f(x)=(x-1)(x-2)(x-3).(x-2010),则f(2010)的导数=?

令：f(x)=(x-1)(x-2)(x-3).(x-2010)=g(x)(x-2010);其中g(x)=(x-1)(x-2)(x-3).(x-2009);则：f '(x)=g '(x)(x-2010)+g(x);所以：f '(2010)=g '(2010)(2010-2010)+g(2010)=0+2009!=2009!;