If $a,b,c,\cdots k,$ are roots of the equation $\displaystyle f(x)= 0$, then the value of $\displaystyle \dfrac{f(x)}{x-a}+\dfrac{f(x)}{x-b}+\cdots+\dfrac{f(x)}{x-k}$, is

Let f(x) = (x-a)(x-b)...(x-k), so differentiating gives f'(x) = sum over all roots of f(x)/(x-r). Therefore, adding f(x)/(x-a) + f(x)/(x-b) + ... + f(x)/(x-k) yields f'(x). The result is f'(x).

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