z=√(x^2+y^2)的二阶偏导数?
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∂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)
∂^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)
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