &= \lim_{x \to c_i} \frac{x - c_i}{p(x)} \\ &= \lim_{x \to c_i} \frac{x - c_i}{p(x) - p(c_i)} \\ &= \frac{1}{p'(c_i)} \end{align*}$
The last is by the definition of the derivative (and $p'(c_i) \ne 0$, by how $p(x)$ is written).