求二阶偏导数

Question

如题所述

Answer 1

z=ln(x²+y²)

∂z/∂x=2x/(x²+y²)

∂z/∂y=2y/(x²+y²)

∂²z/∂x²=[2(x²+y²)-2x·2x]/(x²+y²)²=2(y²-x²)/(x²+y²)²

∂²z/∂x∂y=-4xy/(x²+y²)²


∂²z/∂y²=[2(x²+y²)-2y·2y]/(x²+y²)²=2(x²-y²)/(x²+y²)²