Solution
= Solution
Write the ordered <basis> as $\mathcal B=(1,t,e^t,te^t)$. <Differentiation> gives
$$
F(1)=0,\qquad F(t)=1,\qquad F(e^t)=e^t,\qquad F(te^t)=e^t+te^t.
$$
The corresponding coordinate <vectors> form the columns of the representing <matrix>:
$$
\boxed{[F]_{\mathcal B}=M=
\begin{pmatrix}
0&1&0&0\\
0&0&0&0\\
0&0&1&1\\
0&0&0&1
\end{pmatrix}.}
$$
This is an example of <differentiation on an exponential-polynomial space>.