设函数z=e^y cos(x-y),则-搜答案-搜狗做题网

df(x,y,z)/dx=[d(z^2)/dx]*y*e^x+y*z^2*(de^x/dx)=2zye^x(dz/dx)+y*z^2*e^x另,由x+y+z+xyz=0求dz/dx两边对x求偏导1+0

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

e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^y=1,则y=0所以y'(

两边同时微分zdx+xdz+zdy+ydz+xdy+ydx=0(x+y)dz+(y+z)dx+(z+x)dy=0dz=-[(y+z)dx+(z+x)dy]/(x+y)

对y求导,e^z*z'(y)=xz+xyz'(y),əz/əy=z'(y)=xz/(e^z-xy)

两边微分e^zdz-yzdx-xzdy-xydz=0(e^z-xy)dz=yzdx+xzdy∂z/∂y=xz/(e^z-xy)=xz/（xyz-xy）=z/（yz-y）

e^z-z+xy^3=0偏z/偏x：z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/

两端对x求偏导得：-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得：-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,

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

是e的z次方原式化作e²=x-y-2两边取对数Z=ln(x-y-2)∂z/∂x=1/(x-y-2)∂z/∂y=-1/(x-y-2)再问：是z次

e^z-xyz=0z＝㏑x+㏑y+㏑z[偏z偏x]＝1/x+（1/z）[偏z偏x]（这里y看成常数）[偏z偏x]＝（1/x）/｛1-（1/z）｝＝z/[x(z-1)]

对X的偏导=yz/(e^z-xy)对Y的偏导=xz/(e^z-xy)

再问：非常感谢，还要问大侠一道题面目。曲线y=x³+3x的拐点坐标为？？？再答：y'=3x²+3y''=3x令y"=0，得x=3当x=3时，y=36所以拐点坐标（3，36）

dz=-dx-dy

x+2y-z=3e^(xy-xz)两边对x求导,z看成是x的函数求偏导得,y看成常数,得1-əz/əx=3(y-z-xəz/əx)e^(xy-xz)=><

令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)

dz=[2e^(2x+y)]dx+[e^(2x+y)]dy

x+2y+z=e^(x-y-z)两边对x求偏导注意到z=z(x,y)1+z'=e^(x-y-z)*(1-z')...(1)再对x求偏导z"=e^(x-y-z)(1-z')^2-z"e^(x-y-z).

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