If u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2,th | vedclass

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(A) Given $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$. Differentiating with respect to $x$: $2u \frac{\partial u}{\partial x} = 2(x - a) \implies u \fra

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$\frac{2}{u}$

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(A) Given $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$. Differentiating with respect to $x$: $2u \frac{\partial u}{\partial x} = 2(x - a) \implies u \frac{\partial u}{\partial x} = x - a$. Differentiating again with respect to $x$: $u \frac{\partial^2 u}{\partial x^2} + \left(\frac{\partial u}{\partial x}\right)^2 = 1$. Substituting $\frac{\partial u}{\partial x} = \frac{x - a}{u}$: $u \frac{\partial^2 u}{\partial x^2} = 1 - \left(\frac{x - a}{u}\right)^2 = 1 - \frac{(x - a)^2}{u^2} = \frac{u^2 - (x - a)^2}{u^2}$. Thus,$\frac{\partial^2 u}{\partial x^2} = \frac{1}{u} - \frac{(x - a)^2}{u^3}$. Similarly,$\frac{\partial^2 u}{\partial y^2} = \frac{1}{u} - \frac{(y - b)^2}{u^3}$ and $\frac{\partial^2 u}{\partial z^2} = \frac{1}{u} - \frac{(z - c)^2}{u^3}$. Summing these: $\sum \frac{\partial^2 u}{\partial x^2} = \frac{3}{u} - \frac{1}{u^3} [(x - a)^2 + (y - b)^2 + (z - c)^2]$. Since $(x - a)^2 + (y - b)^2 + (z - c)^2 = u^2$: $\sum \frac{\partial^2 u}{\partial x^2} = \frac{3}{u} - \frac{u^2}{u^3} = \frac{3}{u} - \frac{1}{u} = \frac{2}{u}$.

If $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$,then $\sum \frac{\partial^2 u}{\partial x^2} = $

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