函数z+z(x,y)由方程F(x+y,x+z)=1所确定,其中F有二阶连续偏导数,且F2不等于0,求Ȣ^2z⼀Ȣy^2

2025-01-14 20:03:04
推荐回答(1个)
回答1:

F(x+y,x+z)=1
F1+F2(∂z/∂y)=0 ∂z/∂y= -F1/F2
∂²z/∂y²=[-F2(F11+F12(∂z/∂y))+F1(F21+F22(∂z/∂y)]/F2^2
=[-F2(F11+F12(-F1/F2))+F1(F21+F22(-F1/F2)]/F2^2
=[-F2(F2F11+F12(-F1))+F1(F2F21+F22(-F1)]/F2^3
=[-F2^2F11+2F1F2F12-F1^2F22]/F2^3