Question: This is taken from Palis, Geometric Theory of Dynamical Systems, p.55:

[![enter image description here](https://i.stack.imgur.com/u2uEn.png)](https://i.stack.imgur.com/u2uEn.png)

Here $X$ is a $C^r$ vector field on $M$. What does the notation $DX\_p$ mean?

Answer: 
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 $X(P)=0$.
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.