Proof of Identity: Differentiating $\alpha^ax$ and $\alpha^by$

[matematikuvol]

Homework Statement

\alpha\frac{d}{d\alpha}[f(\alpha^ax,\alpha^by)]|_{\alpha=1}=ax\frac{\partial f}{\partial x}+by\frac{\partial f}{\partial y}

Homework Equations

The Attempt at a Solution

Homework Statement

I'm confused. I don't know what to do here. How to differentiate left side? Thanks for your help.

 

[Office_Shredder]

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

 

[matematikuvol]

For example

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

Right?

 

[HallsofIvy]

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

 

[matematikuvol]

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

 

[SammyS]

Homework Statement

\alpha\frac{d}{d\alpha}[f(\alpha^ax,\alpha^by)]|_{\alpha=1}=ax\frac{\partial f}{\partial x}+by\frac{\partial f}{\partial y}

Homework Equations

The Attempt at a Solution

Homework Statement

I'm confused. I don't know what to do here. How to differentiate left side? Thanks for your help.

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) = (α^ax, α^by) rather than \displaystyle\frac{\partial f(\alpha^a x,\, \alpha^b y)}{\partial x}\,. ​

 

[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=
Please help if you know. I'm confused.

 

[matematikuvol]

It looks a bit for me like this theorem
http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html