求此方程的所有二阶偏导数，要过程

Question

如题所述

Answer 1

z = arctan(y/x),
z'<x> = (-y/x^2)/[1+(y/x)^2] = -y/(x^2+y^2),
z'<y> = (1/x)/[1+(y/x)^2] = x/(x^2+y^2),
z''<xx> = 2x*y/(x^2+y^2)^2 = 2xy/(x^2+y^2)^2,
z''<xy> = -(x^2+y^2-y*2y)/(x^2+y^2)^2 = (y^2-x^2)/(x^2+y^2)^2,
z''<yy> = 2y*(-x)/(x^2+y^2)^2 = -2xy/(x^2+y^2)^2.