# Prove identity

1. Feb 26, 2012

### matematikuvol

1. The problem statement, all variables and given/known data
$$\alpha\frac{d}{d\alpha}[f(\alpha^ax,\alpha^by)]|_{\alpha=1}=ax\frac{\partial f}{\partial x}+by\frac{\partial f}{\partial y}$$

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data
I'm confused. I don't know what to do here. How to differentiate left side? Thanks for your help.

Last edited by a moderator: Feb 26, 2012
2. Feb 26, 2012

### Office_Shredder

Staff Emeritus
You have to use the chain rule for multi-variable functions. Try it out and if you get stuck post what you've done

3. Feb 26, 2012

### matematikuvol

For example

$$\frac{d}{d\alpha}f(\alpha^ax)=ax\alpha^{a-1}\frac{\partial f}{\partial \alpha}$$

Right?

Last edited: Feb 26, 2012
4. Feb 26, 2012

### HallsofIvy

Staff Emeritus
Well,
$$\frac{\partial f}{\alpha}$$
is meaningless. Was that a typo? What did you mean? Perhaps
$$\frac{\partial f}{\partial x}$$?

5. Feb 26, 2012

### matematikuvol

Sorry. I made a mistake. You can see know what I meant. I edit my last message.

6. Feb 26, 2012

### SammyS

Staff Emeritus
It would help if you would give us the whole problem, word for word as it was given to you.

For instance, what is meant by $\displaystyle\frac{\partial f}{\partial x}\,?$

I assume that's $\displaystyle\frac{\partial f(x,\,y)}{\partial x}\,,$ evaluated at (x, y) = (αax, αby) rather than $\displaystyle\frac{\partial f(\alpha^a x,\, \alpha^b y)}{\partial x}\,.$ ​

7. Feb 26, 2012

### matematikuvol

I'm not quite sure. I had only left side. So
$$\alpha\frac{d}{d\alpha}[f(\alpha^a x,\alpha^by)]|\alpha=1=$$