Accordingly to Yang-Mills theories, after the introduction of a covariant derivative such that
$$D_\mu = \partial_\mu - igA_\mu, \tag1$$
you can built the kinetic term for the gauge potential $A_\mu$ as
$${\cal L}_A = -\frac{1}{2}tr\{F_{\mu \nu}F^{\mu \nu}\},\qquad F_{\mu \nu} = \frac{i}{g}[D_\mu, D_\nu]. \tag2$$
The action of the covariant derivative transforms under a local group tranformation $\Omega$ in the following way:
$$D_\mu\psi \rightarrow \Omega D_\mu\psi. \tag3$$
And the gauge field as
$$A_\mu \rightarrow \Omega A_\mu\Omega^\dagger + \frac{i}{g}\Omega\partial_\mu \Omega^\dagger . \tag4$$
Introducing Eq. (4) in Eq. (1) you get that under the group tranformations, the covariant derivate transforms as,
$$D_\mu \rightarrow \partial_\mu - ig\Omega A_\mu\Omega^\dagger + \Omega\partial_\mu\Omega^\dagger = \Omega D_\mu\Omega^\dagger + \partial_\mu . \tag5$$
Eq. (5) into definition of strength tensor in Eq. (2) gives
$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger + \frac{i}{g}(\partial_\mu(\Omega D_\nu\Omega^\dagger) - \partial_\nu(\Omega D_\mu\Omega^\dagger)) . \tag6$$
Taking $\Omega \simeq 1 - g\theta^iT^i$, with $\{T^i\}$ the set of generators of the group, $\theta^i \in \mathbb{R}$ and $f^{ijk}$ the structure constants:
$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger + \frac{i}{g}(\partial_\mu A_\nu^k - \partial_\nu A_\mu^k)T^k + ig^2f^{ajk}T^k[\partial_\mu(\theta^jA_\nu^a) - \partial_\nu(\theta^jA_\mu^a)] . \tag7$$
In Particle Physics books, it is said that $$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger \tag{8},$$ but I don't get that in my calculus (Eq. (7)). What am I doing wrong? This result is important because if $F_{\mu\nu}$ doesn't transforms as books say, the kinetic term in Eq. (2) isn't gauge invariant.
