设xy+lnx-lny=o,则dy/dx=

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设xy+lnx-lny=o,则dy/dx=
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设xy+lnx-lny=o,则dy/dx=
设xy+lnx-lny=o,则dy/dx=

设xy+lnx-lny=o,则dy/dx=
xy+lnx-lny=0
两边对x求导
(y+xy')+1/x-y'/y=0
y+1+(x-1/y)y'=0
y'=(y+1)/(1/y-x)
y'=y(y+1)/(1-xy)

两边取微分 ydx+xdy+1/xdx-1/ydy=0

dxy+dlnx-dlny=0
xdy+ydx+dx/x-dy/y=0
所以dy/dx=(y+1/x)/(1/y-x)
即dy/dx=(xy²+y)/(x-x²y)