derivative of curvature tensor on vector bundle Let $E \to M$ be a riemannian bundle and $D$ a linear connection compatible with the metric. For every pair of vector fields $X$ and $Y$ on $M$ and every section $\xi$ of $E$, the curvature is given by
$$F(X,Y)\xi=D_X D_Y \xi - D_Y D_X \xi - D_{[X,Y]}\xi$$
This is actually a tensor $F \in \Gamma(T^*M \otimes T^*M \otimes \mathrm{End}E)$. I would like to know how the covariant derivative of $F$ looks like.
It should be defined in a way such that a sort of "product rule" works. Considering $F(X,Y) \in \Gamma(\mathrm{End}E)$ such a formula could be
$$D_XF(Y,Z)\xi=D_X(F(Y,Z)\xi-F(Y,Z)D_X \xi$$
but what if I consider $F \in \Gamma(T^*M \otimes T^*M \otimes \mathrm{End}E)$? A connection on $TM$ should appear.
I'm trying to prove the Bianchi identity
$$D_XF(Y,Z)+D_YF(Z,X)+D_ZF(X,Y)=0$$
 A: You're right that to define $DF\in\Gamma(T^*M \otimes T^*M \otimes T^*M \otimes \operatorname{End}(E)),$ you need a connection on $TM$. Once you fix such a connection, you get unique natural connections on any bundle constructed from $TM$ and $E$ by taking duals and tensor products simply by imposing the product rule. 
It turns out that the second Bianchi identity is true for any torsion-free connection on $TM.$ You should be able to prove this with a long but straightforward calculation - the terms like $D_X D_Y D_Z \xi$ all cancel out thanks to the cyclic symmetrization, and the Lie bracket terms should all cancel with the terms involving the connection on $TM$ thanks to the torsion-free assumption.
The usual way to think of this identity without requiring the arbitrary choice of connection is via the exterior covariant derivative: if we think of $F$ as a 2-form taking values in $\operatorname{End}(E)$, then the second Bianchi identity is exactly the equation $d^\nabla F = 0$ where $\nabla$ is the natural connection induced on $\operatorname{End}(E).$
