f(x)=x(x-1)(x-2)......(x-999),求f\'(0)=_____

f(x)=x(x-1)(x-2)......(x-999),
f'(x)=(x-1)(x-2)......(x-999)+x(x-2)......(x-999)+x(x-1)......(x-999)+......+x(x-1)(x-2)......(x-998),
f'(0)=(0-1)(0-2)......(0-999)+0(0-2)......(0-999)+0(0-1)......(0-999)+......+0(0-1)(0-2)......(0-998)
=(0-1)(0-2)......(0-999)=-999!

!表示阶乘

f(x)=x^101+(a1)x^100+(a2)x^99+……+(a99)x^2+(a100)x
所以f(x)在x=0处的导数值为a100
也就是f(x)中x的系数
而f(x)=x(x-1)(x-2)……(x-100)
所以此系数为(-1)×(-2)×(-3)×(-4)×……×(-99)×(-100)=100的阶乘
也即所求为 100！

999!