Multivariable chain rule proof

[mrcheeses]

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

 

[Dick]

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.

 

[mrcheeses]

how would you differentiate with respect to lambda?

Also made a mistake.

Its: f(λx, λy) = (λ^p)f(x,y)

 

[Dick]

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.

 

[mrcheeses]

why do you put x prime=λx and y prime=λy?

 

[Dick]

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.

 

[mrcheeses]

What about the right side? Is p * lamdba^(p-1) f(x,y) correct?

 

[Dick]

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.

 

[mrcheeses]

But lamda is an element of all real numbers right? So how can you prove this if we only choose lambda as 1?

 

[mrcheeses]

I got ∂f/∂x'*x + ∂f/∂y' *y on the left side. What do we do after that?

 

[Dick]

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.

 

[Dick]

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.

 

[mrcheeses]

Got it, thanks!