Covariant derivative of a contravariant derivative This is a pretty basic question, but I am having problems understanding the covariant derivative of contravariant vector.
My understanding is that
$$\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}.\vec{e_{\nu}}+A^{\alpha}\Gamma_{\mu\alpha}^{\nu}.\vec{e_{\nu}}$$
So I decided to write it out longhand for a 3D case. The following is my longhand of the covariant derivative.
$$
(\frac{\partial A^{x}}{\partial x}+A^{\alpha}\Gamma_{0\alpha}^{0}).\vec{e_{0}}+(\frac{\partial A^{y}}{\partial x}+A^{\alpha}\Gamma_{0\alpha}^{1}).\vec{e_{1}}+(\frac{\partial A^{z}}{\partial x}+A^{\alpha}\Gamma_{0\alpha}^{2}).\vec{e_{2}}\\
(\frac{\partial A^{x}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{0}).\vec{e_{0}}+(\frac{\partial A^{y}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{1}).\vec{e_{1}}+(\frac{\partial A^{z}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{2}).\vec{e_{2}}\\
(\frac{\partial A^{x}}{\partial z}+A^{\alpha}\Gamma_{2\alpha}^{0}).\vec{e_{0}}+(\frac{\partial A^{y}}{\partial z}+A^{\alpha}\Gamma_{2\alpha}^{1}).\vec{e_{1}}+(\frac{\partial A^{z}}{\partial z}+A^{\alpha}\Gamma_{2\alpha}^{2}).\vec{e_{2}}
$$
Is this correct?. If not, I would be grateful if someone would give the correct formulae in [longhand please].
On the other hand, If it is correct. Would someone please explain the following.
Some of these terms, e.g. $(\frac{\partial A^{x}}{\partial y}+A^{\alpha}\Gamma_{1\alpha}^{0}).\vec{e_{0}}$   ,are partial derivatives with respect to one direction, but the corresponding basis is in another direction. I would have expect them to be in the same direction,like a gradient of a scalar. For instance assuming we have a 3D orthogonal system, the second terms disappear and if we have two components Ay and Az go to zero, then maybe you should get a scalar-like result.
Looking forward to your comments.
 A: The quantities $\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+A^{\alpha}\,\Gamma_{\mu\alpha}^{\nu}$ are the components of the $1\choose 1$-tensor
$$\tag{1}
(\nabla_{\mu}A^{\nu})(\boldsymbol{\omega}^\mu\otimes \boldsymbol{e}_\nu).
$$
Likewise, $A^\nu$ are the components of the vector $\boldsymbol{A}=A^\mu\boldsymbol{e}_\nu\,.$ If $\varphi$ is a scalar then
$$
\nabla_\mu \varphi=\partial_\mu\phi\,.
$$
These are the components of the covector $\boldsymbol{d}\phi=(\partial_\mu\varphi)\,\boldsymbol{\omega}^\mu\,.$

*

*Your long formula is not correct. If we write out (1) we get
$$\tag{1'}
\left(\partial_\mu A^\nu+
A^0\,\Gamma^\nu_{\mu 0}+
A^1\,\Gamma^\nu_{\mu 1}+
A^2\,\Gamma^\nu_{\mu 2} \right)(\boldsymbol{\omega}^\mu\otimes \boldsymbol{e}_\nu)\,.
$$


*Both, $\nabla_\mu\varphi$ and $\nabla_\mu A^\nu$ contain only the basis covectors $\boldsymbol{\omega}^\mu$ (no other directions), respectively $\boldsymbol{e}^\nu$ (ditto).


*However, $\nabla_\mu A^\nu$ contains all components $A^0,A^1,A^2$ unless the corresponding Christoffel symbol is zero.
