# Partial derivative second order

1. Feb 16, 2017

### Taylor_1989

1. The problem statement, all variables and given/known data
Hi guys, I am have a problem with the question displayed below:

Its 6.1 ii) I am really not sure how I am suppose to approach this. I am new to partials, so any advice would be great.

2. Relevant equations

3. The attempt at a solution
So far I have:
$$\frac{\partial ^2 f}{\partial x^2}=\frac{\partial}{\partial x}\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}\frac{-x}{r^3}$$

using quotient rule:

$$=[\frac{\partial}{\partial x}(-x)(r^3)-(-x)(\frac{\partial}{\partial x}r^3)]/r^6$$

$$\frac{\partial}{\partial x}r^3=3r^2\frac{\partial r}{\partial x}=3r^2*(x/r)=3rx$$

subbing the above into the quotient and simplifying
I get
$$\frac{-r^2+3x^2}{r^5}$$

2. Feb 16, 2017

### BvU

might not be the best thing to do. How did you find $\partial f\over\partial x$ ?
And: was 6 i) allright and clear ? How come 6 ii) is then problematic ?

 Stupid me . Your working is completely correct. Well done...
'Quotient rule' confused me, but it works just fine. My approach would be to use the chain rule -- with, of course, the same result. Matter of preference, not matter of 'best thing to do'.

Last edited: Feb 16, 2017
3. Feb 16, 2017

### Ray Vickson

Your answer is correct. It is a matter of taste whether you leave your final numerator as $3x^2-r^2$, or re-write it as $2x^2-y^2-z^2$.

BTW: Remove all those offensive bold fonts: it looks like you are yelling at us.

4. Feb 16, 2017

### Taylor_1989

thank for the responses. I will remove bold next, i did not even notice it was bold.