e^(x y)-xy=0的微分

我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y

方程两边对x求导,得：y+xy'+y'e^y=2y+2xy'y'e^y-xy'=y得y'=y/(e^y-x)因此dy=ydx/(e^y-x)

两边对x求导：e^xy(y+xy')=1+y'则y'=[ye^(xy)-1]/[1-xe^(xy)]x=0时,代入原方程,得：1=0+y,得：y=1因此y'(0)=[1-1]/[1-0]=0△x=0.

e^(-xy)-2z+e^z=0-ye^(-xy)-2z'(x)+e^zz'(x)=0z'(x)=ye^(-xy)/(e^z-2)-xe^(-xy)-2z'(y)+e^zz'(y)=0z'(y)=xe

Z=e^xy在x处的导函数为ye^(xy)在y处的导函数为xe^(xy)dz=ye^(xy)dx+xe^(xy)dy=2e^2dx+e^2dy

对方程两边求全微分得：(e^z-1)dz+y^3dx+3xy^2dy=0（方法和求导类似）移项,有dz=-(y^3dx+3xy^2dy)/(e^z-1)

1、y=2x/(x+1),求y'（0）y'=2/(x+1)-2x/(x+1)^2=2/(x+1)^2y'（0）=22、y=ln(x^2+3),求dyy'=1/(x^2+3)*2x=2x/(x^2+3)

dz=d(xyln(xy))=xyd(ln(xy))+ln(xy)d(xy)=xyd(xy)/(xy)+ln(xy)d(xy)=d(xy)+ln(xy)d(xy)=(1+ln(xy))d(xy)=(1

把y看做x的函数y=y(x)两边对x求导得y+xy'+y'e^y=0所以y'=-y/(e^y+x)可以继续化简又x=(e-e^y)/y所以dy/dx=-y^2/(ye^y+e-e^y)

该题为隐函数求导.xy+e^(xy)=1则y+xy'+e^(xy)(y+xy')=0解得：y'=-y/x解答完毕.

两端对x求导得y+xy'=e^(xy)*(y+xy')整理即可得dy/dx=y再问：y'=y+e^xy/e^xy-x?再答：是的啊，就是这样啦。

y'+xy=0的通解y.=Ce^(-x).特解y=x^2-2x.通解y=Ce^(-x)+x^2-2x.再问：不好意思啊，之前一直在忙别的。没有及时回复，首先谢谢你的回答。但是我觉得你的回答有点问题。‘

3、e^(xy)=2x+y^3,两边取微分d[e^(xy)]=d[2x+y^3]ye^(xy)dx+xe^(xy)dy=2dx+3y^2dy[xe^(xy)-3y^2]dy=[2-ye^(xy)]dx

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

由已知得：e^(x+y)=xy.de^(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=yd

xy-e^x+e^y=0d(xy-e^x+e^y)/dx=0d(xy)dx=y*dx/dx+x*dy/dx=y+xdy/dxd(e^x)/dx=e^x*dy/dxd(e^y)/dx=e^y*dy/dx

直接求导：e^y*y'+y+x*y'=2y*y'解得y'=dy/dx=(-y)/(e^y+x-2y).

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

取对数的时候要加绝对值x+y=ln(xy)不能直接拆成lnx+lny,因为x,y可以为负再问：书上例题其它题目有ln（x-1）也没有说x大于1啊？再答：写成ln(x-1)的形式自然规定了x>1但这个隐

dz=(y+1/y)dx+(x-x/y^2)dy