MCQ
If ${u^2} = {(x - a)^2} + {(y - b)^2} + {(z - c)^2}$, then $\sum {{{\partial ^2}u} \over {\partial {x^2}}} = $
- ✓${2 \over u}$
- B${3 \over u}$
- C$0$
- D${1 \over u}$
==>$u\,\frac{{\partial u}}{{\partial x}} = x - a$
==> $u.\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} = 1$
==> $u.\frac{{{\partial ^2}u}}{{\partial {x^2}}} = 1 - {\left( {\frac{{x - a}}{u}} \right)^2}$
==> $\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{1}{u} - \frac{{{{(x - a)}^2}}}{{{u^3}}}$
Similarly, $\frac{{{\partial ^2}u}}{{\partial {y^2}}} = \frac{1}{u} - \frac{{{{(y - b)}^2}}}{{{u^3}}}$,
$\frac{{{\partial ^2}u}}{{\partial {z^2}}} = \frac{1}{u} - \frac{{{{(z - b)}^2}}}{{{u^3}}}$
$\therefore $ $\sum \,\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{3}{u} - \frac{1}{{{u^3}}}[{(x - a)^2} + {(y - b)^2} + {(z - c)^2}]$
$= \frac{3}{u} - \frac{1}{{{u^3}}}.({u^2}) = \frac{3}{u} - \frac{1}{u} = \frac{2}{u}$.
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$[A]$ $\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$
$[B]$ $\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
$[C]$ $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$[D]$ $\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$