Question

求二阶偏导数

Answer 1

1. lnz=-1/x-1/y→(∂z/∂x)·1/z=1/x²、(∂z/∂y)·1/z=1/y²

   ∴x²(∂z/∂x)+y²(∂z/∂x)=2z

2. sinz=xy

   cosz·(∂z/∂x)=y→∂z/∂x=y/cosz=y/√(1-x²y²)

   cosz·(∂z/∂y)=x→∂z/∂y=x/cosz=x/√(1-x²y²)

   ∂²z/∂x²=y/-2(1-x²y²)^(3/2)·-2xy²=xy³√(1-x²y²)/(1-x²y²)²

   ∂²z/∂x∂y=[√(1-x²y²)-2x²y²/2√(1-x²y²)]/√(1-x²y²)²=[(1-2x²y²)√(1-x²y²)]/(1-x²y²)²

   ∂²z/∂y²=x/-2(1-x²y²)^(3/2)·-2x²y=xy³√(1-x²y²)/(1-x²y²)²