1-y'=e^(xy)*(1*y+x*y')
y'[xe^(xy)+1]=1-ye^(xy)
dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]
lnx-lny=xy
1/x-（1/y)*y'=xy'+y
得到y'=(y-xy^2)/(x^2*y+x)
我觉得我的方法比较常用来着……
dx/dy=(1-ye^xy)/(xe^xy+1)
设由方程X-Y=e^(xy) 确定由函数Y=f(x),则dy\/dx=?
y'[xe^(xy)+1]=1-ye^(xy)dy\/dx=y'=[1-ye^(xy)]\/[xe^(xy)+1]
设函数y=f(x)由方程x+y=e^y确定,求dy\/dx
1+y'=y'e^y 得dy\/dx=y'=1\/(e^y-1)
设y=f(x) 由方程e^y=xy确定,则dy\/dx=? 谢谢
两边对x求导有 y'e^y =y+xy'整理解得 y‘= dy\/dx = x\/(e^y-x)
设y=f(x) 由方程e^y=xy确定,则dy\/dx=?
两边对x求导有 y'e^y =y+xy'整理解得 y‘= dy\/dx = x\/(e^y-x)
设y=f(x)是由方程y=xy+e^x所确定的函数,求dy\/dx=
图
设方程y=x(e的y次方),确定了函数y=y(X),则dx\/dy=
y=xe^y两边同时求导得，（e^y表示e的y次方，y丶表示对y求导，即dy\/dx）y丶=e^y+(xe^y)y丶即dy\/dx=e^y+(xe^y)dy\/dx。化简得dx\/dy=(1-xe^y)\/e^y
求导:xy=x-e^xy,求dy\/dx
答：xy=x-e^(xy)e^(xy)=x-xy=x(1-y)两边对x求导：(xy)' e^(xy)=1-y-xy'(y+xy')e^(xy)=1-y-xy'ye^(xy)+xy'e^(xy)+xy'=1-y[ 1+e^(xy) ] xy'=1-y-ye^(xy)y'=dy\/dx= [ 1-y -ye^(xy) ] \/ [ x+xe^(xy) ]
y=y(x)是由方程y=e^y-x所确定,则dy\/dx=()?
为了求解dy\/dx，我们可以对y=e^(y-x)关于x求导：y' = (e^(y-x))' = e^(y-x) * (y'-1)移项可得：y' = e^(y-x) \/ (1-e^(y-x))因此，dy\/dx = e^(y-x) \/ (1-e^(y-x))。
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)
