设函数f(x，y)=ln(√x+√y)，求证x(∂f/∂x)+y(∂f/∂y)=1/2．
证明：∵∂f/∂x=1/(√x+√y)•(1/2)x-(1/2)=1/[2(x+√xy)] ∂f/∂y=1/(√x+√y)•(1/2)x-(1/2)=1/[2(√xy+y)] ∴x(∂f/∂x)+y(∂f/∂y)=√x/[2(√x+√y)]+√y/[2(√x+√y)]=1/2
设函数f(x，y)=ln(√x+√y)，求证x(∂f/∂x)+y(∂f/∂y)=1/2．
证明：∵∂f/∂x=1/(√x+√y)•(1/2)x-(1/2)=1/[2(x+√xy)] ∂f/∂y=1/(√x+√y)•(1/2)x-(1/2)=1/[2(√xy+y)] ∴x(∂f/∂x)+y(∂f/∂y)=√x/[2(√x+√y)]+√y/[2(√x+√y)]=1/2
