In the proof of Euler's homogeneous function theorem here
The start with the definition of a homogeneous function $$ f(tx,ty)=t^n f(x,y) $$ and take the derivative of both sides w.r.t $t$ which gives
$$ nt^{n-1}f(x,y) = \frac{\partial f}{\partial tx} \frac{\partial tx}{\partial t} + \frac{\partial f}{\partial ty} \frac{\partial ty}{\partial t} $$
or, letting letting $y'=ty$ and $x'=tx$ we can write this as
$$ nt^{n-1}f(x,y) = \frac{\partial f}{\partial x'} \frac{\partial x'}{\partial t} + \frac{\partial f}{\partial y'} \frac{\partial y'}{\partial t} $$
My questions are:
How should I interpret $\frac{\partial f}{\partial tx}$. For example, if $f(x,y) = x^2 + y^2$ what would be $\frac{\partial f}{\partial tx}$?
- also, Is $\frac{\partial f}{\partial tx}$ evaluated at $t=1$ different than $\frac{\partial f}{\partial x}$? If so, why?
I think you problem resides in understanding the chain rule of differentiation. Suppose that you have a differentiable function $(x,y)\mapsto f(x,y)$ and that you have two diferentiable functions $t\mapsto \phi(t)$ and $t\mapsto\psi(t)$,then what is the derivative of $g:t\mapsto f(\phi(t),\psi(t))$? The chain rule states that $$ g'(t)=\frac{\partial f}{\partial x}(\phi(t),\psi(t))\phi'(t)+ \frac{\partial f}{\partial y}(\phi(t),\psi(t))\psi'(t). $$ Now, we consider a particular case: We fix $(x,y)$ and we consider $\phi:t\mapsto \phi(t)=tx$ and $\psi:t\mapsto \psi(t)=ty$, it follows that $$ \frac{d}{dt}\left(f(tx,ty)\right)=\frac{\partial f}{\partial x}(tx,ty)x+ \frac{\partial f}{\partial y}(tx,ty)y. $$ But since $f(tx,ty)=t^nf(x,y)$, then from the above we conclude that $$ nt^{n-1}f(x,y)=\frac{\partial f}{\partial x}(tx,ty)x+ \frac{\partial f}{\partial y}(tx,ty)y. $$ The final step is to choose $t=1$ and we are done.