Partial derivatives of homogenous functions. Consider a homogenous function $f(x, y)$ of degree $r$. That is,
$f(tx, ty) = t^{r}f(x,y)$.
In all textbooks,
$\frac{t{\partial f(tx, ty)}}{\partial x} = \frac{t^{r}{\partial f(x, y)}}{\partial x}$.
I think it should be written as
$\frac{t{\partial f(tx, ty)}}{\partial (tx)} = \frac{t^{r}{\partial f(x, y)}}{\partial x}$
because of the chain rule, because $f(tx, ty)$ is first a function of $tx$ and $ty$ (and only 'ultimately' of $x$ and $y$) and then $tx$ and $ty$ are functions of $x$ and $y$ respectively. Please help.
 A: To provide a proof of
$$t\frac{\partial f}{\partial x}(tx,ty)=t^r\frac{\partial f}{\partial x}(x,y)$$
it is sufficient to show $\frac{\partial f}{\partial x}$ is homogeneous of degree $r-1$.
By definition
$$\frac{\partial f}{\partial x}(tx,ty)=\lim_{h\to 0}\frac{f(tx+h,ty)-f(tx,ty)}{h}.$$
Using homogeneity, we can rewrite this as
$$t^r\lim_{h\to 0}\frac{f(x+\frac{h}{t},y)-f(x,y)}{h}.$$
Then, as $t$ is independent of $h$, this is equal to
$$t^r\lim_{h/t\to 0}\frac{1}{t}\frac{f(x+\frac{h}{t},y)-f(x,y)}{h/t}=t^{r-1}\lim_{s\to 0}\frac{f(x+s,y)-f(x,y)}{s}$$
where $s=h/t$. But this is precisely $t^{r-1}\frac{\partial f}{\partial x}(x,y)$. The result follows immediately.
A: Consider a function $f$ that is homogeneous of degree $r$, i.e.,
$f(tx, ty) = t^rf(x, y)$ ......$(1)$
Differentiate both sides of equation (1) with respect to $t$. Then by Chain Rule:
$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}x + \frac{\partial f(tx, ty)}{\partial(ty)}y = rt^{r-1}f(x, y) 
\end{equation}$ ......$(2)$
Differentiate both sides of equation $(1)$ with respect to $x$. Then by chain rule:
$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}t = t^r\frac{\partial f(x, y)}{\partial(x)} 
\end{equation}$
$\implies$$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)} = t^{r-1}\frac{\partial f(x, y)}{\partial x} 
\end{equation}$ ......$(3)$
Similarly,
$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(ty)} = t^{r-1}\frac{\partial f(x, y)}{\partial y} 
\end{equation}$ ......$(4)$
Put $(3)$ and $(4)$ in $(2)$ to get the result:
$\begin{equation}
\frac{\partial f(x, y)}{\partial x}x + \frac{\partial f(x, y)}{\partial y}y = rf(x, y) 
\end{equation}$
