Covariant Derivative Homework: Solve ∇_c ({∂}_b X^a)

[MattRob]

Homework Statement

Take the Covariant Derivative

∇_{c} ({∂}_b X^a)

Homework Equations

∇_{c} (X^a) = ∂_c X^a + Γ_{bc}^a X^b
∇_{c} (X^a_b) = ∂_c X^a_b + Γ_{dc}^a X^d_b - Γ^d_{bc} X^a_d

The Attempt at a Solution

Looking straight at
∇_{c} ({∂}_b X^a)
I'm seeing two indices. However, the b is nothing more than a reference to the derivative with respect to what, so I'm not sure whether this counts as a one or two-index object. Namely, do I solve it as

∇_{c} ({∂}_b X^a) = ∂_c {∂}_b X^a + Γ_{dc}^a {∂}_b X^d

since the index in the partial differential operator isn't the same as an index on a tensor, or is it?

So I'm thinking since the indices merely indicate you're working with a variety of objects (x^a = k => x^1 = k, x^2 = k, x^3 = k, ...), then the index on the differential operator should be treated the same, so I can treat {∂}_b X^a as a two-index tensor T^a_b = {∂}_b X^a

∇_{c} ({∂}_b X^a) = ∂_c {∂}_b X^a + Γ_{dc}^a {∂}_b X^d - Γ^d_{bc} {∂}_d X^a

So is it the latter case, former, or something else entirely?

I suppose I have every reason to think it should be the latter case, but I guess I'm just uncertain and want some clarification.

Thanks!

 

[strangerep]

Treat it as a 2-index object.

 

[MattRob]

Treat it as a 2-index object.

Second formula, then?

Thanks!