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

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]

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