导数题：设f(X)=（x-1)(x-2)...(x-101),求f'(1) 过程越详细越好，答得好有加分哦！小女子在此先谢啦

求导公式
f(x)=m(x)*n(x)
则
f'(x) =m'(x)*n(x)+m(x)*n'(x)

将f(x)=（x-1)(x-2)...(x-101)记作f(x)=(x-1)*g(x)
其中g(x)=(x-2)...(x-101)
f'(x)=(x-1)*g'(x)+g(x)
f'(1)=g(1) (不用计算g'(x)！）
=（-1）×（-2）×（-3）×...×(-100）
=100！

f'(x)=[（x-1)(x-2)...(x-101)]'
=(x-1)'(x-2)...(x-101)+(x-1)(x-2)'...(x-101)+(x-1)(x-2)...(x-101)'
=(x-2)(x-3)...(x-101)+(x-1)(x-3)...(x-101)+(x-1)(x-2)...(x-100)
将x=1代入上式,所有含(x-1)的项都为0,因此
f'(1)=(1-2)(1-3)...(1-101)
=(-1)(-2)...(-99)(-100)
=1*2*...99*100
=100!