# Partial Derivative Stickler.

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.

• See my editing, use that as a reference to edit the rest part. Jul 30 '14 at 10:13
• $\sqrt{x^2+y^2+z^2}$ is not the same as $x+y+z$. Try it with $(x,y,z)=(3,4,12)$ if you don't believe me. Jul 30 '14 at 10:14
• I suppose that $\rho$ is $P$. If this is the case, please edit. Jul 30 '14 at 10:18
• @ClaudeLeibovici I was thinking that when editing. But I only edit the text with format not put my own thoughts and possibly change the question :/. It should be edited accordingly. Jul 30 '14 at 10:30

$$\frac{\partial \rho}{\partial x} = \frac{1}{2}\frac{2x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho} \tag{*}$$ 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 $$\begin{eqnarray} \frac{\partial^2\rho}{\partial x^2} + \frac{\partial^2\rho}{\partial y^2} + \frac{\partial^2\rho}{\partial z^2} &=& \frac{3}{\rho} - \frac{x^2+y^2+z^2}{\rho^3}\\ &=& \frac{3}{\rho}-\frac{\rho^2}{\rho^3}\\ &=& \frac{3}{\rho}-\frac{1}{\rho} = \frac{2}{\rho}. \end{eqnarray}$$