z=√（x^2+y^2)的二阶偏导数？

Question

如题所述

Answer 1

Answer 2

∂z/∂x = x/√(x^2+y^2), ∂z/∂y = y/√(x^2+y^2)
∂^2z/∂x^2 = ∂[x(x^2+y^2)^(-1/2)]/∂x

= (x^2+y^2)^(-1/2) + x(-1/2)(x^2+y^2)^(-3/2)2x
= (x^2+y^2)^(-1/2) - x^2(x^2+y^2)^(-3/2)
= y^2/(x^2+y^2)^(3/2)
∂^2z/∂x∂y = ∂[x(x^2+y^2)^(-1/2)]/∂y

= x(-1/2)(x^2+y^2)^(-3/2)2y
= - xy/(x^2+y^2)^(3/2)
∂^2z/∂y^2 = ∂[y(x^2+y^2)^(-1/2)]/∂y

= (x^2+y^2)^(-1/2) + y(-1/2)(x^2+y^2)^(-3/2)2y
= (x^2+y^2)^(-1/2) - y^2(x^2+y^2)^(-3/2)
= x^2/(x^2+y^2)^(3/2)