Question

求教一道高数偏导数题

设r=根号(x^2+y^2+z^2)，证明：
x、y、z的二阶偏导数加在一起等于2/r
我怎么证都算不出，求教学，谢谢！

Answer 1

r = √(x²+y²+z²)
∂r/∂x = 2x / 2√(x²+y²+z²) = x/r
∂²r/∂x² = (1*r - x*x/r)/r² = (r²-x²)/r³ = (y²+z²)/r³
同理
∂²r/∂y² = (x²+z²)/r³
∂²r/∂z² = (y²+x²)/r³
∂²r/∂x²+∂²r/∂y²+∂²r/∂z² = (y²+z²+x²+z²+y²+x²)/r³ = 2r²/r³ = 2/r

Answer 2

r=√(x^2+y^2+z^2 )
∂r/∂x=x/r ∂r/∂y=y/r ∂r/∂z=z/r
(∂^2 r)/(∂^2 x)=(y^2+z^2)/r^□3 (∂^2 r)/(∂^2 y)=(z^2+x^2)/r^□3 (∂^2 r)/(∂^2 z)=(x^2+y^2)/r^□3
(∂^2 r)/(∂^2 x)+(∂^2 r)/(∂^2 y)+(∂^2 r)/(∂^2 z)=((y^2+z^2 )+(z^2+x^2 )+(x^2+y^2 ))/r^□3 =〖2r〗^2/r^□3
(∂^2 r)/(∂^2 x)+(∂^2 r)/(∂^2 y)+(∂^2 r)/(∂^2 z)= 规范

Answer 3

答案如下：
r=√(x^2+y^2+z^2 )
∂r/∂x=x/r ∂r/∂y=y/r ∂r/∂z=z/r
(∂^2 r)/(∂^2 x)=(y^2+z^2)/r^□3 (∂^2 r)/(∂^2 y)=(z^2+x^2)/r^□3 (∂^2 r)/(∂^2 z)=(x^2+y^2)/r^□3
(∂^2 r)/(∂^2 x)+(∂^2 r)/(∂^2 y)+(∂^2 r)/(∂^2 z)=((y^2+z^2 )+(z^2+x^2 )+(x^2+y^2 ))/r^□3 =〖2r〗^2/r^□3
(∂^2 r)/(∂^2 x)+(∂^2 r)/(∂^2 y)+(∂^2 r)/(∂^2 z)=2/r