Formula for $1/f(x)$ where $f$ is a polynomial Let $f$ be a polynomial having $n$ distinct real roots:
  $$f(x)=(x-x_1)(x-x_2)\dots(x-x_n)$$
  Prove that $$\frac{1}{f(x)}=\sum_{k=1}^n \frac{1}{f'(x_k)(x-x_k)}, \: \forall x \in \mathbb{R} - \{x_1,x_2,\dots,x_n \} $$


I don't know much about partial fractions, but this looks very much like them. I tried to use induction for this, but I couldn't really make the jump from $n-1$ to $n$ and I honestly wouldn't have been satisfied even if I had solved it that way, because I would really like to see where this formula comes from. I'm sorry if this is actually very easy, I simply don't know how to approach it
Let $s(x)$ be the summation and notice that 

$$s(x)f(x)$$ is the Lagrangian interpolation polynomial* on the points $\left(x_k,y_k\right)$ where $$y_k:=\frac{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}{f'(x_k)}.$$

But we precisely have

$$f'(x_k)=\prod\limits_{\substack{i\ne k}}(x_k-x_i)$$

so that $$s(x)f(x)$$ is identically $1$.



*Identify with

$$\sum_{k} y_k\frac{\prod\limits_{\substack{i\ne k}}(x-x_i)}{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}.$$

For conciseness the ranges $1, n$ have been left implicit.
Have you tried decomposing $\frac{1}{f(x)}$ into partial fractions. By that i mean writing $\frac{1}{f(x)} = \frac{A_1}{x-x_1}+...+\frac{A_n}{x-x_n}$ and solving for each $A_i$? Then i think the first fraction would look something like $\frac{1}{(x-x_1)(x_1-x_2)(x_1-x_3)...(x_1-x_n)}$. Then calculating $f'(x_1)$ should get you the result you want.