Why is $\partial f(x)/\partial x_i$ homogeneous of degree $k-1$ if $f(x)$ is homogeneous of degree $k \geq 1$? It's as (simple) as the title says, but I can't figure out why that is.
How can I show (I think I need small, tiny steps) that
$$\frac{\partial f(tx)}{\partial (tx_i)}t=t^k\frac{\partial f(x)}{\partial x_i}$$?
My textbook says "If $f(x)$ is homogeneous of degree $k \geq 1$, then $\partial f(x)/\partial x_i$ is homogeneous of degree $k-1$". How come?
Where f is a function $f:\mathbb{R}_+^n \mapsto \mathbb{R}$
 A: Fix $t>0$. Then
$$\frac{\partial f}{\partial x}(tx)=\lim_{\Delta x\to 0}\frac{f(tx+\Delta x)}{\Delta x}=\lim_{\Delta x\to 0 }\frac{f(t(x+\Delta x))}{t\Delta x}=t^{k-1}\lim_{\Delta x\to 0}\frac{f(x+\Delta x)}{\Delta x}=t^{k-1}\frac{\partial f}{\partial x}(x)$$
A: Since $f$ is homogenenous of degree $k\geq 1$, the following must be true for any $t>0$ and $x\in\mathbb R^n_+$: $$f(t x)=t^kf(x).$$ Fix $t>0$ and differentiate both sides with respect to $x_i$ (where $i\in\{1,\ldots,n\}$). By the chain rule, you obtain $$f_{x_i}(t x)\cdot t=t^k f_{x_i}(x),$$ where the subscripts denote partial derivatives. Canceling $t$ yields the desired result: $$f_{x_i}(tx)=t^{k-1}f_{x_i}(x).$$
A: As you have a polynomial of degree $k$, each term consists of a total of $k$ factors. If you now take the derivative by one of the $x_i$, each term which has an $x_i^{r_i}$ ($r_i\geq 1$) as a factor will convert to $r_i x_i^{r_i-1}$, i.e. the degree is one less than before.
All other terms that has no $x_i$ will yield $0$ in the derivative.
Now coming to the first part of the question:
As $f$ is homogeneous of degree $k$ we can conclude that $f(tx)=t^k f(x)$. As factors can be taken outside of the derivative we get:
$$
\frac{\partial f(tx)}{\partial (tx_i)} t = \frac{t^k \partial f(x)}{t \partial x_i}t = t^k \frac{\partial f(x)}{\partial x_i}
$$
