I am reading the proof of second Bianchi identity on wiki. In the proof, it says the following condition must satisfy:
$$((D_X R) (Y,Z)) (W) + R (D_XY,Z) W + R(Y,D_XZ) W + R(Y,Z) D_X W = D_X (R(Y,Z)W)$$
I don't quite understand the reason. I know by definition, $R(Y,Z)=D_{[Y,Z]}-[D_Y,D_Z]$, and I also know $D_X$ is a derivation, which satisfies the Leibniz rule, that is, for two tensor fields $K_1,K_2$, $$D_X(K_1\otimes K_2)=(D_XK_1)\otimes K_2+K_1\otimes (D_X K_2)$$
After some reading, I found the result may due to the fact that covariant derivative commutes with tensor contraction. That is, view $R(Y,Z)W$ as a composition of three contractions, such that $$\otimes^3 TM\otimes \otimes^4 T^*M\overset{C_1}{\rightarrow}\otimes^2 TM\otimes \otimes^3 T^*M\overset{C_2}{\rightarrow}\otimes^1 TM\otimes \otimes^2 T^*M\overset{C_3}{\rightarrow} T^*M$$ and each time uses the $(0,4)$ tensor $R$ to act on one vector.
But I am not sure whether my understanding is correct since usually contraction map only combines one tangent vector and one cotangent vector at each time and leaves other components fixed, while in my explanation, I use the whole $(0,4)$ tensor $R$ to act on a vector and produces a $(0,3)$ tensor and I am not sure whether covariant derivative still commutes with such map.
Thanks for your advice.