Covariant derivative of $Ricc^2$ How to derivate (covariant derivative) the expressions $R\cdot Ric$ and $Ric^2$ where $Ric^2$ means $Ric \circ Ric$? Here, $Ric$ is the Ricci tensor seen as a operator and $R$ is the scalar curvature of a Riemannian manifold.
 A: We want to compute
$$(\nabla(R\cdot Rc))(X)$$
Connections commute with contractions, so we start by considering the expression (where we contract $X$ with $Rc$, obtaining $Rc(X)$)
$$\nabla(R\cdot Rc\otimes X) =  (\nabla (R\cdot Rc))\otimes X + R\cdot Rc\otimes (\nabla X)$$
Taking the contraction and moving terms around, we get
$$(\nabla(R\cdot Rc))(X) = \nabla(R\cdot Rc(X)) - R\cdot Rc(\nabla X)$$
For the second one, notice that
$$Rc\circ Rc = Rc\otimes Rc$$
after contraction, so we proceed as before and start with
$$\nabla[Rc\otimes Rc\otimes Y] = [\nabla(Rc\otimes Rc)]\otimes Y + Rc\otimes Rc\otimes \nabla Y$$
We contract and move terms to get
$$[\nabla(Rc\circ Rc)](Y) = \nabla(Rc(Rc(Y))) - Rc(Rc(\nabla Y))$$
Now notice that using twice the first equality we derived we obtain
$$\nabla(Rc(Rc(Y))) = (\nabla Rc)(Rc(Y)) + Rc((\nabla Rc(Y))) =$$
$$= (\nabla Rc)(Rc(Y)) + Rc((\nabla Rc)(Y) +Rc(\nabla Y)))$$
Putting all together we obtain
$$\nabla(Rc\circ Rc) = (\nabla Rc)\circ Rc + Rc\circ(\nabla Rc)$$

This was an explicit derivation of everything. You can also use the various rules on the connections on the bundles associated to the tangent bundle to get directly
$$\nabla(R\cdot Rc) = dR\otimes Rc + R\nabla Rc$$
(which agrees with the first result, if you work it out a bit) and
$$\nabla(Rc\circ Rc) = \nabla(Rc\otimes Rc) = (\nabla Rc)\otimes Rc + Rc\otimes(\nabla Rc) = (\nabla Rc)\circ Rc + Rc\circ(\nabla Rc)$$
where there is a contraction of the tensors in the intermediate steps.
