# Question #f6339

Apr 19, 2017

#### Explanation:

$V = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}} = {\left({x}^{2} + {y}^{2} + {z}^{2}\right)}^{\frac{1}{2}}$

By taking the partial derivative w.r.t. $x$,

${V}_{x} = \frac{1}{\cancel{2}} {\left({x}^{2} + {y}^{2} + {z}^{2}\right)}^{- \frac{1}{2}} \cdot \left(\cancel{2} x\right) = \frac{x}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}$

By taking another partial derivative w.r.t. $x$,

${V}_{x x} = \frac{1 \cdot \sqrt{{x}^{2} + {y}^{2} + {z}^{2}} - x \cdot \frac{x}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}}{\sqrt{{x}^{2} + {y}^{2} + {y}^{2}}} ^ 2 = \frac{V - {x}^{2} / V}{V} ^ 2$

By multiplying the numerator and the denominator by $V$,

$R i g h t a r r o w {V}_{x x} = \frac{{V}^{2} - {x}^{2}}{V} ^ 3$

By using symmetries,

$R i g h t a r r o w {V}_{y y} = \frac{{V}^{2} - {y}^{2}}{V} ^ 3$

and

$R i g h t a r r o w {V}_{z z} = \frac{{V}^{2} - {z}^{2}}{V} ^ 3$

Hence,

${V}_{x x} + {V}_{y y} + {V}_{z z} = \frac{{V}^{2} - {x}^{2}}{V} ^ 3 + \frac{{V}^{2} - {y}^{2}}{V} ^ 3 + \frac{{V}^{2} - {z}^{2}}{V} ^ 3$

$= \frac{3 {V}^{2} - \left({x}^{2} + {y}^{2} + {z}^{2}\right)}{V} ^ 3 = \frac{3 {V}^{2} - {V}^{2}}{V} ^ 3 = \frac{2 {V}^{2}}{V} ^ 3 = \frac{2}{V}$

I hope that this was clear.