# What is the covariant derivative of a wedge product?

If we have a covariant derivative for vector fields given by an affine connection on a manifold, can we extend that to a covariant derivative for k-vectors by assuming that the product rule hold for wedge products of vectors? In other words, does it make sense to assume that:

$$\nabla_v(a\wedge b) = (\nabla_va)\wedge b+a\wedge (\nabla_vb)$$

where $$v$$ is a vector field, $$a$$ and $$b$$ are $$p$$-vector and $$q$$-vector fields, and $$\nabla_va$$ agrees with the covariant derivative of vectors when $$a$$ is a vector field?

(I am wondering since the product rule for the exterior derivative has a grade-dependent sign in it, but I think that should not be the case here.)

Yes, $$\nabla_v (a\wedge b)=(\nabla_v a)\wedge b+a\wedge(\nabla_v b)$$ for any $$v\in TM$$ and $$a,b\in\mathcal{T}^{\bullet,0}M$$. This is the Leibniz rule for tensor product, projected down to the alternating parts. See, for example, this question.
(The $$(-1)^k$$ sign for exterior derivative comes from "moving $$v$$ pass other vectors" when impose the constraint $$v$$ also participate in the antisymmetry.)