Multivariable chain rule proof

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Homework Help Overview

The discussion revolves around proving a relationship involving a function of two variables, specifically using the multivariable chain rule. The original poster presents a problem where a function f(x, y) satisfies a scaling property with respect to a constant p and a variable λ, leading to a specific equation involving partial derivatives.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to apply the multivariable chain rule to differentiate both sides of the given equation with respect to λ. There are questions about the setup of variables and the implications of choosing λ=1 for the proof.

Discussion Status

Some participants have suggested specific steps to differentiate the equation, while others are clarifying the reasoning behind variable substitutions and the choice of λ. There is an ongoing exploration of how to handle both sides of the equation after differentiation.

Contextual Notes

Participants note that the function has continuous first order partial derivatives and that the proof must hold for all real numbers λ, raising questions about the implications of evaluating at λ=1.

mrcheeses
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Homework Statement



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

Homework Equations



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

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?

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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mrcheeses said:

Homework Statement



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

Homework Equations



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


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?

Take the derivative of both sides of your defining relation with respect to λ. Then put λ=1.
 
how would you differentiate with respect to lambda?

Also made a mistake.

Its: f(λx, λy) = (λ^p)f(x,y)
 
mrcheeses said:
how would you differentiate with respect to lambda?

Also made a mistake.

Its: f(λx, λy) = (λ^p)f(x,y)

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.
 
why do you put x prime=λx and y prime=λy?
 
mrcheeses said:
why do you put x prime=λx and y prime=λy?

Just apply the multivariable chain rule to ∂f(x',y')/∂λ. You'll see.
 
What about the right side? Is p * lamdba^(p-1) f(x,y) correct?
 
mrcheeses said:
What about the right side? Is p * lamdba^(p-1) f(x,y) correct?

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

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.
 
  • #12
mrcheeses said:
I got ∂f/∂x'*x + ∂f/∂y' *y on the left side. What do we do after that?

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

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