函数z=x²y y²的全微分是
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![函数z=x²y y²的全微分是](/uploads/image/f/2328266-2-6.jpg?t=%E5%87%BD%E6%95%B0z%3Dx%C2%B2y+y%C2%B2%E7%9A%84%E5%85%A8%E5%BE%AE%E5%88%86%E6%98%AF)
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
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
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)
z=1/2*ln(x^2+y^2+4)Z'x=1/2*1/(x^2+y^2+4)*(2x)=x/(x^2+y^2+4)Z'y=1/2*1/(x^2+y^2+4)*(2y)=y/(x^2+y^2+4)所
dz=1/y/(1+x^2/y^2)*dx-x/y^2/(1+x^2/y^2)*dy
dz=1/(1+(x/1+y^2)^2)*(dx/1+y^2)-1/(1+(x/1+y^2)^2)*x*(2ydy/1+y^2)^2
先求出z对x和y的偏导数分别是1/y,-x/y^2所以dz=(1/y)*dx-(x/y^2)*dy
2^(x^2+y^2)*In2*(2xdx+2ydy)
对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】
dz=x的偏导数乘德尔塔x+y的偏导数乘德尔塔y这是最基础的题呀,直接套公式啊
先求偏导数:zx=ycos(x-y)zy=sin(x-y)-ycos(x-y)明显,两偏导数都连续故全微分存在dz=zxdx+zydy=ycos(x-y)dx+[sin(x-y)-ycos(x-y)]
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