已知函数f(x)=x(x-1)(x-2)… (x-10),则f'(1)=

记f(x)=(x-1)x(x-2)...(x-10)=u(x)*v(x),其中u(x)=x-1,v(x)=x(x-2)...(x-10).则f'(x)=u'(x)*v(x)+u(x)*v'(x) 而后一项u(x)*v'(x)当x=1时为0.故f'(1)=u'(1)*v(1)=1*(1-2)*(1-3)...(1-10)=-9!=-362880