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求有方程y=x+ln y所确定的函数y=y(x)的微分dy
人气:211 ℃ 时间:2019-08-20 20:44:38
解答
F(x,y)=x+lny-y=0
dF(x,y)=0=(∂F(x,y)dx/∂x)+(∂F(x,y)dy/∂y)
dy/dx=-(∂F(x,y)/∂x)/(∂F(x,y)/∂y)
=-1/(1/y-1)
=y/(y-1)
dy=ydx/(y-1)
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