I'm currently following this article to cosntruct a gauge invariant energy stress tensor for pure Yang-Mills gauge:
$$ \mathcal{L} = -\frac{1}{4}F_{\mu\nu}^aF_{\mu\nu}^a, \qquad F_{\mu\nu}^a = \partial_\mu A^a_\nu - \partial_\nu A^a_\mu + g\,C^{abc}A^b_\mu A^c_\nu, $$
where $C^{abc}$ are antisymemtric structure constants. I have trouble finding the correct derivation for the result 2.18. So I'm interested in:
$$ \frac{\partial F^b_{\rho\sigma}}{\partial A_\mu^a} = \frac{\partial}{\partial A^a_\mu}\left[ \dots + g\,C^{bca}A^c_\rho A^a_\sigma \right] = -2\,C^{bca}A_\rho^c\; \delta^\sigma_\mu. $$
I can't seem to find a factor of 2 and a minus sign. I tried using some of the antisymmetric properties of $F_{\mu\nu}$ and $C^{abc}$ but without success. Any help is much appriciated!
EDIT: My best guess was to try something I used in variation of strength field tensor, namely:
$$ \begin{split} \delta F^b_{\sigma\rho} &= \dots + g\,C^{bca} \delta A_\sigma^c A_\rho^a + g\,C^{bca} A_\sigma^c \delta A_\rho^a = \dots + g\,C^{bca} \delta A_\sigma^c A_\rho^a - g\,C^{bca} A_\rho^c \delta A_\sigma^a \\ &= \dots + g\,C^{bca} \delta A_\sigma^c A_\rho^a + g\,C^{bca} A_\rho^a \delta A_\sigma^c \end{split} $$$$ \begin{split} \delta F^b_{\rho\sigma} &= \dots + g\,C^{bca} \delta A_\rho^c A_\sigma^a + g\,C^{bca} A_\rho^c \delta A_\sigma^a = \dots + g\,C^{bca} \delta A_\rho^c A_\sigma^a - g\,C^{bca} A_\sigma^c \delta A_\rho^a \\ &= \dots + g\,C^{bca} \delta A_\rho^c A_\sigma^a + g\,C^{bca} A_\sigma^a \delta A_\rho^c \end{split} $$
where I get factor 2 as a result of antisymmetric properties, but I can't find the minus sign.