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Multivariable chain rule proof

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Take a constant p ≥ 1 and f(x, y) a function of two variables with continuous
    first order partial derivatives. If, f(λx, λy) = (γ^p)f(x,y) for λ ε ℝ, prove that

    x(∂f/∂x) + y(∂f/∂y) = pf

    2. Relevant equations

    x(∂f/∂x) + y(∂f/∂y) = pf
    f(λx, λy) = (λ^p)f(x,y)


    3. The attempt at a solution

    I just don't know how to start. I know you have to use multivariable chain rule. Can someone point me in the right direction?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jul 2, 2013
  2. jcsd
  3. Jul 2, 2013 #2

    Dick

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    Take the derivative of both sides of your defining relation with respect to λ. Then put λ=1.
     
  4. Jul 2, 2013 #3
    how would you differentiate with respect to lambda?

    Also made a mistake.

    Its: f(λx, λy) = (λ^p)f(x,y)
     
  5. Jul 2, 2013 #4

    Dick

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    Put x'=λx and y'=λy, so f(λx, λy)=f(x',y'). Apply the multivariable chain rule to find ∂f(x',y')/∂λ. Then differentiate the other side with respect to λ as well.
     
  6. Jul 2, 2013 #5
    why do you put x prime=λx and y prime=λy?
     
  7. Jul 2, 2013 #6

    Dick

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    Just apply the multivariable chain rule to ∂f(x',y')/∂λ. You'll see.
     
  8. Jul 2, 2013 #7
    What about the right side? Is p * lamdba^(p-1) f(x,y) correct?
     
  9. Jul 2, 2013 #8

    Dick

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    Yes, now do the left side, that's the chain rule part. Remember in the end you are going to put λ=1.
     
  10. Jul 2, 2013 #9
    But lamda is an element of all real numbers right? So how can you prove this if we only choose lambda as 1?
     
  11. Jul 2, 2013 #10
    I got ∂f/∂x'*x + ∂f/∂y' *y on the left side. What do we do after that?
     
  12. Jul 2, 2013 #11

    Dick

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    You can get a more complicated relation that's true for general λ. But the specific result they want you to prove doesn't have a λ in it. That's why I suggest putting λ=1 after you do the differentiations.
     
  13. Jul 2, 2013 #12

    Dick

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    Convince yourself that e.g. ∂f(x',y')/∂x'=∂f(x,y)/∂x=∂f/∂x when λ=1.
     
  14. Jul 2, 2013 #13
    Got it, thanks!
     
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