Question

设方程xy-e的x次方+e的y次方=0确定了函数y=y(x),求dx分之dy.

Answer 1

xy-e^x+e^y=0

对x求导

则(xy)'=1*y+x*y'

(e^x)'=e^x

(e^y)=e^y*y'

所以y-e^x+(x+e^y)y'=0

y'=(e^x-y)/(x+e^y)

所以dy/dx=(e^x-y)/(x+e^y)