Question: If there exists a covariant derivative, then why doesn't there also exist a "contravariant derivative"? Why are all or most forms of differentiation "covariant", or rather why do all or most forms of differentiation transform covariantly? What aspect of differentiation makes it intrinsically "covariant" and intrinsically "not contravariant"? Why isn't the notion of differentiation agnostic to "co/contra-variance"?
To me it is unclear (on an intuitive, i.e. stupid/lazy, level) how notions of differentiation could be restrained to being either "covariant" or "contravariant", since any notion of differentiation should be linear*, and the dual of any vector space is exactly as linear as the original vector space, i.e. vector space operations in the dual vector space still commute with linear functions and operators, they same way they commute with such linear objects in the original vector space.
So to the extent that the notion of linearity is "agnostic" to whether we are working with objects from a vector space or from its dual vector space, so I would have expected any notion of differentiation to be similarly "agnostic". Perhaps a better word would be "symmetric" -- naively, I would have expected that if a notion of "covariant differentiation" exists, then a notion of "contravariant differentiation" should also exist, because naively I would have expected one to exist if and only if the other exists.
However, it appears that no such thing as "contravariant derivative" exists (see here on Math.SE, also these two posts [a][b] on PhysicsForums), whereas obviously a notion of "covariant derivative" is used very frequently and profitably in differential geometry. Even differential operators besides the so-called "covariant derivative" seemingly transform covariantly, see this post for a discussion revolving around this property for the gradient. I don't understand why this is the case.
(* I think)