Question

求高手解答z=ln(x^2+y^2)的二阶偏导数

Answer 1

z=ln(x²+y²)
∂z/∂x=2x/(x²+y²)
∂²z/∂x²=2(y²-x²)/(x²+y²)²
∂z/∂y=2y/(x²+y²)
∂²z/∂y²=2(x²-y²)/(x²+y²)²
∂²z/∂x∂y=-4xy/(x²+y²)²