Am really not sure how to prove this question, does

$\displaystyle \nabla . (\frac{1}{r}r)$= $\displaystyle \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} \dot$ (\frac{1}{x^{2}+y^{2}+z^{2})^{1/2} ( x+y+z) [/tex]

$\dfrac {\vec{r}} {\| \vec{r} \|} = \left\{ \dfrac {x} {\sqrt{x^2+y^2+z^2}},\dfrac {y} {\sqrt{x^2+y^2+z^2}},\dfrac {z} {\sqrt{x^2+y^2+z^2}} \right \}$

Take the divergence and then take the gradient and plow through the algebra. It's not that bad. You end up with the right hand side when you're done.

Originally Posted by romsek
$\dfrac {\vec{r}} {\| \vec{r} \|} = \left\{ \dfrac {x} {\sqrt{x^2+y^2+z^2}},\dfrac {y} {\sqrt{x^2+y^2+z^2}},\dfrac {z} {\sqrt{x^2+y^2+z^2}} \right \}$

Take the divergence and then take the gradient and plow through the algebra. It's not that bad. You end up with the right hand side when you're done.
so is it,

$\displaystyle \frac{\partial}{\partial x }( \{ \dfrac {x} {\sqrt{x^2+y^2+z^2}}) + \frac{\partial}{\partial y } \dfrac {y} {\sqrt{x^2+y^2+z^2}}, + \frac{\partial}{\partial z}\dfrac {z} {\sqrt{x^2+y^2+z^2}}$

than use the quotient rule for each one?

Originally Posted by Tweety
so is it,

$\displaystyle \frac{\partial}{\partial x }( \{ \dfrac {x} {\sqrt{x^2+y^2+z^2}}) + \frac{\partial}{\partial y } \dfrac {y} {\sqrt{x^2+y^2+z^2}}, + \frac{\partial}{\partial z}\dfrac {z} {\sqrt{x^2+y^2+z^2}}$
if you move the delx operator inside those braces, then what you have there is the div operator. Once you find that you'll have a scalar function in 3 variables, (x, y, z). Take the gradient of that function, simplify it, substitute r in where appropriate and you'll end up with the answer.

Originally Posted by Tweety
so is it,

$\displaystyle \frac{\partial}{\partial x }( \{ \dfrac {x} {\sqrt{x^2+y^2+z^2}}) + \frac{\partial}{\partial y } \dfrac {y} {\sqrt{x^2+y^2+z^2}}, + \frac{\partial}{\partial z}\dfrac {z} {\sqrt{x^2+y^2+z^2}}$

than use the quotient rule for each one?
You might find it simpler to write $\displaystyle x(x^2+ y^2+ z^2)^{-1/2}$, etc. and use the product rule.