I am having trouble with a question with partial derivatives.
Here is the question:
Let $\rho = \sqrt{x^2 + y^2 + z^2}$
Show that $$\frac{\partial ^2\rho}{\partial x^2} + \frac{\partial ^2\rho}{\partial y^2}+\frac{\partial ^2\rho}{\partial z^2} = \frac{2}{\rho}$$
All I am getting is that the LHS is equal to O.
A square root is something to the power of 0.5 $$ \begin{eqnarray} \rho &=& \left(x^2\right)^{1/2} + \left(y^2\right)^{1/2} + \left(z^2\right)^{1/2}\\ &=&x^1 + y^1 + z^1\\ &=& x + y + z. \end{eqnarray} $$
Therefore, $$ \frac{\partial \rho}{\partial x} = \frac{\partial \rho}{\partial y} = \frac{\partial \rho}{\partial z} = 1 $$
Furthermore, $$ \frac{\partial^2 \rho}{\partial x^2} = \frac{\partial^2 \rho}{\partial y^2} = \frac{\partial^2 \rho}{\partial z^2} = 0 $$
But this isn't equal to $2/ \rho$.
Any help here is appreciated. Apologies for the bad formatting. I don't know how to make it any nicer.
$$ \frac{\partial \rho}{\partial x} = \frac{1}{2}\frac{2x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho} \tag{*} $$ similarly for the other derivatives $$ \begin{eqnarray} \frac{\partial \rho}{\partial y} &=& \frac{y}{\sqrt{x^2+y^2+z^2}} = \frac{y}{\rho},\\ \frac{\partial \rho}{\partial z} &=& \frac{z}{\sqrt{x^2+y^2+z^2}} = \frac{z}{\rho}. \end{eqnarray} $$
then taking the second derivative for the x. We use the previous result from Eq. (*) as follows $$ \frac{\partial^2\rho}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{x}{\rho}\right) $$ this leads to $$ \frac{\partial^2\rho}{\partial x^2} = \frac{1}{\rho} - \frac{x}{\rho^2}\frac{x}{\rho} = \frac{1}{\rho} - \frac{x^2}{\rho^3}. $$ similary we have $$ \begin{eqnarray} \frac{\partial^2 \rho}{\partial y^2} &=& \frac{1}{\rho} - \frac{y^2}{\rho^3},\\ \frac{\partial^2 \rho}{\partial z^2} &=& \frac{1}{\rho} - \frac{z^2}{\rho^3}. \end{eqnarray} $$ all together $$ \begin{eqnarray} \frac{\partial^2\rho}{\partial x^2} + \frac{\partial^2\rho}{\partial y^2} + \frac{\partial^2\rho}{\partial z^2} &=& \frac{3}{\rho} - \frac{x^2+y^2+z^2}{\rho^3}\\ &=& \frac{3}{\rho}-\frac{\rho^2}{\rho^3}\\ &=& \frac{3}{\rho}-\frac{1}{\rho} = \frac{2}{\rho}. \end{eqnarray} $$