e^(x y)-xy=0的微分
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/03 05:21:08
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