已知隐函数XY=e(X+Y)次方，求dy

Question

如题所述

Answer 1

先将等式两边同时取对数：ln（xy）=x+y
两边同时求导：（y+xy')/xy=1+y'
整理得y'=（y-xy）/(xy-1)
因为y'=dy/dx
所以dy=[（y-xy）/(xy-1)]dx

Answer 2

这种题很简单啊！！
前提是不要紧张
函数两边对x求导数就可以了e^(xy)=x+y+e-2；等式两边对x求导
得左边为d（e^(xy)）=e^(xy)*y*dx+e^(xy)*x*dy
右边=dx+dy，则有e^(xy)*y*dx+e^(xy)*x*dy=dx+dy整理即可解出dy/dx；

Answer 3

xy=e^(x+y),
∵xy=e^(x+y)
∴d(xy)=d[e^(x+y)]
∴y+xdy/dx=d(x+y)e^(x+y)=(1+dy/dx)e^(x+y)
∴(x-e^(x+y))dy/dx=e^(x+y)-y
∴dy/dx=[e^(x+y)-y] / [x-e^(x+y)]
dy=[e^(x+y)-y] / [x-e^(x+y)]dx

Answer 4

x y = e^(x+y)
求导：y + x * y' = e^(x+y) * (1 + y')
即： y + x * y' = x y * (1 + y')
解得： y' = (xy - y) / (x - xy)
dy = [(xy - y) / (x - xy)] * dx

Answer 5

e^y-xy=e
e^y·dy/dx-(y+x·dy/dx)=0
e^y·dy/dx-y-x·dy/dx=0
(e^y-x)·dy/dx=y
dy/dx=y/(e^y-x)

dy/dx不能叫做dx分之dy,因为这是个导数符号,而不是分数!