Question

求u=xyz的二阶偏导数,其中x>0,y>0,z>0 且1/x+1/y+1/x=1/a

Answer 1

对

1/x+1/y+1/z = 1/a

两端求微分,得

(-1/x^2)dx + (-1/y^2)dy + (-1/z^2)dz = 0,

改写成

dz = ----dx + ----dy

的形式,可得

Dz/Dx = ……,Dz/Dy = ……,（关键）

于是

Du/Dx = yz + xy(Dz/Dx) = ……,

Du/Dy = xz + xy(Dz/Dy) = ……,

D(Du/Dx)/Dx = ……,

D(Du/Dx)/Dy = ……,

D(Du/Dy)/Dx = ……,

D(Du/Dy)/Dy = …….