计算题第二题，详细的过程
1个回答
展开全部
e^y=acos(x+y)
则，(e^y)·y'=a[-sin(x+y)]·(x+y)'
==> (e^y)·y'=-asin(x+y)·(1+y')
==> (e^y)·y'=-asin(x+y)-asin(x+y)y'
==> [(e^y)+asin(x+y)]y'=-asin(x+y)
==> y'=-asin(x+y)/[(e^y)+asin(x+y)]
则，(e^y)·y'=a[-sin(x+y)]·(x+y)'
==> (e^y)·y'=-asin(x+y)·(1+y')
==> (e^y)·y'=-asin(x+y)-asin(x+y)y'
==> [(e^y)+asin(x+y)]y'=-asin(x+y)
==> y'=-asin(x+y)/[(e^y)+asin(x+y)]
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
