Property of a covariant derivative of a vector field along a curve I am studying the covariant derivative of a vector field $X$ along a curve $\gamma$ ( $\nabla_{\dot \gamma}X:=\frac{\nabla X}{dt}$) and I have read the following property:
$$\nabla_{\dot \gamma}X=\Big(\frac{d \xi^k}{dt}+\Gamma_{ij}^k\frac{dx^i}{dt}\xi^j\Big)\frac{\partial}{\partial_k}$$
In order to prove this above I have thought that if $X=\xi^j \frac{\partial}{\partial x^j}$ and $\dot \gamma= \frac{dx^i}{dt} \frac{\partial}{\partial x^i}$, we have that:
since $\nabla X=\Big(\frac{\partial \xi^k}{\partial x^j}+\Gamma_{ji}^k\xi^j\Big)\frac{\partial}{\partial_k}$ we obtain:
$$\nabla_{\dot \gamma}X=\nabla_{\frac{dx^i}{dt}\frac{\partial}{\partial x^i}}(\xi^j \frac{\partial}{\partial x^j})=\frac{dx^i}{dt}\Big(\nabla_{\frac{\partial}{\partial x^k}}(\xi^j \frac{\partial}{\partial x^j})\Big)=\frac{dx^i}{dt}\Big(\xi^j\nabla_{\frac{\partial}{\partial x^i}}( \frac{\partial}{\partial x^j})\Big)+\frac{dx^i}{dt}\Big(\xi^j\frac{\partial}{\partial x^i}\Big)\frac{\partial}{\partial x^j}=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}+\color{red}{\frac{dx^i}{dt}(\frac{\partial}{\partial x^i})}\frac{\partial}{\partial x^j}\Big)=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}+\color{red}{\frac{d\xi^j}{dt}}\frac{\partial}{\partial x^j}\Big)=\Big(\frac{dx^i}{dt}\xi^j\Gamma_{ij}^k+{\frac{d\xi^k}{dt}}\Big)\frac{\partial}{\partial x^k}$$
where the last equality is obtained considering the index $j$ in the second term as $k$.
$\textbf{My question:}$ do you think that the passages that I have done are correct? In particular it is valid the equality that involves the red quantities (I am not sure that is mathematically rigorous)?
 A: This is a pretty straightforward application of the properties of the covariant derivative. Remember that the Christoffel symbols are defined as a measure of how the covariant derivative acts on the basis vectors:
$$\nabla_{j}\mathrm{e}_i=\Gamma^k_{ij}\mathrm{e}_k$$
And also that the covariant derivative is linear:
$$\nabla_\mathrm{u}\mathrm{v}=\nabla_{u^j\mathrm{e}_j}\mathrm{v}=u^j\nabla_{j}\mathrm{v}=u^j\nabla_j(v^i\mathrm{e}_i)$$
Now we use the other important property of the covariant derivative, the product rule
$$u^j\nabla_j(v^i\mathrm{e}_i)=u^j(\mathrm{e}_i\nabla_j(v^i)+v^i\nabla_j(\mathrm{e}_i))$$
Since covariant differentiation on a scalar is the same as partial differentiation, we get
$$=u^j(\mathrm{e}_i\partial_jv^i+v^i\Gamma^k_{ij}\mathrm{e}_k)$$
$$=\left(u^j\partial_jv^k+v^iu^j\Gamma^k_{ij}\right)\mathrm{e}_k$$
Replacing $\mathrm{u}$ with $\dot{\gamma}$ with components $\{\dot{x}^i\}$ and replacing $\mathrm{v}$ with $\mathrm{X}$ with components $\{\xi^j\}$, using $\frac{\partial}{\partial x^k}$ instead of the more linear algebra type notation $\mathrm{e}_k$ and renaming some indices will give your desired expression.
ADDITION:
It is indeed correct that
$$\frac{\mathrm{d}\xi^i}{\mathrm{d}t}=\frac{\partial \xi^i}{\partial x^j}\frac{\mathrm{d}x^j}{\mathrm{d}t}$$
See exact differential.
A: Here is the correct computation (which you can find in any number of texts).
\begin{align*}
\nabla_{\dot\gamma}X &= \nabla_{\dot\gamma}\big(\xi^j\frac{\partial}{\partial x^j}\big) \\
&= \big(\nabla_{\dot\gamma}\xi^j\big)\frac{\partial}{\partial x^j} + \xi^j\nabla_{\dot\gamma}\frac{\partial}{\partial x^j} \\
&= \frac{d\xi^j}{dt}\frac{\partial}{\partial x^j} + \xi^j\nabla_{\frac{dx^i}{dt}\frac{\partial}{\partial x^i}} \frac{\partial}{\partial x^j} \\
&= \frac{dx^j}{dt}\frac{\partial}{\partial x^j} + \xi^j\frac{dx^i}{dt} \Gamma^k_{ij}\frac{\partial}{\partial x^k}.
\end{align*}
Can you finish now?
Note that because a priori $X$ is a vector field only along the curve (this can be fixed, but it's not worth the trouble), $\xi^j$ are functions only of $t$, not of $x^1,\dots,x^n$. So this is why we cannot change $\nabla_{\dot\gamma}$ immediately. Indeed, $d\xi^j/dt$ is typical shorthand for $d(\xi^j\circ\gamma)/dt$.
