Let me define $G:=\left\{a(t)= \begin{pmatrix} a &b \\ 0 & a^{-1} \end{pmatrix}, a>0, b\in\Bbb{R}\right\}$ and view it as a submanifold of $M_2(\Bbb{R})$. Then define the map $\phi:\Bbb{R}\times \Bbb{R}\rightarrow G, (t,x)\mapsto \begin{pmatrix} e^t &xe^t \\ 0 & e^{-t} \end{pmatrix}$. I would like to compute $\frac{\partial \phi}{\partial t}(\phi(t,x))$
I know that by definition $\frac{\partial \phi}{\partial t}(\phi(t,x)):=D(\phi)_{\phi^{-1}(\phi(t,x))}(e_1)=D(\phi)_{(t,x)}(e_1)$ which is a map from $\Bbb{R}^2\rightarrow T_{(t,x)}G$. Now my problem is that I don't know how to compute $D(\phi)_{(t,x)}(e_1)$, since I cannot compute the Jacobian matrix and multiply by $e_1$. Can maybe somone help me to understand how one can compute this?
The Lie algebra $\mathfrak{g}$ of $G$ is the tangent space at the identity of $G$ which consists of matrices of the form
$$\begin{pmatrix} a & b\\ 0 & -a \end{pmatrix} $$
This space has a basis
$$\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} $$
To find the tangent space at $\begin{pmatrix} a & b\\ 0 & a^{-1} \end{pmatrix}$ you can just translate this basis, so a basis at $\begin{pmatrix} a & b\\ 0 & a^{-1} \end{pmatrix}$ is
$$\begin{pmatrix} a & -b\\ 0 & -a^{-1} \end{pmatrix},\begin{pmatrix} 0 & a\\ 0 & 0 \end{pmatrix}$$
In your case the corresponding basis at $\phi(t,x)$ is $\begin{pmatrix} e^t & -xe^t\\ 0 & -e^{-t} \end{pmatrix}$,$\begin{pmatrix} 0 & e^t\\ 0 & 0 \end{pmatrix}$
The $t$ and $x$ partial derivatives of your map are given by
$$\begin{pmatrix} e^t & xe^t\\ 0 & -e^{-t} \end{pmatrix},\begin{pmatrix} 0 & e^t\\ 0 & 0 \end{pmatrix}$$.
The coordinates of these partial derivatives in the above basis are $(1,1+x)$ and $(0,1)$
Putting this together gives that with respect to these bases the Jacobian at $(t,x)$ is
$$\begin{pmatrix} 1 & 0\\ 1+x & 1 \end{pmatrix}$$