2个回答
展开全部
e^x -e^y -arctany =6
(e^x -e^y -arctany)' =0
e^x -e^y. y' - y'/(1+y^2) =0
[e^y +1/(1+y^2) ]y' = e^x
y' = e^x/[e^y +1/(1+y^2)]
(e^x -e^y -arctany)' =0
e^x -e^y. y' - y'/(1+y^2) =0
[e^y +1/(1+y^2) ]y' = e^x
y' = e^x/[e^y +1/(1+y^2)]
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