已知隐函数XY=e(X+Y)次方，求dy

Question

如题所述

Answer 1

解法一：∵xy=e^(x+y) ==>d(xy)=d(e^(x+y)) (两端取微分)
==>xdy+ydx=e^(x+y)(dx+dy)
==>xdy+ydx=e^(x+y)dx+e^(x+y)dy
==>xdy-e^(x+y)dy=e^(x+y)dx-ydx
==>(x-e^(x+y))dy=(e^(x+y)-y)dx
∴dy=[(e^(x+y)-y)/(x-e^(x+y))]dx；
解法二：∵xy=e^(x+y) ==>xy=e^x*e^y
==>ye^(-y)=e^x/x
==>d(ye^(-y))=d(e^x/x) (两端取微分)
==>(e^(-y)-ye^(-y))dy=((xe^x-e^x)/x²)dx
==>dy=[(xe^x-e^x)/(x²(e^(-y)-ye^(-y)))]dx
==>dy=[(xe^(x+y)-e^(x+y))/(x(x-xy))]dx (分子分母同乘e^y)
==>dy=[(xe^(x+y)-xy)/(x(x-e^(x+y)))]dx (∵e^(x+y)=xy)
==>dy=[x(e^(x+y)-y)/(x(x-e^(x+y)))]dx
∴dy=[e^(x+y)-y)/(x-e^(x+y))]dx。

Answer 2

一：∵xy=e^(x+y) ==>d(xy)=d(e^(x+y)) (两端取微分)
==>xdy+ydx=e^(x+y)(dx+dy)
==>xdy+ydx=e^(x+y)dx+e^(x+y)dy
==>xdy-e^(x+y)dy=e^(x+y)dx-ydx
==>(x-e^(x+y))dy=(e^(x+y)-y)dx
∴dy=[(e^(x+y)-y)/(x-e^(x+y))]dx；本回答被提问者采纳

Answer 3

函数XY=e^(X+Y)
对x求导
y+xy'=e^(X+Y)（1+y‘）
y+xy'=e^(X+Y)+y‘e^(X+Y)
xy'-y‘e^(X+Y)=e^(X+Y)-y
y'【x--e^(X+Y)】=e^(X+Y)1-y
y' =【e^(X+Y)-y】/【x--e^(X+Y)】
所以 dy={【e^(X+Y)-y】/【x--e^(X+Y)】}dx
希望对你有所帮助，祝你学习进步！

Answer 4

原式变为：ln(xy)＝ln（e的x+y次方）
lnx+lny=x+y
两边对x求导得：1/x+dy/ydx=1+dy/dx
解得：dy/dx=y(x-1)/x(1-y)

Answer 5

见图，将dx移过去即可