STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

==>$u\,\frac{{\partial u}}{{\partial x}} = x - a$

==> $u.\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} = 1$

==> $u.\frac{{{\partial ^2}u}}{{\partial {x^2}}} = 1 - {\left( {\frac{{x - a}}{u}} \right)^2}$

==> $\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{1}{u} - \frac{{{{(x - a)}^2}}}{{{u^3}}}$

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 $ $\sum \,\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{3}{u} - \frac{1}{{{u^3}}}[{(x - a)^2} + {(y - b)^2} + {(z - c)^2}]$

$= \frac{3}{u} - \frac{1}{{{u^3}}}.({u^2}) = \frac{3}{u} - \frac{1}{u} = \frac{2}{u}$.