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
问题描述:
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
可以看成z=z(x,y),则u=xyz(x,y)=f(x,y)
标题中的条件是1/x+1/y+1/z=1/a
答
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 = …….