求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,是xyz + (x2 + y2 + z2)^(1/2) =2^(1/2)

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求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,是xyz + (x2 + y2 + z2)^(1/2) =2^(1/2)

求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,是xyz + (x2 + y2 + z2)^(1/2) =2^(1/2)
求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,
是xyz + (x2 + y2 + z2)^(1/2) =2^(1/2)

求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,是xyz + (x2 + y2 + z2)^(1/2) =2^(1/2)
为方便,记p=√(x^2+y^2+z^2)
对x求导:yz+xyz'x+(x+zz'x)/p=0, 得:z'x=-(yz+x/p)/(xy+z/p)
同样,对y求导,得:z'y=-(xz+y/p)/(xy+z/p)
所以在(1,0,-1)处,
有p=√2
z'x=-(1/p)/(-1/p)=1
z'y=-(-1)/(-1/p)=-1/p=-√2/2
所以dz=z'xdx+z'ydy=dx-√2/2*dy

我不知道补充是什么意思,就原题而言,解之如下:
对x、y、z各自求导,得
yzdz+xzdy+xydz+2xdx+2ydy+2zdz=0,从而求得
dz=-(yzdx+xzdy+2xdx+2ydy)/(xy+2z),
将x=1、y=0、z=-1代入上式,得
dz=(2dx-dy)/2