Leibnitz's rule for covariant derivatives I have realized that one must be very careful when defining the notion of covariant derivative on a vector bundle, and I want to know whether there has been any discussion on this or how to proceed with care when switching from one book to another.
I'll clarify with an example. Suppose you define a covariant derivative on a vector bundle $E\rightarrow X$ as a linear operator
$$ D: \Omega^0(X,E) \rightarrow \Omega^1(X,E)$$
satisfying the following properties:

*

*$D(fs) = df \otimes s + f Ds$ for any function $f\in\Omega^0(X,\mathbb{F})$ and section $s\in\Omega^0(X,E)$

*$D(c s) = c Ds$ for any constant $c\in\mathbb{F}$.

This definition is equivalent to that in which the order of $f$ and $s$ is inverted ($fs = sf$ as sections). However, one usually extends covariant derivatives to exterior covariant derivatives
$$ D: \Omega^p(X,E) \rightarrow\Omega^{p+1}(X,E)$$
and in this case I have seen both choices of the Leibnitz rule in different textbooks:

*

*Option 1: $D(\omega\otimes s) = d\omega \otimes s + (-1)^p \omega \wedge Ds$

*Option 2: $D(s\otimes \omega) = Ds \wedge \omega + (-1)^q s \otimes d\omega$
Here $\omega \in \Omega^p(X), s\in \Omega^q(X,E)$
I think this has a lot of implications on formulas to be derived in many subfields.
Therefore I must ask: is there a preferred definition? Why is it better?
 A: I just started learning this stuff involving vector bundles, but it makes the most sense to me to simply avoid using the $\otimes$ symbol, and rather define everything in terms of $\wedge$ appropriately so that both the rules you state become equivalent.
To do so, we fix a vector bundle $(E,\pi,X)$ and for each integer $q\geq 0$, we shall view an $E$-valued $q$-form on $X$ to be a smooth map over the identity, $s: (TX)^{\oplus q}\to E$ which is fiberwise alternating and multilinear. This is of course nothing but a slight shift in perspective to considering $s$ to be a smooth section of the vector bundle $(\bigwedge^qT^*X)\otimes E$.
Now, given a $p$-form $\omega$ on $X$ (in the usual sense) we can define a wedge product $s\wedge \omega$, which will be an $E$-valued $(p+q)$-form on $X$. The definition is just as we expect: given tangent vectors $h_{1,x},\dots h_{p+q,x}\in T_xX$, we define
\begin{align}
(s\wedge \omega)(h_{1,x},\dots h_{p+q,x}):=\frac{1}{p!q!}\sum_{\sigma\in S_{p+q}}s(h_{\sigma(1),x},\cdots, h_{\sigma(p),x})\cdot \omega_x(h_{\sigma(p+1),x},\cdots, h_{\sigma(p+q),x})
\end{align}
So, this is really just the same definition of the wedge product of usual forms, except now the addition and scalar multiplication on the RHS are taking place in the vector space $E_x$. Similarly, we can define $\omega\wedge s$, and it will turn out as usual that $s\wedge \omega=(-1)^{pq}\omega\wedge s$ (just follow the usual proof, and note that if $c$ is a scalar and $v$ is a vector in any vector space, it doesn't matter whether we write $cv$ or $vc$).
So, now with this understanding of $\wedge$ (both on the left and the right), the two Leibniz rules can be written as (for a usual $p$-form $\omega$ and $E$-valued $q$-form $s$)

*

*$d_{D}(\omega\wedge s)=(d\omega) \wedge s+(-1)^p\omega \wedge (d_{D}s)$

*$d_{D}(s\wedge \omega)=(d_{D}s) \wedge \omega + (-1)^{q} s \wedge d\omega$
To get from one to the other simply multiply throughout by $(-1)^{pq}$ and use the "anti-commutativity" of the wedge product.
