How to differentiate a tensor like $ A = A^{\alpha\beta} e_{\alpha} ⊗ e_{\beta} $ wrt $ x^{\mu} $?

Question

Given a tensor object: $$ A(x) = A^{\alpha\beta}(x)\;e_{\alpha}(x) ⊗ e_{\beta}(x) $$ Its derivative wrt. $x^\mu$ is: $$ \frac{\partial A(x)}{\partial x^\mu} = e_\alpha(x) \;\frac{\partial A^{\alpha\beta}(x)}{\partial x^\mu}\;+\;A^{\alpha\beta}(x) \; \frac{\partial (e_{\alpha}(x) \;⊗\; e_{\beta}(x))}{\partial x^\mu} $$

How do you differentiate the second term?

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Brief Motivation:

For a first order tensor, $$ \frac{\partial e_\alpha}{\partial x^\mu} := \Gamma_{\mu \alpha}^\kappa \;e_\kappa $$

Which invokes the notion of the affine connection, $\Gamma_{\mu \alpha}^\kappa $.

Answer 1

Just apply the derivative of the product: $\partial_\mu (e_\alpha \otimes e_\beta) = \partial_\mu (e_\alpha) \otimes e_\beta+e_\alpha \otimes \partial_\mu (e_\beta).$

It works because $\nabla_v (T \otimes S) = \nabla_v T \otimes S + T \otimes \nabla_v S$ for any tensors $T,S$.

Answer 2

Covariant derivative is Leibnizian so for $A= A^{\alpha\beta}\;e_{\alpha}\otimes e_{\beta}$ you've $$\nabla_{e_{\mu}}A= (A^{\alpha\beta}{}_{,\mu})\;e_{\alpha}\otimes e_{\beta}+ A^{\alpha\beta}\Gamma^{\nu}{}_{\alpha\mu}e_{\nu}\otimes e_{\beta}+ A^{\alpha\beta}\Gamma^{\nu}{}_{\beta\mu}e_{\alpha}\otimes e_{\nu}$$ which rearranges into $$\nabla_{e_{\mu}}A= (A^{\alpha\beta}{}_{,\mu}+ A^{\nu\beta}\Gamma^{\alpha}{}_{\nu\mu}+ A^{\alpha\nu}\Gamma^{\beta}{}_{\nu\mu})\ e_{\alpha}\otimes e_{\beta}.$$