Question about connections and usual derivative. Let $\nabla $ a covariant derivative. What does mean "in the normal coordinate, $\nabla $ is equivalent to the usual derivative". 
I recall that the normal coordinate is coordinate system on a normal neighborhood given by the diffeomorphism $$\exp_p: V\to U$$ 
where $V$ is a subspace of the tangent vector space $T_pM$ (but in fact, is not very important). 
I know what is a covariant derivative, but I don't understand what is a usual derivative.
 A: Assume you are given two vector fields $X = (X_1,\dots,X_n) = X^i e_i$ and $Y = (Y_1,\dots,Y_n) = Y^j e_j$ in $\mathbb{R}^n$. At a point $p \in \mathbb{R}^n$, we can define the directional derivative of $Y$ in the direction of $X$ by mimicking the standard definition of a directional derivative
$$ (\overline{\nabla}_X Y)(p) = \lim_{t \to 0} \frac{Y(p+tX(p)) - Y(p)}{t} = \lim_{t \to 0} \left( \frac{Y_1(p + tX(p)) - Y_1(p)}{t}, \dots, \frac{Y_n(p + tX(p)) - Y_n(p)}{t} \right) = ((\nabla Y_1)(p) \cdot X(p), \dots, (\nabla Y_n)(p) \cdot X(p)) = X^i \frac{Y^j}{\partial x^i}  e_j.$$
On a general manifold, this limit doesn't make sense and one introduces the notion of a connection in order to differentiate vector field with respect to vector fields. You can always choose some coordinate system and take the derivative of the representations of the vector fields in the sense above but this won't be "invariant". 
Now, in coordinates, the covariant derivative is given by
$$ \nabla_X Y = \nabla_{X^i \partial_i} (Y^j \partial_j) = X^i \frac{Y^j}{\partial x^i} \partial_j + X^i Y^j \Gamma^k_{ij} \partial_k.  $$
Here, $(x^1, \dots, x^n)$ are coordinates around $p \in M$, $\partial_i = \frac{\partial}{\partial x^i}$ and $X = X^i \partial_i, Y = Y^j \partial_j$. The first expression is precisely the "usual derivative" $\overline{\nabla}_X Y$ while the second expression is a "correction term" that makes the expression covariant. In a normal coordinate system around $p$, the Christoffel symbols $\Gamma^k_{ij}(p)$ all vanish and so the covariant derivative reduces to the regular derivative (only at $p$!). 
