Answer
Using the result from Exercise 47, we prove that
$\frac{{{\partial ^2}g}}{{\partial {x_1}^2}} + ... + \frac{{{\partial ^2}g}}{{\partial {x_n}^2}} = {g_{rr}} + \frac{{n - 1}}{r}{g_r}$
Work Step by Step
Let $r = \sqrt {{x_1}^2 + ... + {x_n}^2} $ and let $g\left( r \right)$ be a function of $r$ as is given in Exercise 47.
From Exercise 47, we obtain
$\frac{{{\partial ^2}g}}{{\partial {x_i}^2}} = \frac{{{x_i}^2}}{{{r^2}}}{g_{rr}} + \frac{{{r^2} - {x_i}^2}}{{{r^3}}}{g_r}$, ${\ \ }$ for $i = 1,2,...,n$
So,
$\frac{{{\partial ^2}g}}{{\partial {x_1}^2}} + ... + \frac{{{\partial ^2}g}}{{\partial {x_n}^2}} = \left( {\frac{{{x_1}^2}}{{{r^2}}} + ... + \frac{{{x_n}^2}}{{{r^2}}}} \right){g_{rr}}$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$ $ + \left( {\frac{{{r^2} - {x_1}^2}}{{{r^3}}} + ... + \frac{{{r^2} - {x_n}^2}}{{{r^3}}}} \right){g_r}$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$ $ = \frac{{{x_1}^2 + ... + {x_n}^2}}{{{r^2}}}{g_{rr}} + \frac{{n{r^2} - \left( {{x_1}^2 + ... + {x_n}^2} \right)}}{{{r^3}}}{g_r}$
Since ${r^2} = {x_1}^2 + ... + {x_n}^2$, so
$\frac{{{\partial ^2}g}}{{\partial {x_1}^2}} + ... + \frac{{{\partial ^2}g}}{{\partial {x_n}^2}} = {g_{rr}} + \frac{{n{r^2} - {r^2}}}{{{r^3}}}{g_r}$
Hence,
$\frac{{{\partial ^2}g}}{{\partial {x_1}^2}} + ... + \frac{{{\partial ^2}g}}{{\partial {x_n}^2}} = {g_{rr}} + \frac{{n - 1}}{r}{g_r}$
