Question

设e^(x+y+z)-xyz=e，则全微分dz为

Answer 1

对x求偏导
e^(x+y+z) * (1+∂z/∂x) - (yz + xy∂z/∂x)=0
e^(x+y+z) + e^(x+y+z) * ∂z/∂x - yz -xy∂z/∂x=0
[e^(x+y+z) - xy]*∂z/∂x=yz- e^(x+y+z)
∂z/∂x=[yz- e^(x+y+z)]/[e^(x+y+z) - xy]
对y求偏导
e^(x+y+z) * (1+∂z/∂y) - (xz + xy∂z/∂y)=0
e^(x+y+z) + e^(x+y+z) * ∂z/∂y - xz -xy∂z/∂y=0
[e^(x+y+z) - xy]*∂z/∂y=xz- e^(x+y+z)
∂z/∂x=[xz- e^(x+y+z)]/[e^(x+y+z) - xy]
全微分
dz=[yz- e^(x+y+z)]/[e^(x+y+z) - xy]dx + [xz- e^(x+y+z)]/[e^(x+y+z) - xy]dy