Covariant derivative of a covariant tensor wrt superscript Is it true that when you take the covariant derivative of a covariant tensor, do you always have to do with a subscript? What if you do it wrt a superscript?Does the first term (with the partial derivative) take a minus sign? More specifically, is this true?
$$\nabla^{\mu}R_{\mu\nu} = -\frac{\partial R_{\mu\nu}}{{\partial x_{\mu}}} + \text{(Christoffels)}$$
Where does the minus sign come from? Is there a proof for this, or is it just a definition?
Also, is there a change in the signs for Christoffel symbols(not the change if the the tensor's indices change position, but the change when the index of the covariant differential changes)?
I want to know the PROOF/REASON behind the minus sign. 
 A: No.  The subscript is the defined thing.  If you have the superscript, you just assume raising with the metric tensor, so:
$$\nabla^{\mu}R_{\mu\nu} \equiv g^{\mu\alpha}\nabla_{\alpha}R_{\mu\nu}$$
which you expand normally with partial derivatives and Christoffels.  Of course, since we know that $\nabla^{a}\left(R_{ab} - \frac{1}{2}Rg_{ab} \right)= 0$, we know right away that we can simplify $\nabla^{\mu}R_{\mu\nu}$ to $\frac{1}{2}\nabla_{\nu}R$
A: From your comments, I will try to answer what confuses you. Let us take a metric signature:
$$ \eta_{\mu \nu} = \mathrm{diag}(-1,1,1,1) $$
and let us consider some general $x^\mu$. We will denote the time component of $x^\mu$ by $x^0$. If we want to lower the index of $x^0$, we get:
$$x^0 = \eta^{0 \mu} x_\mu = \eta^{0 0} x_0 + \eta^{0 1} x_1 + \eta^{0 2} x_2  + \eta^{0 3} x_3 \tag{1}$$
Since $\eta^{0 0} = -1$ and $\eta^{0 1} = \eta^{0 2} = \eta^{0 3} = 0$, equation $(1)$ becomes:
$$x^0 = - x_0$$
and so we get the minus sign.
Note that if we only consider the spatial component $x^i$ (where $i$ is either the $1$st, the $2$nd or the $3$rd component), then we lower the index again as:
$$ x^i = \eta^{i \mu} x_\mu = \eta^{i0} x_0 + \eta^{i1} x_1 + \eta^{i2} x_2 + \eta^{i3} x_3 = x_i$$
and so we don't get a minus sign.
