# Multivariable chain rule proof

1. Jul 2, 2013

### mrcheeses

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. Jul 2, 2013

### Dick

Take the derivative of both sides of your defining relation with respect to λ. Then put λ=1.

3. Jul 2, 2013

### mrcheeses

how would you differentiate with respect to lambda?

Its: f(λx, λy) = (λ^p)f(x,y)

4. Jul 2, 2013

### Dick

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.

5. Jul 2, 2013

### mrcheeses

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

6. Jul 2, 2013

### Dick

Just apply the multivariable chain rule to ∂f(x',y')/∂λ. You'll see.

7. Jul 2, 2013

### mrcheeses

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

8. Jul 2, 2013

### Dick

Yes, now do the left side, that's the chain rule part. Remember in the end you are going to put λ=1.

9. Jul 2, 2013

### mrcheeses

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

10. Jul 2, 2013

### mrcheeses

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

11. Jul 2, 2013

### Dick

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. Jul 2, 2013

### Dick

Convince yourself that e.g. ∂f(x',y')/∂x'=∂f(x,y)/∂x=∂f/∂x when λ=1.

13. Jul 2, 2013

### mrcheeses

Got it, thanks!