Proving $\nabla_X (\text{tr}\ F) = \text{tr} (\nabla_XF)$. Let $F\in\mathcal T^{k}_{l}(M)$ be a smooth section of a $(k,l)$-tensor bundle over the manifold $M$. The trace operator
$$
\text{tr}:\mathcal T^{k}_{l}(M) \to\mathcal T^{k-1}_{l-1}(M)
$$
is defined by
$$
(\text{tr}\ F)(\omega^1,\dots,\omega^{l-1},Y_1,\dots,Y_{k-1}):=\text{tr}\left( F(\omega^1,\dots,\omega^{l-1},\omega,Y_1,\dots,Y_{k-1},Y) \right)
$$
where trace on the RHS is the regular trace of the $(1,1)$-tensor in variables $\omega$ and $Y$.

How do I show that $$\nabla_X (\text{tr} F) =  \text{tr} (\nabla_XF)$$
  holds?

Writing it out in coordinate seems like a real pain, is there a nicer way to prove this? I tried to do it in local coordinate and it looks quite messy.
EDIT: The definition I am using is the computational one, i.e
$$\begin{align}
\nabla_X F(\omega^1,\dots,\omega^l,Y_1,\dots,Y_k):&=XF(\omega^1,\dots,Y_k)\\&-F(\nabla_X\omega^1,\omega^2,\dots,Y_k)-\dots -F(\omega^1,\dots,Y_{k-1},\nabla_X Y_k)
\end{align}$$
 A: Hint: It is known that $\nabla_XF$ is a tensor (if you haven't already, it may be a good idea to verify that first). This means that the value of $$\nabla_XF\left(\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k\right)$$at a point $p$ only depends on the values of $\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k$ at $p$. Consequently, you may assume that $\omega^1,\ldots,\omega^l,Y_1,\ldots,Y_k$ are all parallel in the direction $X$. This will cause most of the terms in your formula vanish and make the computation considerably less messy.
Edit: What I meant to say is that, given $\omega^1|_p,\ldots,Y_k|_p,$ we may extend all those vectors and covectors in any way we choose. In particular, we may assume $$\nabla_X\omega^1=\ldots=\nabla_XY_k=0.$$ This would leave us with $$\nabla_XF\left(\omega^1,\ldots,Y_k\right)=X\left(F\left(\omega^1,\ldots,Y_k\right)\right),$$which is much more convenient to work with.
Edit #$2$: We wish to show $$\nabla_X\mathrm{tr}F=\mathrm{tr}\nabla_XF.$$This equality holds if and only if it holds at every point, so let $p\in M$, and let $$\omega^1,\ldots,\omega^{l-1}\in T^*_pM,\quad Y_1,\ldots,Y_{k-1}\in T_pM.$$As explained above, we may extend the $\omega$'s and the $Y$'s to global vector fields and $1$-forms in whatever manner we choose; different choices yield different summands in the expression for $\nabla_XF$, but the overall sum remains the same. So we assume $$\nabla_X\omega^1=\ldots=\nabla_XY_{k-1}=0.$$ Let $e_1,\ldots,e_n$ be a local frame of $TM$. Once again, we may assume $$\nabla_Xe_1=\ldots=\nabla_Xe_n=0.$$Let $\varphi^1,\ldots,\varphi^n$ be the dual local frame of $T^*M$. It automatically follows (why??) that $$\nabla_X\varphi^1=\ldots=\nabla_X\varphi^n=0.$$ Finally, due to all our assumptions, $$\begin{align}\nabla_X\mathrm{tr}F\left(\omega^1,\ldots,Y_{k-1}\right)&=X\left(\mathrm{tr}F\left(\omega^1,\ldots,Y_{k-1}\right)\right)\\&=X\left(\sum_iF\left(\omega^1,\ldots,\omega^{l-1},\varphi^i,Y_1,\ldots,Y_{k-1},e_i\right)\right)\\&=\sum_iX\left(F\left(\omega^1,\ldots,\omega^{l-1},\varphi^i,Y_1,\ldots,Y_{k-1},e_i\right)\right)\\&=\sum_i\nabla_XF\left(\omega^1,\ldots,\omega^{l-1},\varphi^i,Y_1,\ldots,Y_{k-1},e_i\right)\\&=\mathrm{tr}\nabla_XF\left(\omega^1,\ldots,Y_{k-1}\right),\end{align}$$as desired.
