f'(x)=(x-1)'*(x-2)(x-3)……(x-100)+(x-1)(x-2)'*(x-3)……(x-100)+(x-1)(x-2)(x-3)'*……(x-100)+……+(x-1)(x-2)(x-3)……(x-100)'
除了(x-1)(x-2)(x-3)'*……(x-100)这一项外,其他都有x-3这一项
所以x=3是等于0
而(x-1)(x-2)(x-3)'*……(x-100)=(x-1)(x-2)(x-4)……(x-100)
所以f'(3)=2*1*(-1)*(-2)*……*(-97)
=-2*97!