设y=y(x)由方程xy+lnx+lny=1所确定,求dy/dx.
问题描述：
设y=y(x)由方程xy+lnx+lny=1所确定,求dy/dx.
答
两边求导
xdy+ydx+dx/x+dy/y=0
同除dx得
xdy/dx+y+1/x+(dy/dx)/y=0
整理得
dy/dx=-(xy^2+y)/x^2y+x
答
xy+lnx+lny=1对x求导
y+xy'+1/x+y'/y=0 (其中y'表示dy/dx)
所以
y'=(-1/x-y)/(x+1/y)
=-(y+xy^2)/(x^2y+x)