Question

求二元函数Z=ln(x^2+y^2)+x/y的一阶,二阶偏导

Answer 1

先求一阶导数，对x求导，y看成常数，对y求导，x看成常数
∂z/∂x=1/(x^2+y^2)*2x +1/y
∂z/∂y=1/(x^2+y^2)*2y -x/y^2
再求二阶导数，如下(再一阶导数的基础上，继续对x,y求导
∂^2z/∂x^2=1/(x^2+y^2)*2-1/(x^2+y^2)^2*2x*2x
=2(y^2-x^2)/(x^2+y^2)^2
∂^2z/∂x∂y=-1/(x^2+y^2)^2*2x*2y-1/y^2
=-4xy/(x^2+y^2)^2-1/y^2
∂^2z/∂y^2=1/(x^2+y^2)*2-1/(x^2+y^2)^2*2y*2y+2x/y^3
=(x^2-y^2)/(x^2+y^2)^2+2x/y^3