= Solution
If $E$ has local frame transition matrices $g_{ab}$ and the cotangent bundle has transitions $J_{ab}^{-T}$, then $E\otimes\Lambda^rT^*B$ has local trivializations with transitions
$$
g_{ab}\otimes\Lambda^r(J_{ab}^{-T}),
$$
which satisfy the cocycle condition. Thus it is a well-defined <tensor product of vector bundles>.
A <connection on a vector bundle> is a linear map
$$
\nabla:\Gamma(E)\to\Omega^1(B;E)
$$
satisfying $\nabla(fs)=df\otimes s+f\nabla s$. Contracting with a vector field gives the covariant derivative $\nabla_Xs$. In a local frame, $\nabla=d+A$ for a matrix-valued one-form $A$; under a frame change $g$ the matrix transforms as
$$
A'=g^{-1}Ag+g^{-1}dg.
$$
Its <covariant exterior derivative> is defined by
$$
d_A(s\otimes\alpha)=\nabla s\wedge\alpha+s\otimes d\alpha,
$$
and locally
$$
d_A\eta=d\eta+A\wedge\eta.
$$
This formula and the graded Leibniz rule show that definitions in different frames agree.
The <curvature form of a connection> is $F(A)=d_A^2$. Locally,
$$
F(A)=dA+A\wedge A.
$$
Its covariant derivative satisfies the <Bianchi identity>
$$
d_AF=dF+A\wedge F-F\wedge A=0.
$$
Indeed, substituting $F=dA+A\wedge A$, using $d^2=0$, and applying the graded Leibniz rule leaves equal and opposite terms.
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