y(x)是由方程xy=ln(x+y)确定的隐函数 求dy

Question

如题所述

Answer 1

两边对x求导得
y+xy'=(1+y')/(x+y)
y(x+y)+x(x+y)y'=1+y'
y'[x(x+y)-1]=1-y(x+y)
y'=[1-y(x+y)]/[x(x+y)-1]
dy=[1-y(x+y)]/[x(x+y)-1]dx