z=e^xy的全微分-搜答案-智慧作业帮

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

全微分公式dz=(偏z/偏x)dx+(偏z/偏y)dy求偏导时发现是复合函数求导=[e^(y/x)*偏(y/x)/偏x]dx+[e^(y/x)*偏(y/x)/偏y]dy=[e^(y/x)*(-y/x^

z'x=2e^(2x+y)z'y=e^(2x+y)所以dz=2e^(2x+y)dx+e^(2x+y)dy

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)

dz=[sin(xy)+xycosxy]dx+(x^2cosxy)dydz|(1,1)=(sin1+cos1)dx+cos1dy再问：先求dx，dy，详细过程谢谢再答：=sin(xy)+xycosxy

(y^2+2xy-cos(y+z))/(e^z+cos(y+z))再问：没有过程吗？再答：求导：e^z*dz-y^2-2xy+cos(y+z)(1+dz)=0把含有dz的项移到一起：(e^z+cos(

解；z(x)=2x+2y²z(y)=4xy+12y²dz=(2x+2y²)dx+(4xy+12y²)dy

他说的方法对但算的好像不对,高数扔好久了,我试试哈,dz=y*(1/x^2)*e^(y/x)*dx+(1/x)*e^(y/x)*dy.另外,我不知道是不是你手误,我给出的答案是按照z=e^(y/x)算

代入：2z-2z+lnz=0--->z=1,所以z'(y)=-z/y从而dz=z'(x)dx+z'(y)dy=(e^x-yz)/(xy)

两边即对数得：lnz=xy*ln(lnu),不妨记u=x^2+y^2z'x/z=yln(lnu)+2x^2y/lnu,z'x=z[yln(lnu)+2x^2y/lnu]z'y/z=xln(lnu)+2

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+(&#

我帮你做一步下面的你应该就会了,

这里需要用到隐函数定理.令F(x,y,z)=xy-yz+xz-e^z.记Fx,Fy,Fz表示对x,y,z求偏导,则：dz/dx=-Fx/Fz=-(y+z)/(x-y-e^z),dz/dy=-Fy/Fz

dz=2e^(2x+y^2)dx+2ye^(2x+y^2)dy把对x和对y的偏导分别求了出来再乘以各自的微分项即可.

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

再问：就是这个吗？再答：是的。如还有不懂请追问，懂了请采纳。再问：还有这三题