e^(x+y)=xy
e^(x+y)*(1+y')=y+xy'
e^(x+y) +e^(x+y)y'=y+xy'
e^(x+y)y'-xy'=y-e^(x+y)
y'=[y-e^(x+y)]/[e^(x+y)-x]本回答被提问者和网友采纳
急急急!已知e^(x+y)-xy=0,求y'
e^(x+y)-xy=0 e^(x+y)=xy e^(x+y)*(1+y')=y+xy'e^(x+y) +e^(x+y)y'=y+xy'e^(x+y)y'-xy'=y-e^(x+y)y'=[y-e^(x+y)]\/[e^(x+y)-x]
求由方程e^(x+y)-xy=0所确定的隐函数y=f(x)的微分dy
由已知得：e^(x+y)=xy.d e^(x+y)=dxy.e^(x+y)*d(x+y)=(ydx+xdy).e^(x+y)*(dx+dy)=ydx+xdy.e^(x+y)dx+e^(x+y) dy=ydx+xdy.[(e^(x+y)]dy-xdy=[y-e^(x+y)]dx.dy={[y-e^(x+y)]\/[e^(x+y)-x]}dx.
高等数学 方程e^x+y-xy=0确定隐函数y=f(x),求dy\/dx
解；F(x,y)=e^(x+y)-xy=0 Fx=e^(x+y)-y Fy=e^(x+y)-x dy\/dx=-Fx\/Fy=-[e^(x+y)-y]\/[e^(x+y)-x]=[y-e^(x+y)]\/[e^(x+y)-x]
设函数y=y(x)由方程e^y+xy-x=0确定,求y''(0)
两边同时对x求导得：e^y·y '+y+xy '=0 得y '=-y\/(x+e^y)y ''=（y'）'=太长了,自己算.当x=0时,e^y=e,得y=1,y'=-1\/e,代入y''得答案为1\/e∧2
X+Y–XY=0 求y的二阶导数
e^(x+y)*(1+y')+y+xy'=0,解得：y'=-(e^(x+y)+y)\/((e^(x+y)+x))=(xy-y)\/(x-xy)e^(x+y)*(1+y')+y+xy'=0两边对x求导得：e^(x+y)*(1+y')^2+y''e^(x+y)+2y'+xy''=0 解得：y''=-(e^(x+y)*(1+y')^2+2y')\/((e^(x+y)+x))=(xy(1...
e^(x+y)-xy=1,求y''(0)
原式可得x=0,y=0 两边同时对x求导，得：e^(x+y)(1+y')-y'-xy'=0（1） 把x=0,y=0代入（1）得到y'(0)=-1 把（1）继续对x求导，可得：e^(x+y)(x+y)^2+e^(x+y)y''-y'-y'-xy''=0（2） 把x=0.y=0,y'=-1代入（2）得到y''（0）=-2 ...
已知y= e^(x+ y),求y'.
方法如下，请作参考：
e^(x+y)-xy=1,求y''(0)
原式可得x=0,y=0 两边同时对x求导，得：e^(x+y)(1+y')-y'-xy'=0（1）把x=0,y=0代入（1）得到y'(0)=-1 把（1）继续对x求导，可得：e^(x+y)(x+y)^2+e^(x+y)y''-y'-y'-xy''=0（2）把x=0.y=0,y'=-1代入（2）得到y''（0）=-2 ...
e^(x\/y)-xy=0求dy
两边对x求导：e^(x\/y)*(1*y-xy')\/y^2-(1*y+xy')=0e^(x\/y)\/y-xe^(x\/y)\/y^2*y'-y-xy'=0y'*x(e^(x\/y)\/y^2+1)=y-e^(x\/y)\/yy'=(y-e^(x\/y)\/y)\/(xe^(x\/y)\/y^2+x)dy=(y-e^(x\/y)\/y)\/(xe^(x\/y)\/y^2+x)*dx ...
方程ex+y次方-xy=0确定隐函数y=f(x),求dy\/dx,求完整的答案,这是一道计...
xy=e^(x+y)两边对x求导,得：y+xy’=(1+y’)e^(x+y)移项,得：[x-e^(x+y)]y’=e^(x+y)-y 整理得：y’=[e^(x+y)-y]\/[x-e^(x+y)]将xy=e^(x+y)代入,即把e^(x+y)换成xy,得：y’=(xy-y)\/(x-xy)所以dy\/dx=(xy-y)\/(x-xy)
