[tex]\displaystyle f(x) = (x^2+x+1)^{100}\\ \\ f(x) = (x-x_1)(x-x_2)(x-x_3)\cdot...\cdot(x-x_n) \\ \\\text{Folosind regula produsului, derivata va fi:}\\ \\f'(x) = \sum\limits_{k=1}^{200} \dfrac{f(x)}{x-x_k}\\ \\ f'(-1) = \sum\limits_{k=1}^{200} \dfrac{f(-1)}{-1-x_k} = -\sum\limits_{k=1}^{200} \dfrac{f(-1)}{1+x_k} =[/tex]
[tex]\displaystyle = - \sum\limits_{k=1}^{200} \dfrac{1}{1+x_k}\\ \\ \Rightarrow \sum\limits_{k=1}^{200} \dfrac{1}{1+x_k} = -f'(-1) \\ \\ f'(x) = 100(x^2+x+1)(2x+1) \\ \\ \sum\limits_{k=1}^{200} \dfrac{1}{1+x_k} = -100(1-1+1)(-2+1)=\\ \\ = \boxed{100}[/tex]
