设y=y(x)是有方程cos(x+y)+e^y=1确定的隐函数,求dy

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设y=y(x)是有方程cos(x+y)+e^y=1确定的隐函数,求dy
设y=y(x)是有方程cos(x+y)+e^y=1确定的隐函数,求dy

设y=y(x)是有方程cos(x+y)+e^y=1确定的隐函数,求dy
两边微分
-sin(x+y)(dx+dy)+e^y*dy=0
[e^y-sin(x+y)]dy=sin(x+y)dx
dy=sin(x+y)dx/[e^y-sin(x+y)]

两边对x求导有
-sin(x+y)·(1+y')+y'e^y=0
整理有 y'= sin(x+y)/ (e^y-sin(x+y))
所以 dy = sin(x+y)/ (e^y-sin(x+y)) dx