求下列方程所确定的隐函数y=f（x）的导数dy/dx

题目

求下列方程所确定的隐函数y=f（x）的导数dy/dx
e^(x+y)-xy=1

答案

两边微分e^(x+y)（dx+dy）-xdy-ydx=0→
[e^(x+y)-y]dx=[x-e^(x+y)]dy
dy/dx=[x-e^(x+y)]/[e^(x+y)-y]