I'm reading Riemannian Geometry and Geometric Analysis by Jost and the way he defines the exterior covariant derivative comes naturally using the definition of the induced connection on the tensor product.
Definition 3.1.4 Let $E_{1}, E_{2}$ be vector bundles over M with connections D1, D2 resp. The induced connection D on $E=E_{1}\otimes E_{2}$ is defined by the requirement $$D(\mu_{1}\otimes \mu_{2})=D_{1}\mu_{1}\otimes \mu_{2}+\mu_{1}\otimes D_{2}\mu_{2}$$ for $\mu_{i}\in\Gamma(E_{i})$, $i=1,2$.
So, in order to extend a connection $D$, defined on a vector bundle $E$, from $\Gamma(E)$ to $\Gamma(E)\otimes\Omega^{p}(M)$, $0\leq p \leq \dim M$, he requires $$D(\mu\otimes \omega )=D\mu\wedge \omega+\mu\otimes d\omega$$ for $\mu\in\Gamma(E)$, $\omega \in \Omega^{p}(M)$. But in the bibliography I came across with the Leibniz rule for the exterior covariant derivative $$D(\omega\otimes \mu )=d\omega\otimes\mu+(-1)^{p}\omega\wedge D\mu$$ and I am wondering, if we were to extend D from $\Gamma(E)$ to $\Omega^{p}(M)\otimes\Gamma(E)$ using again the definition for the tensor product, then wouldn't we require $$D(\omega\otimes \mu )=d\omega\otimes\mu+\omega\wedge D\mu \ ?$$ In other words, were does the $(-1)^{p}$ sign hide when we use the above definition? Thanks in advance.