What is the meaning of $DX_p$ for $X$ a vector field on a manifold? This is taken from Palis, Geometric Theory of Dynamical Systems, p.55:

Here $X$ is a $C^r$ vector field on $M$. What does the notation $DX_p$ mean?
 A: This is often called the intrinsic derivative. (This makes sense, more generally, for the section of any vector bundle at a zero.) It is well-defined at a zero of $X$. Think in local coordinates of $X$ as a map from $\Bbb R^n$ to $\Bbb R^n$, and compute its derivative at $0$ (corresponding to $P$). You can check that you get a well-defined map $T_PM\to T_PM$ precisely because $X(P)=0$.
A: This is a fleshed out version of Ted's answer. Given coordinates $\phi = (x^1 ,\ldots,x^n)$ about $p$ such that $\phi(p) = 0$, write $X = a^i \frac{\partial}{\partial x^i}$ and define $DX_p:T_pM \to T_pM$ by
$$ DX_p\left( \frac{\partial}{\partial x^j}\lvert_p \right) = \frac{\partial \widehat{a^i}}{\partial x^j}(0) \frac{\partial}{\partial x^i}\lvert_p, $$
where $\widehat{a^i} = a^i \circ \phi^{-1}$ is the coordinate representation of $a^i$. To see that this is independent of the chosen coordinate system, we must check that the change of basis and $DX_p$ operations commute.
Let $(y_1,\ldots,y_n)$ be another coordinate system, with associated coordinate functions $\widetilde{{a^i}}$. On one hand, we have
$$\begin{align}
DX_p\left(\frac{\partial}{\partial y^j}\lvert_p\right) &= DX_p\left( \frac{\partial x^k}{\partial y^j}(0) \frac{\partial}{\partial x^k}\lvert_p\right) \\
&= \frac{\partial x^k}{\partial y^j}(0) \frac{\partial \widehat{a^i}}{\partial x^k}(0) \frac{\partial}{\partial x^i}\lvert_p.
\end{align} $$
On the other hand, since $X = a^i \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}$, we compute
$$\begin{align}
DX_p\left( \frac{\partial}{\partial y^j}\lvert_p\right) &= \left( \frac{\partial}{\partial y^j} \left( \widetilde{a^i} \frac{\partial y^k}{\partial x^i}\right)(0) \right)\frac{\partial}{\partial y^k}\lvert_p \\
&= \left( \frac{\partial \widetilde{a^i}}{\partial y^j}(0) \frac{\partial y^k}{\partial x^i}(0) \right) \frac{\partial}{\partial y^k}\lvert_p \qquad \text{by the product rule and since } a^i(p) = 0 \\
&= \frac{\partial \widetilde{a^i}}{\partial y^j}(0) \frac{\partial}{\partial x_i}\lvert_p \\
&= \frac{\partial \widehat{a^i}}{\partial x^k}(0) \frac{\partial x^k}{\partial y^j}(0) \frac{\partial}{\partial x^i}\lvert_p,
\end{align}$$
which is what we got above.
