求方程e∧z=x y xyz所确定的隐函数z=f(x,y)的偏导数-搜答案-大师作文网

这是隐函数,把z看成是x,y的函数.两边对x求导,得：e^z*z'x=yz+xy*z'x,这样得：z'x=yz/(e^z-xy)=yz/(xyz-xy)=z/(xz-x)两边对y求导,得：e^z*z'

dz=y*x^(y-1)/cosz*dx+x^y*lnx/cosz*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'(

z=x/ln(y/2)z′（x）=1/ln(y/2)z′（y）=-x/ln(y/2)^2*(1/(y/2))*1/2=-2x/(y*ln(y/2）^2)

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）

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)

x=z(lny-lnz)对x求导1=∂z/∂x*(lny-lnz)+z*(0-1/z*∂z/∂x)1=∂z/∂x(lny-lnz

对方程e^(-xy)+2z-e^z=2两边微分,有：e^(-xy)*d(-xy)+2*dz-e^z*dz=0-e^(-xy)*(x*dy+y*dx)+2*dz-e^z*dz=0移项,得：(e^z-2)

方程两边对x求偏导：yz+xyəz/əx=(z+xəz/əx)e^xz得：əz/əx=(ze^xz-yz)/(xy-xe^xz)方程两边对y

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,

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

先对x求偏导数得z'(x)cosz=yz+z'(x)y所以z'(x)=yz/(cosz-y)同理对y求偏导数得z'(y)=xz/(cosz-x)所以dz=yz/(cosz-y)dx+xz/(cosz-

我帮你做一步下面的你应该就会了,

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).

your answer here

两边求微分的2xdx+2zdz=2e^zdy+2ye^zdz解得dz=（2e^zdy-2xdx）/(2z-2ye^z)=(e^zdy-xdx）/(z-ye^z)

F(x,y,z)=xy+e^xz-zlny-1.Fx=y+ze^xzFy=x-z/yFz=xe^xz-lnyz对x的偏导：-Fx/Fz=-(y+ze^xz）/（xe^xz-lny）z对y的偏导：-Fy

e^(-xy)-x^2*y+e^z=z,令F(x,y,z)=e^(-xy)-x^2*y+e^z-z=0分别对F取x,y,z的偏导数,可得əF/əx=e^(-xy)*(-y)-2xy