Change of coordinates between charts Let $M$ be a differentiable manifold, $(U, \phi)$ and $(V,\psi)$ two coordinate charts and $p$ a point of $M$. Let $\{ \frac{\partial}{\partial \phi_{1}} (p), \ldots, \frac{\partial}{\partial \phi_{n}} (p) \}$ and $\{ \frac{\partial}{\partial \psi_{1}}(p), \ldots, \frac{\partial}{\partial \psi_{n}}(p) \}$ two basis of $T_{p} M$
I want to find the change of basis matrix between the two charts.
My try: 
$$\frac{\partial}{\partial \phi_{i}} (p) =  \sum_{k=1}^{n} \frac{\partial}{\partial (\psi \circ \phi^{-1})_{k}} \frac{\partial (\psi \circ \phi^{-1})_{k}}{\partial \phi_{i}} (p)$$
Well i think is this, but now how do i put this into a matrix?
Thanks in advance
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$Let $V$ be an $n$-dimensional vector space with bases $B = (\Basis_{j})_{j=1}^{n}$ and $B' = (\Basis_{i}')_{i=1}^{n}$. If
$$
\Basis_{j} = \sum_{i=1}^{n} A_{ij} \Basis_{i}'\quad\text{for all $j$,}
$$
the change of basis matrix from $B$ to $B'$ is $A = [A_{ij}]$.
In your situation, assume $\phi$ and $\psi$ map $p$ to the origin, and let $(x_{1}, \dots, x_{n})$ and $(y_{1}, \dots, y_{n})$ denote the respective coordinate representations, so that $(y_{1}, \dots, y_{n}) = \psi \circ \phi^{-1}(x_{1}, \dots, x_{n})$.
In classical notation you get the change of basis matrix for bases of $\Reals^{n} = T_{0} \Reals^{n}$:
$$
\Basis_{j} = \frac{\dd}{\dd x_{j}},\quad
%
\Basis_{i}' = \frac{\dd}{\dd y_{i}},\quad
%
A_{ij} = \frac{\dd y_{i}}{\dd x_{j}};
$$
the chain rule reads
$$
\frac{\dd}{\dd x_{j}}
  = \sum_{i=1}^{n} \frac{\dd y_{i}}{\dd x_{j}}\, \frac{\dd}{\dd y_{i}}.
\tag{1}
$$
To express this as a change of basis in $T_{p}M$, write the images of the standard basis as
$$
\frac{\dd}{\dd \phi}
  = (\phi^{-1})_{*} \frac{\dd}{\dd x},\qquad
\frac{\dd}{\dd \psi}
  = (\psi^{-1})_{*} \frac{\dd}{\dd x},
$$
and express the chain rule as
$$
\frac{\dd}{\dd \phi}
  = (\phi^{-1} \circ \psi)_{*}\; (\psi^{-1})_{*}\frac{\dd}{\dd x}
  = (\phi^{-1} \circ \psi)_{*} \frac{\dd}{\dd \psi}.
$$
In terms of individual basis elements,
$$
\frac{\dd}{\dd \phi_{j}}
  = \sum_{i=1}^{n} \frac{\dd (\phi^{-1} \circ \psi)_{i}}{\dd \phi_{j}}\, \frac{\dd}{\dd \psi_{i}}.
\tag{2}
$$
(The last is not quite what you have; it's possible I've misconstrued your notation. The important thing is, of course, to express elements of one basis of $T_{p}M$ as linear combinations of elements of another basis.)
