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高数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

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')e^(x+y) xy'-y'e^(x+y)=e^(x+y)-y 解出:y'=[e^(x+y)-y]/[x-e^(x+y)] 两种方法结果是一样的.

直接求,两边对x求导 e^(x+y) * (1+y') = y + xy' 这里e^(x+y)=xy的 所以可以写成 xy(1+y')=y+xy' 这样就和两边取对数再求一样的形式了

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

已知xy=e^(x+y),求dy/dx.解一:将原式写成F(x,y)=xy-e^(x+y)=0 则dy/dx=-(F/x)/(F/y)=-[y-e^(x+y)]/[x-e^(x+y)]=-(y-xy)/(x-xy)=(xy-y)/(x-xy);解二:直接求导:y+xy′=[e^(x+y)](1+y′)=xy(1+y′)=xy+xyy′,(x-xy)y′=xy-y;故y′=(xy-y)/(x-xy);解三:两边取对数后再求导:lnx+lny=x+y;(1/x)+y′/y=1+y′;y+xy′=xy+xyy′;(x-xy)y′=xy-y;故y′=(xy-y)/(x-xy).三种方法都可以.

xdy+ydx=e^(x+y)(dx+dy)

d(xy)=de^(x+y)xdy+ydx=e^(x+y)d(x+y)xdy+ydx=e^(x+y)(dx+dy)xdy+ydx=e^(x+y)dx+e^(x+y)dyxdy-e^(x+y)dy=e^(x+y)dx-ydx[x-e^(x+y)]dy=[e^(x+y)-y]dxdy=[e^(x+y)-y]dx/[x-e^(x+y)]

直接带入公式就可以了啊

ydx/dy+x=(e^x)(e^y)dx/dy+(e^x)(e^y) dx/dy=[(e^x)(e^y)-x]/[y-(e^x)(e^y)] dx/dy=(xy-x)/(y-xy) dx/dy=x(y-1)/y(1-x) 忘光了,不知道对不..

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].

xy=e^(x+y) x.dy/dx + 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) ]

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