作业帮一道偏倒数题目e^z-xyz=0求az^2/axay
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一道偏倒数题目e^z-xyz=0求az^2/axay
一道偏倒数题目
e^z-xyz=0求az^2/axay
一道偏倒数题目e^z-xyz=0求az^2/axay
e^z - xyz = 0
∂(e^z)/∂x - y · ∂(xz)/∂x = 0
e^z · ∂z/∂x - y · (z · 1 + x · ∂z/∂x) = 0
e^z · ∂z/∂x - yz - xy · ∂z/∂x = 0
(e^z - xy) · ∂z/∂x = yz
∂z/∂x = yz/(e^z - xy)
= yz/(xyz - xy)
= z/(xz - x)
e^z - xyz = 0
e^z · ∂z/∂y - x · (z · 1 + y · ∂z/∂y) = 0
(e^z - xy) · ∂z/∂y = xz
∂z/∂y = xz/(e^z - xy)
= xz/(xyz - xy)
= z/(yz - y)
∂²z/∂x∂y = ∂(∂z/∂x)/∂y
= ∂[z/(xz - x)]/∂y
= [(xz - x) · ∂z/∂y - z · (x · ∂z/∂y - 0)]/(xz - x)²
= [(xz - x) · z/(yz - y) - z · (x · z/(yz - y))]/(xz - x)²
= z/(yz - y) · [(xz - x) - xz]/(xz - x)²
= z/[y(z - 1)] · (- x)/[x²(z - 1)²]
= - z/[xy(z - 1)³]
两边对X求导得：e^z*z'(x)-(yz+xyz'(x))=0，求得：z'(x)=yz/(e^z-xy)=z/(xz-x)， z'(y)=z/(yz-y)。
所以：z''(xy)={z'(y)(xz-x)-zx*z'(y)} /(xz-x)^2=(-x*z/(yz-y) )/(xz-x)^2 = -z/xy(z-1)^3
e^zdz/dx-xydz/dx-yz=0,dz/dx=(yz)/(e^z-xy);
e^zdz/dy-xydz/dy-xz,dz/dy=(xz)/(e^z-xy)
e^zdz^2/dxdy-xydz^2/dxdy-xdz/dx-z-ydz/dy=0
dz^2/dxdy=(xdz/dx+z+ydz/dy)/(e^z-xy)
=2xyz