Im interested in calculating the exterior derivative of a 2-form defined by $\omega(x,y) = g(x,Ay)$ where $A \in \Gamma(End(T) )$ is skew and $g$ is some metric.
I hope to reach some coordinate-independent expression involving the covariant derivative of $A$ and possibly the curvature of $g$.
With respect to coordinates, $A$ and $g$ can be written as \begin{align*} A(\partial_k) &= A_k^p\partial_p\\ g(\partial_i,\partial_j) &= g_{ij}. \end{align*} Let $$ A_{jk} = g_{ij}A^i_k. $$ That $A$ is skew (with respect to $g$) means $$A_{jk} = - A_{kj}.$$ Moreover, it is easy to check that $$ \omega(\partial_j,\partial_k) = A_{jk} $$ and therefore $$\omega = A_{jk}\,dx^j\wedge dx^k. $$ So $$ d\omega = \partial_iA_{jk}\,dx^i\wedge dx^j\wedge dx^k. $$ Now, given a point $x_0$, choose coordinates so that $\partial_ig_{jk}(x_0) = 0$. With respect to those coordinates, $$ \partial_iA_{jk}(x_0) = \nabla_iA_{jk}(x_0). $$ It therefore follows that with respect to any coordinates, $$ d\omega = \nabla_iA_{jk}\,dx^i\wedge dx^j\wedge dx^k.$$