二元函数z=x^3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

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

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=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]dy

z偏x=-sin3xy*3y+1/(x+y+1)z偏y=-sin3xy*3x+1/(x+y+1)dz=[-sin3xy*3y+1/(x+y+1)]dx+[sin3xy*3x+1/(x+y+1)]dy

z=3x²y+x/yzx=6xy+1/yzy=3x²-x/y²所以dz=zxdx+zydy=(6xy+1/y)dx+(3x²-x/y²)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

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

看图,AB段的方程为y=0将y=0代入积分后,对于dy来说,由于y是常数,dy就是0,因此这个积分为0,不用计算；对于dx这个积分来说,由于前面乘了个y,因此y=0代入后结果也为0,所以AB段的积分为

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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,

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.

u'x=2x/(x^2+y^2+z^2)u'y=2y/(x^2+y^2+z^2)u'z=2z/(x^2+y^2+z^2)du=2xdx/(x^2+y^2+z^2)+2ydy/(x^2+y^2+z^2)

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

z=x^3y^2dz=3x^2*y^2dx+2x^3*ydy,在（1,1）处的全微分dz=3dx+2dy.

对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】

dz=2xydx+x^2dy再问：有全过程吗再答：en我想知道这里的X^2Y是指的X得平方乘以Y吗？如果是过程如下：dz/dx=2xydz/dy=x^2dz=2xydx+x^2dy再问：是X的2Y次方

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