If \[{{u}^{2}}={{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}\], then \[\sum \frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\] [Tamilnadu (Engg.) 2002]

If \[{{u}^{2}}={{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}\], then \[\su

1.	If \[{{u}^{2}}={{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}\], then \[\sum \frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\] [Tamilnadu (Engg.) 2002]
A.	\[\frac{2}{u}\]
B.	\[\frac{3}{u}\]
C.	0
D.	\[\frac{1}{u}\]
Answer» B. \[\frac{3}{u}\]

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If \[{{u}^{2}}={{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}\], then \[\sum \frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\] [Tamilnadu (Engg.) 2002]

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