dXY=de(X+Y)次方
xdy+ydx=e(X+Y)次方*d(x+y)
xdy+ydx=e(X+Y)次方*dx+e(X+Y)次方*dy
dy(x-e(X+Y)次方)=(e(X+Y)次方-y)dx
dy=)=[(e(X+Y)次方-y)dx]/(x-e(X+Y)次方）
微分不变性
Y+Xdy=(1+dy)*e^(X+Y)
dy=(y-e^(X+Y))/[e^(X+Y)-x]
ln(xy)=ln(e^(x+y))
d(lnx+lny)=d(x+y)
dx/x+dy/y=dx+dy
(1/y-1)dy=(1-1/x)dx
dy=(1-1/x)/(1/y-1)dx
dy=(xy-y)/(x-xy)dx
dy=de(X+Y)/X
dy=de+dey/x
（y+xy')/xy=1+y'
y'=（y-xy）/(xy-1)
y'=dy/dx
dy=[（y-xy）/(xy-1)]dx
已知隐函数XY=e(X+Y)次方,求dy
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...
怎么求微分方程的通解?
步骤：xy=e^(x+y)，微分得ydx+xdy=e^(x+y)*(dx+dy)，整理得[y-e^(x+y)]dx=[e^(x+y)-x]dy，所以dy\/dx=[y-e^(x+y)]\/[e^(x+y)-x]。已知隐函数XY=e(X+Y)次方，求dy。x y = e^(x+y)。求导：y + x * y' = e^(x+y) * (1 + y')。即： y + x * ...
已知隐函数XY=e(X+Y)次方,求y'和dy,在线等,急!!!
x y = e^(x+y) => y dx + x dy = e^(x+y) (dx+dy) 即 y dx + x dy = (x y) (dx+dy)=> dy = [(y - xy) \/ (xy-x)] dx y ' = (y - xy) \/ (xy-x)
已知隐函数XY=e(X+Y)次方,求y'和dy,
y+xy'=（1+y'）e^（x+y）则y'=（y-e^(x+y)）\/(e^(x+y)-x),dy=（y-e^(x+y)\/(e^(x+y)-x)dx
求由方程xy=e的(x+y)次方所确定的隐函数y=y(x)的导数dy\/dx
xy=e^(x+y)(y+xy')=e^(x+y)*(x+y)'y+xy'=e^(x+y)(1+y')y+xy'=e^(x+y)+e^(x+y)(1+y')所以：dy\/dx=y'=[e^(x+y)-y]\/[x-e^(x+y)].
求这个隐函数的微分 xy=e的(x+y)次 的微分dy
xy=e^(x+y)两边对x求导得：y+xy'=e^(x+y)(1+y')解得：y'=(e^(x+y)-y)\/(x-e^(x+y))=(xy-y)\/(x-xy)dy=[(xy-y)\/(x-xy)]dx
xy=e^(x+y)求dy\/dx 谢谢 我是不明白为什么方法不一样 答案不一样呢_百...
xy=e^(x+y)求dy\/dx 这是隐函数求导问题：正统方法是用：隐函数存在定理来做；另一方法是等式两边对x求导，再解出y'来：方法1：f(x,y)=xy-e^(x+y)=0 dy\/dx=-f'x\/f'y f'x=y-e^(x+y) f'y=x-e^(x+y)dy\/dx=-[y-e^(x+y)]\/[x-e^(x+y)]方法2：y+xy'=(1+y'...
求隐函数xy等于e的x+y次方的微分dy
Y+xY'=e^(X+Y)*(1+Y')Y'(x-e^(X+Y))=e^(X+Y)-Y dy=[e^(X+Y)-Y]\/[(x-e^(X+Y))]*dx
由方程xy=e^(x+y)所确定的隐函数的导数dy\/dx=?
解：两边对x求导得：y+xy' = e^(x+y)(1+y')=xy(1+y')即： dy\/dx=y' =(y-xy)\/(xy-x)
xy=e^x+y 确定隐函数y的导数dy\/dx?
xy = e^x +y xy' + y = e^x + y'y'(1-x) = y -e^x y' = (y-e^x)\/(1-x),1,∵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)]...
