展开全部
本题关键在于将 (x-1)(x-2)…(x-100) 视为一个函数 p(x),
对 f(x) = x*p(x) 乘积求导,计算f'(0),无需实际求出 p'(x).
f(x) = x * [(x-1)(x-2)…(x-100)]
f'(x) = (x)' * [(x-1)(x-2)…(x-100)] + x * [(x-1)(x-2)…(x-100)]'
= [(x-1)(x-2)…(x-100)] + x * [(x-1)(x-2)…(x-100)]'
f'(0) = [(-1)(-2)…(-100)] + 0 * [(x-1)(x-2)…(x-100)]'|x=0
= (-1)(-2)…(-100)
= 100!
对 f(x) = x*p(x) 乘积求导,计算f'(0),无需实际求出 p'(x).
f(x) = x * [(x-1)(x-2)…(x-100)]
f'(x) = (x)' * [(x-1)(x-2)…(x-100)] + x * [(x-1)(x-2)…(x-100)]'
= [(x-1)(x-2)…(x-100)] + x * [(x-1)(x-2)…(x-100)]'
f'(0) = [(-1)(-2)…(-100)] + 0 * [(x-1)(x-2)…(x-100)]'|x=0
= (-1)(-2)…(-100)
= 100!
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
f'(x)=x'*(x-1)(x-2)…(x-100)+x(x-1)'*(x-2)…(x-100)+……+x(x-1)(x-2)…(x-100)'
=(x-1)(x-2)…(x-100)+x(x-2)…(x-100)+……+x(x-1)(x-2)…(x-99)
后面每项都有x
则x=0时等于0
所以f'(0)=-1*(-2)*……(-100)=100!
=(x-1)(x-2)…(x-100)+x(x-2)…(x-100)+……+x(x-1)(x-2)…(x-99)
后面每项都有x
则x=0时等于0
所以f'(0)=-1*(-2)*……(-100)=100!
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
f(x)=x^101+100次项+……+二次项+(-1)*(-2)*……(-100)x
f'(x)=x高次项+……+一次项+100!
f’(0)=100!
f'(x)=x高次项+……+一次项+100!
f’(0)=100!
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询