Having a problem calculating covariant derivative So if we have a curve $\gamma(t): I \rightarrow M$ and riemannian connection $\triangledown$, i'm trying to calculate $\triangledown_{\gamma'} \gamma'$
Where $\gamma$' is the velocity vector of $\gamma$. Write $\gamma' = v^i \delta_i$, using Einstein notation. Then we have:
$\triangledown_{v^i \delta_i} \gamma'$
$=v^i\triangledown_{\delta_i} \gamma'$ since $\triangledown$ is linear over $C^{\infty}(M)$ in it's lower variable
$=v^i\triangledown_{\delta_i} v^j \delta_j$ 
$=v^i(\delta_i v^j \delta_j+ v^j \triangledown_{\delta_i} \delta_j$
$=v^i\delta_i v^j \delta_j+ v^iv^j \triangledown_{\delta_i} \delta_j$
$=(v^i\delta_i v^j \delta_j+ v^iv^j \Gamma_{i,j}^k) \delta_k$
This is supposed to equal
$=(\gamma''^k+ v^iv^j \Gamma_{i,j}^k) \delta_k$
where $\gamma''^k$ are the component functions of the acceleration of $\gamma$.
Can somebody please find where I went wrong or how to finish? Thanks!
 A: What's misleading here is that you have $v^i(t)$ and $\partial_i(\alpha(t))$, so you're losing track of $t$ derivatives and the chain rule. (For some reason, you're writing $\delta$ instead of $\partial$.)
I don't like the notation, but they're writing $\gamma''{}^k$ to mean $(v^k)'$. For any function $f$ defined along $\alpha(t)$, we pretend that it is defined on a neighborhood of the curve so that $\partial_if$ makes sense; then the chain rule tells us that
$f' = (f\circ\gamma)'(t) = (\partial_i f)v^i$, since $\gamma'(t) = \sum v^i\partial_i$. In your case, this gives
$$(v^k)' = v^i\partial_i v^k,$$
which is the term you were trying to get.
A: I suppose you are using $\delta$ to keep track of the abstract basis vectors. 
In this case, the notation commonly used is $\nabla_{\delta_i} = \nabla_{i}$.
The definition of connection is such that $\Gamma_{ij}^k \delta_k = \nabla_i \delta_j$, so 
$$
\nabla_{\gamma'} \gamma' = v^a \nabla_a (v^b \delta_b) = (v^a \partial_a v^b) \delta_b +  v^b v^a \nabla_a  \delta_b 
$$
Now, use the definition of connection in the last term and obtain the final expression:
$$
\nabla_{\gamma'} \gamma' = (v^a \partial_a v^b) \delta_b + \Gamma^k_{ab} v^b v^a   \delta_k  =(v^a \partial_a v^k  +  \Gamma^k_{ab} v^a v^b )\delta_k
$$
