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Partial derivative second order

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I am have a problem with the question displayed below:

    upload_2017-2-15_23-33-6.png


    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. jcsd
  3. Feb 16, 2017 #2

    BvU

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    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 ?

    [edit] Stupid me o:) . 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
  4. Feb 16, 2017 #3

    Ray Vickson

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    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.
     
  5. Feb 16, 2017 #4
    thank for the responses. I will remove bold next, i did not even notice it was bold.
     
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