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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.

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    $\begingroup$ See my editing, use that as a reference to edit the rest part. $\endgroup$
    – Tunk-Fey
    Jul 30 '14 at 10:13
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    $\begingroup$ $\sqrt{x^2+y^2+z^2}$ is not the same as $x+y+z$. Try it with $(x,y,z)=(3,4,12)$ if you don't believe me. $\endgroup$
    – TonyK
    Jul 30 '14 at 10:14
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    $\begingroup$ I suppose that $\rho$ is $P$. If this is the case, please edit. $\endgroup$ Jul 30 '14 at 10:18
  • $\begingroup$ @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. $\endgroup$
    – Chinny84
    Jul 30 '14 at 10:30
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$$ \frac{\partial \rho}{\partial x} = \frac{1}{2}\frac{2x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho} \tag{*} $$ 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 $$ \begin{eqnarray} \frac{\partial^2\rho}{\partial x^2} + \frac{\partial^2\rho}{\partial y^2} + \frac{\partial^2\rho}{\partial z^2} &=& \frac{3}{\rho} - \frac{x^2+y^2+z^2}{\rho^3}\\ &=& \frac{3}{\rho}-\frac{\rho^2}{\rho^3}\\ &=& \frac{3}{\rho}-\frac{1}{\rho} = \frac{2}{\rho}. \end{eqnarray} $$

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  • $\begingroup$ I tried it myself via the chain rule and I can't see how you got p as the denominator. Can you explain the differentiation again please? $\endgroup$ Jul 30 '14 at 10:50
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    $\begingroup$ Now that is a rather open question ;). But yeah sure go for it. $\endgroup$
    – Chinny84
    Jul 30 '14 at 10:54
  • $\begingroup$ Very sorry! I can see my mistake! I wrote it wrong on the page. Thanks a lot for all your help. It is very much appreciated! $\endgroup$ Jul 30 '14 at 11:04
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    $\begingroup$ No worries :). Also modify your original question by subbing P for "\rho". Also take a look at how the mathjax looks in the revised edits. $\endgroup$
    – Chinny84
    Jul 30 '14 at 11:15

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