Question

求该方程所确定的隐函数y的导数dx分之dy

Answer 1

两边对x求导：

(y-xy')/y^2=(y+xy')/(xy)

即：x(y-xy')=y(y+xy')

xy-x^2y'=y^2+xyy'

y'=(xy-y^2)/(x^2+xy)