高等数学，求详细过程
展开全部
f'(x) = {[(x-1)(x-2)...(x-n)]'[(x+1)(x+2)...(x+n)] - [(x-1)(x-2)...(x-n)][(x+1)(x+2)
...(x+n)]'}/[(x+1)(x+2)...(x+n)]^2
= {[(x-2)(x-3)...(x-n)+(x-1)(x-3)...(x-n)+...+((x-1)(x-2)...(x-n+1)][(x+1)(x+2)...(x+n)]
- [(x-1)(x-2)...(x-n)][(x+1)(x+2)...(x+n)]'}/[(x+1)(x+2)...(x+n)]^2
f'(1) = [(1-2)(1-3)...(1-n)]/[(1+1)(1+2)...(1+n)]
= (-1)^(n-1)*(n-1)!/(n+1)! = (-1)^(n-1)/[n(n+1)]
...(x+n)]'}/[(x+1)(x+2)...(x+n)]^2
= {[(x-2)(x-3)...(x-n)+(x-1)(x-3)...(x-n)+...+((x-1)(x-2)...(x-n+1)][(x+1)(x+2)...(x+n)]
- [(x-1)(x-2)...(x-n)][(x+1)(x+2)...(x+n)]'}/[(x+1)(x+2)...(x+n)]^2
f'(1) = [(1-2)(1-3)...(1-n)]/[(1+1)(1+2)...(1+n)]
= (-1)^(n-1)*(n-1)!/(n+1)! = (-1)^(n-1)/[n(n+1)]
更多追问追答
追问
请问f'(1)是怎么得到的
f'(x)那么长怎么一下就得到f'(1)的
本回答由提问者推荐
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
