u'=2x，u ' ' = 2； When n≥3, u(n-order) =0 u(n-order) represents the n-order derivative of u, the same below.

V'= 1/( 1+x), when n≥2, v(n-order) = (-1) * n! /( 1+x)^n

F(n order) (x)=∑(k from 0 to n) c (n, k) u (n-k order) v(k order)

∴f(n order) (0)=C(n, n-2)*u(n-(n-2 order)) (0)*v(n-2 order) (0) +v(n order) In other words, the corresponding derivative of u is 0.

=n(n- 1)/2*2*(- 1)^(n-3)*(n-2)！ /( 1 )^(n-2)+(- 1)^(n- 1)*n！ /( 1)^n

=(- 1)^(n- 1)*(2n！ )

That is, f(n order) (0) = (- 1) (n- 1) * (2n! ), where n≥3