Show that $\omega = dF$ when $\omega = \sum_{i=1}^n f_i(x)dx_i$ and $F = \frac{1}{p+1} \sum_{i=1}^n x_if_i(x)$ Let $f_i : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}$ be a homogeneous function of degree $p \neq -1$ for every $1 \le i \le n$. Namely, $f_i(tx) = t^pf_i(x)$ for every $
 x \in\mathbb{R}^n \setminus \{0\}$ and $t >0$.
We define $\omega = \sum_{i=1}^n f_i(x)dx_i$ and  assume it is closed. 
I want to prove that $\omega = dF$ when $F = \frac{1}{p+1} \sum_{i=1}^n x_if_i(x)$
Basically want we want to show is that $\frac{\partial F}{\partial x_i} = f_i(x)$.
By straightforward deriving I get: 
$$ \frac{\partial F}{\partial x_i} = \frac{1}{p+1} (\sum_{j=1}^n x_j\frac{\partial f_j(x)}{\partial x_i} + f_i(x))$$ 
Here I got stuck. I don't know how to continue. I also don't see how the information that $\omega$ is closed (namely, $\frac{\partial f_i(x)}{\partial x_j} = \frac{\partial f_j(x)}{\partial x_i}$ for $i \neq j$) helps me here. I also don't know how to use the fact thar $f_i$ are homogeneous because it is not given that $x_i > 0$.
Help would be appreciated.
 A: Hint: ${d\over{dt}}_{t=1}f_i(tx)={d\over{dt}}_{t=1}t^pf_i(x)=pf_i(x)$.
You can also compute it as follows
${d\over{dt}}_{t=1}f_i(tx)=(\sum_{j=1}^{j=n}x_j{{\partial f_i}\over{\partial x_j}})_{t=1}$
$=\sum_{j=1}^{j=n}x_j{{\partial f_i}\over{\partial x_j}}$.
If you replace the second formula by the first in your computations, you obtain the result.
A: Assume $x_1$, $x_2$, $\ldots$, $x_n$ are all positive. Then, for fixed $i$, we can write
$$
f_i(x_1,\ldots, x_i, \ldots, x_n) = x_i^{p} f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\,.
$$
Now, 
$$
F(x)=\sum_{i=1}^n x_i f_i(x)=\sum_{i=1}^n x_i^{p+1}f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\,.
$$
Consider, for fixed $j$,
$$
\frac{\partial}{\partial x_j}F=
(p+1)x_j^{p+1}f_j\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\\
+\sum_{i=1}^n x_i^{p+1}\frac{\partial}{\partial x_j}f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\,.
$$
We recognize the first term on the right-hand side as $(p+1)f_j(x)$. The second term, appropriately separating the terms in the sum, gives
$$
\sum_{i\neq j} x_i^{p+1}\frac{\partial }{\partial x_j} f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\\
+
x_j^{p+1} \frac{\partial }{\partial x_j}f_j\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\\
=
\sum_{i\neq j} \frac{x_i^{p+1}}{x_i}\frac{\partial f_i}{\partial x_j}\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\\
+\sum_{k\neq j} \frac{-x_j^{p+1}x_k}{x_j^2} \frac{\partial f_j}{\partial x_k}\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\\
=\sum_{k\neq j} x_k\left(\frac{\partial }{\partial x_j}f_k-\frac{\partial}{\partial x_k}f_j\right)=0\,.
$$
In the last step we used the closedness property; note  that, for $k\neq j$,
$$
\frac{\partial}{\partial x_k}f_j(x)=x_j^{p-1} \frac{\partial f_j}{\partial x_k}\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\,.
$$
When some $x_i$ is allowed to be negative, the argument can be repeated by replacing the $1$ with $-1$ appropriately.
