A calculus problem by Abhinav Raichur

given y=1/[x+x^{2}] hence y(x+x^{2}) = 1. (asterix denotes multiplication)

differentiating once we get

$y^{(1)}$ * (x+x^{2}) + y * (2x+1) =0 ( first derivative relation )

differentiating again we get ....

$y^{(2)}$ * (x+x^{2}) + 2 * $y^{(1)}$ (2x+1) + 2 * y =0 ( second derivative relation )

going on like that ... we get the general form to be ........

(x+x^{2}) * $y^{(n)}$ + (2x+1) * $y^{(n-1)}$ * n + n * (n-1) * $y^{(n-2)}$ = 0

we need the derivative at x=0 ........... plugging in x=0 and n=100000 we get

100000 * $y^{(99999)}$ + 9999900000 * $y^{(99998)}$ = 0. hence '0' is the corect answer.