Question

若y=x(x–1)(x–2)(x–3)，则y'(0)=多少呢

Answer 1

用乘法法则，F=xyz,F'=x'yz+xy'z+xyz'

这题因为求y'(0)

y'=x'(x-1)(x-2)(x-3)+x(x-1)'(x-2)(x-3)+x(x-1)(x-2)'(x-3)+x(x-1)(x-2)(x-3)'

带x项的都可以忽略

y'(0)=1*(0-1)*(0-2)*(0-3)=-6