设f(x)=x(x-1)(x-2)(x-3)...(x-10),则f'(0)等于多少?

来源：学生作业学帮网编辑：学帮网时间：2024/06/04 10:38:50

f′(x)=（x）′*(x-1)(x-2)(x-3)...(x-10)+x*[(x-1)(x-2)(x-3)...(x-10)]′
故f'(0)=(0-1)(0-2)(0-3)...(0-10)=10!=3628800

f'(0)=10!