Question

设z=z(x,y)由方程f(xz,x+y)=0确定，且x=1,y=0对应z=1,f(u,v)具有连

设z=z(x,y)由方程f(xz,x+y)=0确定，且x=1,y=0对应z=1,f(u,v)具有连续偏导数，fu(1,1)=fv(1,1)不等于零，求grad z(1,0)

Answer 1

令u=xz, v=x+y
则∂u/∂x=z+x*∂z/∂x
∂u/∂y=x*∂z/∂y
∂v/∂x=1, ∂v/∂y=1
∂f/∂x=∂f/∂u*∂u/∂x+∂f/∂v*∂v/∂x=0
∂f/∂y=∂f/∂u*∂u/∂y+∂f/∂v*∂v/∂y=0
∴ ∂f/∂x=∂f/∂u*(z+x*∂z/∂x)+∂f/∂v*1=0
∂f/∂y=∂f/∂u*x*∂z/∂y+∂f/∂v*1=0
代入fu(1,1)=fv(1,1), z(1,0)=1
∴∂z/∂x|(1,0)=-2
∂z/∂y|(1,0)=-1
∵grad z(x,y)={∂z/∂x, ∂z/∂y}
∴grad z(1,0)={-2,-1}