Question

第七第八题，求解

Answer 1

7、我打不出偏导数符号，用df/dx或df/dy代替，能看懂吧
xdf/dx+ydf/dy=x·(1/2√x)/(√x+√y) + y·(1/2√y)/(√x+√y)=√x/2(√x+√y) +√y/2(√x+√y)=1/2
8、d²u/dx²=[2x/√(x²+y²+z²)]'=[2√(x²+y²+z²)-4x²/√(x²+y²+z²)]/(x²+y²+z²)=2(y²+z²-x²)/[(x²+y²+z²)√(x²+y²+z²)]，d²u/dy²=2(x²+z²-y²)/[(x²+y²+z²)√(x²+y²+z²)]，d²u/dz²=2(x²+y²-z²)/[(x²+y²+z²)√(x²+y²+z²)]，
三者的和=2(y²+z²-x²+x²+z²-y²+x²+y²-z²)/[(x²+y²+z²)√(x²+y²+z²)]=2(x²+y²+z²)/[(x²+y²+z²)√(x²+y²+z²)]=2/√(x²+y²+z²)=2/u

追问

第八题，

追答

8、d²u/dx²=[2x/√(x²+y²+z²)]'=[2√(x²+y²+z²)-4x²/√(x²+y²+z²)]/(x²+y²+z²)=2(y²+z²-x²)/[(x²+y²+z²)√(x²+y²+z²)]，d²u/dy²=2(x²+z²-y²)/[(x²+y²+z²)√(x²+y²+z²)]，d²u/dz²=2(x²+y²-z²)/[(x²+y²+z²)√(x²+y²+z²)]，
三者的和=2(y²+z²-x²+x²+z²-y²+x²+y²-z²)/[(x²+y²+z²)√(x²+y²+z²)]=2(x²+y²+z²)/[(x²+y²+z²)√(x²+y²+z²)]=2/√(x²+y²+z²)=2/u

Answer 2