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Partial derivative

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    If [itex]f[/itex] is homogeneous of degree [itex]n[/itex], show that [itex]f_{x}(tx,ty)=t^{n-1}f_{x}(x,y)[/itex].

    2. Relevant equations



    3. The attempt at a solution

    There are many solutions out there, and here's one of them:
    The proof is nice, but I just don't get it why from step 1 to step 2, [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex] and then it's rewritten as [itex]{f_x}(tx,ty)[/itex]. Any help is very much appreciated, thanks!
     
  2. jcsd
  3. Mar 29, 2013 #2

    Fredrik

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    It's not entirely clear to me what you're saying that the steps are (I can't make sense of what you wrote), but if you just want to apply ##\partial/\partial x## to both sides of ##f(tx,ty)=t^nf(x,y)##, then you need to use the chain rule when you deal with the left-hand side.

    What you need to know is that the x in the denominator of
    $$\frac{\partial}{\partial x}(\text{something that involves x})$$ tells us among other things that the function that you're supposed to take a partial derivative of takes x and some other variables to the "something that involves x". For example, the x in
    $$\frac{\partial}{\partial x}f(tx,ty)$$ tells us that the function that you're supposed to take a partial derivative of isn't ##(x,y)\mapsto f(x,y)## (i.e. f), but ##(x,y)\mapsto f(tx,ty)##. Since ##f_x## denotes the derivative of f with respect to the first variable, we have
    $$\frac{\partial}{\partial x}f(tx,ty)\neq f_x(tx,ty) =\frac{\partial}{\partial (tx)}f(tx,ty).$$ By the way, if you want all of what you're saying to be quotable, use indent tags instead of quote tags.
    Like this.​
     
    Last edited: Mar 29, 2013
  4. Mar 29, 2013 #3

    HallsofIvy

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    I don't know what you mean by "step 1 to step 2". There is NO statement that "[itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex]" in what you post and it is certainly not true.
     
  5. Mar 29, 2013 #4
    Thanks for your replies.
    Sorry for not making it clear. I meant, from [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty).\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty).\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}}[/itex] to [itex]t{f_x}(tx,ty) = {t^n}{f_x}(x,y)[/itex], doesn't it assume [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex] or I've overlooked something?
     
  6. Mar 29, 2013 #5

    Fredrik

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    ##\frac{\partial }{{\partial (tx)}}f(tx,ty)## isn't a statement, so it can't imply anything or be implied by anything. Same thing with ##\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty)##. Each step should be a statement that follows from the statement in the previous step.
     
  7. Mar 29, 2013 #6

    Fredrik

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    Aaah...the periods denote multiplication. You should probably avoid that notation. :smile: OK, give me a second.
     
  8. Mar 29, 2013 #7

    Fredrik

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    I have thought about it now. No, it doesn't assume that. In fact it contradicts it.

    It's best to not use any symbol at all for multiplication. If you want to use a dot, use the LaTeX code \cdot.
     
  9. Mar 29, 2013 #8
    Ok. So is it a valid proof?
    I thought [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty)\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty)\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}}[/itex] only implies
    [itex]t\frac{\partial }
    {{\partial (tx)}}f(tx,ty) = {t^n}\frac{{\partial f(x,y)}}
    {{\partial x}}[/itex]?
     
  10. Mar 29, 2013 #9

    Fredrik

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    That's right. Now you just cancel a factor of t from both sides and you're done.
     
  11. Mar 29, 2013 #10
    Then [itex]\frac{\partial }
    {{\partial (tx)}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}
    {{\partial x}}[/itex], but the question asks us to prove [itex]\frac{\partial }
    {{\partial x}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}
    {{\partial x}}[/itex].
     
  12. Mar 29, 2013 #11

    Fredrik

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    No, it doesn't. See my first post in this thread.
     
  13. Mar 29, 2013 #12
    Ahh ok. Thanks for clearing that up for me. :)
     
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