Question

f(x)=x(x-1)(x-2)…(x-100),则f’(0)等于多少?

Answer 1

f'(x)=(x-1)(x-2)...(x-100)+x(x-2).(x-100)+x(x-1)(x-3)...(x-100)+...

后面各项依次少一个因式,但都有x

所以f'(x)=(-1)*(-2)*.*(-100)+0 (因都有x,乘积为0)

=1*2*.*100

=100!(100的阶乘)