Question

求全微分 求由方程xyz+√(x^2+y^2+z^2)=√2所确定函数z=z(x,y)在点(1，0，

Answer 1

xyz + √(x^2+y^2+z^2) = √2, x = 1, y = 0 时，z = ±1.
记 F = xyz + √(x^2+y^2+z^2) - √2， 则
Fx = yz + x/√(x^2+y^2+z^2) ；Fy = xz + y/√(x^2+y^2+z^2),
Fz = xy + z/√(x^2+y^2+z^2)
∂z/∂x = -Fx/Fz = -[yz + x/√(x^2+y^2+z^2)]/[xy + z/√(x^2+y^2+z^2)]
= - [yz√(x^2+y^2+z^2) + x]/[xy√(x^2+y^2+z^2) + z]
∂z/∂y = -Fy/Fz = -[xz + y/√(x^2+y^2+z^2)]/[xy + z/√(x^2+y^2+z^2)]
= - [xz√(x^2+y^2+z^2) + y]/[xy√(x^2+y^2+z^2) + z]
在点 (1, 0, 1), ∂z/∂x = -1, ∂z/∂y = -√2, dz = - dx - √2dy ;
在点 (1, 0, -1), ∂z/∂x = 1, ∂z/∂y = -√2, dz = dx - √2dy .

Answer 2

dz（1,0）=dx-（根2）dy