设xy-lny=a 求dy/dx
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设xy-lny=a 求dy/dx
设xy-lny=a 求dy/dx
设xy-lny=a 求dy/dx
这是隐函数求导
xy-lny=a,对两边求偏导,有xdy+ydx-1/ydy=0.整理得dy/dx=y2/(1-xy)
方程两边对x求导得
y+x*dy/dx-1/y*(dy/dx)=0
整理得
dy/dx=(y*y)/(1-xy)
y+x*dy/dx-1/y*dy/dx=0
y=(1/y-x)dy/dx dy/dx=y/(1/y -x)=y^2/(1-xy)
xy-lny=a
y+xdy/dx-1/y*dy/dx=0
(x-1/y)dy/dx=-y
dy/dx=y/(1/y-x)
x+xy'--y'/y=0 所以:dy/dx=y^2/1-xy
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