Showing $\sum_{i=1}^{n}\frac{\partial^2 v}{\partial x_i^2}=\frac{\partial^2v}{\partial r^2}+\frac{n-1}{r}\frac{\partial v}{\partial r}$ 
If $v=v(r)$, where $\displaystyle r^2=\sum_{i=1}^nx_i^2$, show that:$$\sum_{i=1}^{n}\dfrac{\partial^2 v}{\partial x_i^2}=\dfrac{\partial^2v}{\partial r^2}+\dfrac{n-1}{r}\dfrac{\partial v}{\partial r}$$

My tries:
$$\begin{align}\dfrac{\partial v}{\partial x_1}&=\dfrac{\partial v}{\partial r}\cdot\dfrac{\partial r}{\partial x_1}\tag{chain rule}\\&=\dfrac{\partial v}{\partial r}\dfrac{x_1}{r}\\\dfrac{\partial^2 v}{\partial x_1^2}&=\dfrac{1}{r}\cdot\dfrac{\partial v}{\partial r}+\dfrac{x_1}{r}\cdot\dfrac{\partial ^2v}{\partial x_1 \partial r}\\\sum_{i=1}^{n}\dfrac{\partial^2 v}{\partial x_i^2}&=\dfrac{n}{r}\cdot\dfrac{\partial v}{\partial r}+\dfrac{1}{r}\sum_{i=1}^nx_i\dfrac{\partial^2v}{\partial x_i\partial r}\end{align}$$
That last term $\color{red}{\displaystyle\dfrac{1}{r}\sum_{i=1}^nx_i\dfrac{\partial^2v}{\partial x_i\partial r}}$ scares me, please help. How to proceed further?
 A: It appears to Yours Truly that this problem is much easier addressed using a spherical coordinate system; we note we must insist that 
$r \ne 0; \tag 0$
throughout the following I make extensive use of the well-known identity from vector calculus:
$\nabla \cdot (f \vec X) = \nabla f \cdot \vec X + f \nabla \cdot \vec X, \tag{0.5}$
where $f$ and $\vec X$ are a differentiable scalar and vector field, respectively; this formula is in fact a generalization of the Leibniz product rule $(uv)' = u'v + uv'$ from elementary calculus; we let $\vec e_r$ be the unit vector field in the $r$-direction; then we may write
$\nabla v(r) = v'(r) \vec e_r, \tag 1$
from which, via (0.5),
$\nabla^2 v(r) = \nabla \cdot (\nabla v(r)) = \nabla \cdot (v'(r) \vec e_r)$
$= \nabla v'(r) \cdot \vec e_r + v'(r) \nabla \cdot \vec e_r = v''(r) \vec e_r \cdot \vec e_r + v'(r)\nabla \cdot \vec e_r =v''(r) + v'(r)\nabla \cdot \vec e_r,  \tag 2$
since
$\vec e_r \cdot \vec e_r = 1; \tag 3$
now with
$r = \sqrt{ \displaystyle \sum_1^n x_i^2 } \tag 4$
we have
$\vec e_r = \dfrac{1}{r}(x_1, x_2, \ldots, x_n), \tag 5$
whence, again using (0.5),
$\nabla \cdot {\vec e_r} = \left (\nabla \dfrac{1}{r} \right ) \cdot (x_1, x_2, \ldots, x_n) + \dfrac{1}{r} \nabla \cdot (x_1, x_2, \ldots, x_n)$
$= -\dfrac{1}{r^2} \vec e_r \cdot (x_1, x_2, \ldots, x_n) + \dfrac{n}{r}, \tag 6$
since
$\nabla \cdot (x_1. x_2, \ldots, x_n) = \displaystyle \sum_1^n \dfrac{\partial x_i}{\partial x_i} = \displaystyle \sum_1^n 1 = n; \tag{6.5}$
from (5),
$\vec e_r \cdot (x_1, x_2, \ldots, x_n) = \dfrac{1}{r}\displaystyle \sum_1^n x_i^2  = \dfrac{r^2}{r} = r; \tag 7$
thus,
$\nabla \cdot {\vec e_r} = -\dfrac{1}{r} + \dfrac{n}{r} = \dfrac{n - 1}{r}; \tag 8$
substituting this into (2) yields
$\nabla^2 v(r) = v''(r) + \dfrac{n -
1}{r} v'(r),  \tag 9$
the desired result.  $OE\Delta$.
