Leibniz Rule for Covariant derivatives I recently came across a video by prof Fredrick Schuller on general relativity where he defines the leibniz rule to be,
$\nabla_X (T(\omega,Y))=\nabla_XT(\omega,Y)+T(\nabla_{X} \omega,Y)+T(\omega,\nabla_{X} Y)$
Where $X$ and $Y$ are vector fields, $\omega$ is a covector field and $T$ is a $(1,1)$ tensor. The rule can be generalised for $(p,q)$ tensors similarly.
I cannot find a way to show that it is equivalent to the leibniz rule expressed as $\nabla_X$ acting on tensor product of two tensor fields.
i.e. $\nabla_{X}(T \otimes S) = \nabla_XT\otimes S + T\otimes \nabla_XS$
How do I proceed to show the equivalence between the two?
 A: I would think of it like this, using the fact that covariant derivative commutes with contractions (use C for contraction) and Liebniz rule
$\nabla_X (T(\omega,Y))
=\nabla_X (C C ( T\otimes \omega \otimes Y))
= CC \nabla_X(T\otimes \omega \otimes Y)
= CC((\nabla_X T)\otimes \omega \otimes Y+
T\otimes (\nabla_X\omega) \otimes Y+
T\otimes \omega \otimes( \nabla_X Y))
=
\nabla_XT(\omega,Y)+T(\nabla_{X} \omega,Y)+T(\omega,\nabla_{X} Y)$
NOTE: one might have to define what slots C contracts first but I think this just works (maybe some care is needed if you use the clifford product or have antisymmetric tensors?)
A: The covariant derivative is defined to obey the Leibnitz rule. 
If the ${\bf e}_i$ are a vielbein basis then We define the action of $\nabla_X$ on any function $f(x)$  by
$$
\nabla_Xf= Xf = X^\mu \partial_\mu f,
$$
and on the elements ${\bf e}_i$ of a vielbein basis by
$$
\nabla_X {\bf e}_i = {\bf e}_j {\omega^j}_{i\mu}X^\mu.
$$
We  extend to any other object by demanding that both linearity and  Liebnitz rule hold. So, on a vector field $Y= Y^i {\bf e}_i$, we have 
$$
\nabla_X Y= (\nabla_X Y^i){\bf  e}_i +  Y^i (\nabla_X {\bf e}_i)\\
= (X^\mu\partial_\mu Y^i) {\bf  e}_i+ Y^i ({\bf e}_j{\omega^j}_{i\mu}X^\mu)\\
= X^\mu (\partial_\mu Y^i + Y^j {\omega^i}_{j\mu}){\bf e}_i.
$$
Note that the position-dependent numerical components $Y^i(x)$ of a vector are still just functions.
We do the same for a tensor 
$$
\nabla_X (T^{ij}{\bf e}_i\otimes {\bf e}_j)= (\nabla_X T^{ij}){\bf e}_i\otimes {\bf e}_j+  T^{ij}(\nabla_X{\bf e}_i)\otimes {\bf e}_j+
T^{ij}{\bf e}_i\otimes (\nabla_X {\bf e}_j)\\
=X^\mu (\partial_\mu T^{ij}+ {\omega^i}_{k\mu} T^{kj} + {\omega^j}_{k\mu} T^{ik}){\bf e}_{i}\otimes {\bf e}_j
$$
It should now be clear that for Liebnitz holds for any tensor product or contraction. 
