第2个回答 2020-04-23
w = (1/2)ln(x^2+y^2+z^2)
得 ∂w/∂x = x/(x^2+y^2+z^2)
∂^2w/∂x^2 = [(x^2+y^2+z^2)-2x^2]/(x^2+y^2+z^2)^2
= (-x^2+y^2+z^2)/(x^2+y^2+z^2)^2
同理 ∂^2w/∂y^2 = (x^2-y^2+z^2)/(x^2+y^2+z^2)^2
∂^2w/∂z^2 = (x^2-y^2-z^2)/(x^2+y^2+z^2)^2
则 ∂^2w/∂x^2 +∂^2w/∂y^2+∂^2w/∂z^2 = 0