$\displaystyle r^2 = (x - a)^2 + (y - b)^2 + (z -c)^2$

Find $\displaystyle \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}$

Got the following for $\displaystyle \frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}$

$\displaystyle 2rdr = 2(x - a) + 2(y - b) + 2(z - c)$

$\displaystyle dr = \frac{(x - a) + (y - b) + (z - c)}{r}$

Here, I go astray. The answer provided to me for $\displaystyle d^2r$ is $\displaystyle \frac{2}{r}$.

I get $\displaystyle \frac{\partial^2 r}{\partial x^2} = \frac{(1)r - ((x - a) + (y - b) + (z - c))(1)}{r^2}$

Likewise for the other partials leads to :

$\displaystyle d^2r = \frac{3(r - ((x - a) + (y - b) + (z - c))}{r^2}$

Obviously not doing something right.

-Scott