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[latentcorpse]

Edit: Solved
Don't know how to delete thread though!

 

[fzero]

A covariant derivative is an example of a derivation, that is a differential operator that satisfies the Leibnitz (product rule):

$$D ( v \otimes w ) = (Dv)\otimes w + v \otimes (Dw).$$

Therefore the part that you proved above follows. Now, since $$l^2$$ is a scalar, it's covariant derivative reduces to the ordinary derivative. You can see this by explicit computation, since the signs in front of the Christoffel symbols are opposite when acting on covariant and contravariant vectors.

 

[dextercioby]

Judging from a textbook's perspective, your problem is ill-posed, because the $l^2$ for l a 4-vector is a genuine scalar, and the way the covariant derivative acts on the scalars is used as an axiom in deriving how the covariant derivative acts on covectors, knowing how it acts on scalars and 4-vectors.

 

[berkeman]

Edit: Solved
Don't know how to delete thread though!

You don't delete your post after you receive help here. Check your PMs.