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Prove identity

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\alpha\frac{d}{d\alpha}[f(\alpha^ax,\alpha^by)]|_{\alpha=1}=ax\frac{\partial f}{\partial x}+by\frac{\partial f}{\partial y}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data
    I'm confused. I don't know what to do here. How to differentiate left side? Thanks for your help.
     
    Last edited by a moderator: Feb 26, 2012
  2. jcsd
  3. Feb 26, 2012 #2

    Office_Shredder

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    You have to use the chain rule for multi-variable functions. Try it out and if you get stuck post what you've done
     
  4. Feb 26, 2012 #3
    For example

    [tex]\frac{d}{d\alpha}f(\alpha^ax)=ax\alpha^{a-1}\frac{\partial f}{\partial \alpha}[/tex]

    Right?
     
    Last edited: Feb 26, 2012
  5. Feb 26, 2012 #4

    HallsofIvy

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    Well,
    [tex]\frac{\partial f}{\alpha}[/tex]
    is meaningless. Was that a typo? What did you mean? Perhaps
    [tex]\frac{\partial f}{\partial x}[/tex]?
     
  6. Feb 26, 2012 #5
    Sorry. I made a mistake. You can see know what I meant. I edit my last message.
     
  7. Feb 26, 2012 #6

    SammyS

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    It would help if you would give us the whole problem, word for word as it was given to you.

    For instance, what is meant by [itex]\displaystyle\frac{\partial f}{\partial x}\,?[/itex]

    I assume that's [itex]\displaystyle\frac{\partial f(x,\,y)}{\partial x}\,,[/itex] evaluated at (x, y) = (αax, αby) rather than [itex]\displaystyle\frac{\partial f(\alpha^a x,\, \alpha^b y)}{\partial x}\,.[/itex] ​
     
  8. Feb 26, 2012 #7
    I'm not quite sure. I had only left side. So
    [tex]\alpha\frac{d}{d\alpha}[f(\alpha^a x,\alpha^by)]|\alpha=1=[/tex]
    Please help if you know. I'm confused.
     
  9. Feb 26, 2012 #8
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