求全微分 z=根号x y-搜答案-拍题作业网

设u=xy,v=lnx+g(xy),则x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下：dz=(∂f/

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

因为z=z(x,y),所以全微分是dz=P(x,y)dx+Q(x,y)dy的形式,其中P(x,y)=∂z/∂x,Q(x,y)=∂z/∂y等式两边同时对x

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

x^2+y^2+z^2+4z=02xdx+2ydy+2zdz+4dz=0(2z+4)dz-2xdx-2ydydz=(-2xdx-2ydy)/(2z+4)

设F(x,y,z)=z^2-2xyz-1则Fx=-2yz,Fy=-2xz,Fz=2z-2xyαz/αx=-Fx/Fz=-(-2yz)/(2z-2xy)=yz/(z-xy)αz/αy=-Fy/Fz=xz

Zxe^z=YZ+XYZx,Zx=YZ/(e^z-XY)Zy=XZ/(e^z-XY)dZ=Zxdx+Zydy=(ydx+xdy)Z/(e^z-xy)再问：设F(x,y,z)=e^z-xyzə

偏z/偏x=1/2根号(1-x^2-y^2)×(-2x)偏z/偏y=1/2根号(1-x^2-y^2)×(-2y)所以dz=[1/2根号(1-x^2-y^2)×(-2x)]dx+[1/2根号(1-x^2

是∫(x^2-2yz)dx+∫(y^2-2xz)dy+∫(z^2-2xy)dz=x³/3+y³/3+z³/3-2xyz+C=(x³+y³+z³

zx=[√（x²+y²）-x²/√（x²+y²）]/（x²+y²）=y²/(x²+y²)^(3/2)

dz/dx=-3/2*(x^2+y^2)^(-3/2)*2x=-3x*(x^2+y^2)^(-3/2)dz/(dxdy)=-3x*(-3/2)*(x^2+y^2)^(-5/2)*2y=9xy*(x^2

zx=1/y,代入y=1得zx=1zy=-（x/y^2)代入x=2,y=1得zy=-2所以dz=dx-2dy

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

dz/dx=1/y,在(2,1)的值是1dz/dy=-x/y^2,在(2,1)的值是-2所以dz|(2,1)=dx-2dy

zx=1/(1+(x/y)²)*1/y=y/(x²+y²)zy=1/(1+(x/y)²)*(-x/y²)=-x/(x²+y²)所以

dz=[yIn(xy)+y]dx+[xIn(xy)+x]dy分开求导

z=arctanx/y+ln√(x^2+y^2)编微分的符号打不出来,只有用d代替了dz/dx=1/(1+(x/y)^2)*1/y+1/√(x^2+y^2)*1/2√(x^2+y^2)*2x=y/(x

2zdz+zdy+ydz=-sinydx-xcosydydz=[-sinydx-(xcosy+z)dy]/(2z+y)再问：不是先等式两边同时对x求偏微分再对y求偏微分吗？再答：偏微分和全微分的概念不

对左右两边求导：(1+ez)dz=ydx+xdy.dz=1/（1+ez).（ydx+xdy）.

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