VC3.35

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\nabla ^{2}({\frac  {x}{r^{2}}})=\sum {\frac  {\partial ^{2}}{\partial x^{2}}}[{\frac  {x}{r^{2}}}]\, --(1)

Now,{\frac  {\partial }{\partial x}}({\frac  {x}{r^{2}}})={\frac  {1}{r^{2}}}-{\frac  {2x}{r^{3}}}{\frac  {\partial r}{\partial x}}={\frac  {1}{r^{2}}}-{\frac  {2x}{r^{3}}}[{\frac  {x}{r}}]={\frac  {1}{r^{2}}}-{\frac  {2x^{2}}{r^{4}}}\, --(2) [Since r^{2}=x^{2}+y^{2}+z^{2}\, and {\frac  {\partial r}{\partial x}}={\frac  {x}{r}}\,]

Therefore,{\frac  {\partial ^{2}}{\partial r^{2}}}={\frac  {\partial }{\partial x}}[{\frac  {\partial }{\partial x}}({\frac  {x}{r^{2}}})]={\frac  {\partial }{\partial x}}[{\frac  {1}{r^{2}}}-{\frac  {2x^{2}}{r^{4}}}]\,, by using (2)

=-{\frac  {2}{r^{3}}}{\frac  {\partial r}{\partial x}}-[{\frac  {4x}{r^{4}}}-{\frac  {8x^{2}}{r^{5}}}{\frac  {\partial r}{\partial x}}]=-{\frac  {2}{r^{3}}}({\frac  {x}{r}})-{\frac  {4x}{r^{4}}}+{\frac  {8x^{2}}{r^{5}}}{\frac  {x}{r}}\,

Therefore,{\frac  {\partial ^{2}}{\partial x^{2}}}[{\frac  {x}{r^{2}}}]={\frac  {8x^{3}}{r^{6}}}-{\frac  {6x}{r^{4}}}\, --(3)

Now,{\frac  {\partial }{\partial y}}({\frac  {x}{r^{2}}})=-{\frac  {2x}{r^{3}}}({\frac  {y}{r}})\,

{\frac  {\partial ^{2}}{\partial y^{2}}}={\frac  {\partial }{\partial y}}[-{\frac  {2xy}{r^{4}}}]\,

=-2x[{\frac  {1}{r^{4}}}-{\frac  {4y}{r^{5}}}{\frac  {\partial r}{\partial y}}]=-2x[{\frac  {1}{r^{4}}}-{\frac  {4y^{2}}{r^{6}}}]={\frac  {8xy^{2}}{r^{6}}}-{\frac  {2x}{r^{4}}}\, --(4)

Similarly,{\frac  {\partial ^{2}}{\partial z^{2}}}[{\frac  {x}{r^{2}}}]={\frac  {8xz^{2}}{r^{6}}}-{\frac  {2x}{r^{4}}}\, --(5)

Adding (3),(4),(5), we have

\sum {\frac  {\partial ^{2}}{\partial x^{2}}}={\frac  {8x}{r^{6}}}[x^{2}+y^{2}+z^{2}]-{\frac  {10x}{r^{4}}}=-{\frac  {2x}{r^{4}}}\,

or \nabla ^{2}[{\frac  {x}{r^{2}}}]=-{\frac  {2x}{r^{4}}}\,, by using (1)

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