Covariant vs partial derivative Heads up: this is a duplicate from physics S.E. where it is being treated as off topic. I hope its not a problem that I repost here ? 
Question
$\partial_\mu \vec{V} = 0$, does this imply that $\nabla_\mu V^\kappa = 0$ ?
My argument
For any vector $\vec{V}$ we can write that $V=V^\kappa \vec{g_\kappa}$ with $\vec{g_\kappa} = \partial \vec{r}/\partial r^\kappa$ such that:
$$0= \partial_\mu \vec{V} = \partial_\mu(V^\kappa \vec{g}_\kappa) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \partial_\mu(\vec{g_\kappa}) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \Gamma_{\mu\kappa}^\rho \vec{g_\rho} = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\rho \Gamma_{\mu\rho}^\kappa \vec{g_\kappa} = \nabla_\mu (V^\kappa) \vec{g_\kappa}$$
projecting onto the different basis vectors than gives the result that $\nabla_\mu V^\kappa = 0$
Additional question
If this is true, and we consider a translation in space $\vec{x} \rightarrow \vec{x} + \vec{a}$ with $\vec{a}$ constant. Do we than find that $\nabla_\mu a_\nu = 0$?
 A: Not quite, and for two reasons:
1) The quantity $\partial_\mu V $ is not well-defined for $V$ a vector, that is, it transforms in a non-tensorial way under a change of coordinates. 
Since $\partial _\mu V= (\partial _\mu V^\nu) \partial _\nu$ it is enough to check that $\partial_\mu V^\nu$ does not transform tensorially. To see this consider the following example. Let $V$ be the vector field in $\mathbb{R}^2$, $V=\partial _x$ where $(x,y)$ are the usual Cartesian coordinates. With respect to the coordinate basis $\{\partial_x,\partial_y\}$ we have $V^x=1,V^y=0$ hence all the terms $\partial_\mu V^\nu$, $\mu,\nu\in\{x,y\}$ vanish. Consider however the same vector field in polar coordinates,
\begin{equation}
\partial_x=  \cos\theta\  \partial_r -(\sin\theta/r)\ \partial_\theta.
\end{equation}
With respect to the coordinate basis $\{\partial_r,\partial_\theta\}$ e.g. $\partial_\theta V^r=-\sin\theta \neq 0$ hence it cannot hold $\partial_\theta V^r = J_\theta ^\mu J^r _\nu\partial _\mu V^\nu$, where $J^\mu_\theta =\partial_\theta V^\mu$, $J^r_\nu = \partial_\nu V^r$, $\mu,\nu\in\{x,y\}$ are (entries of) the Jacobian of the coordinate transformation and its inverse. 
You can still define and use $\partial_\mu V$ in  a fixed coordinate system but that is  not very useful.
In fact this is the reason for introducing the covariant derivative.
Instead, the partial derivative of a function transforms tensorially as we have, by the chain rule (or the transformation law for 1-forms if you prefer) 
\begin{equation*}
\partial_a f = J^\mu_a \partial_\mu f,
\end{equation*}
so if it vanishes in a coordinate system it vanishes in all. If it is non-zero it transforms in a controlled way, namely by the Jacobian of the change of bases. This is why we want to work with quantities which transform tensorially.
2) The covariant derivative of a vector field is well-defined, that is independent of the coordinate system. Applied to a function (as opposed to a tensor) however, it is simply the ordinary partial derivative, so
\begin{equation*}
\nabla_\mu V^\nu=\partial_\mu V^\nu \neq \nabla_\mu V=\nabla_\mu (V ^\nu \partial_\nu)
=(\partial_\mu V^\nu)\partial_\nu +V^\nu \nabla_\mu \partial_\nu 
=\Big( \partial_\mu V^\rho + V^\nu \Gamma^\rho_{\mu\nu}\Big) \partial_\rho
\end{equation*}
In fact, the extra term in the covariant derivative
\begin{equation}
\Gamma^\rho_{\mu\nu}=  \mathrm{d}x^\rho \Big(\nabla_\mu (\partial_\nu)\Big)
\end{equation}
is the correction required to obtain an expression which transforms in a tensorial way under changes of coordinates.
