Why does this matrix give the derivative of a function?
If $$ A = \begin{bmatrix} a & 1 \\ 0 & a \end{bmatrix} $$ then by induction you can prove that $$ A^n = \begin{bmatrix} a^n & n a^{n-1} \\ 0 & a^n \end{bmatrix} \tag 1 $$ for $n \ge 1 $. If $f$ can be developed into a power series $$ f(z) = \sum_{n=0}^\infty c_n z^n $$ then $$ f'(z) = \sum_{n=1}^\infty n c_n z^{n-1} $$ and it follows that $$ f(A) = \sum_{n=0}^\infty c_n A^n = I + \sum_{n=1}^\infty c_n \begin{bmatrix} a^n & n a^{n-1} \\ 0 & a^n \end{bmatrix} = \begin{bmatrix} f(a) & f'(a) \\ 0 & f(a) \end{bmatrix} \tag 2 $$ From $(1)$ and $$ A^{-1} = \begin{bmatrix} a^{-1} & -a^{-2} \\ 0 & a^{-1} \end{bmatrix} $$ one gets $$ A^{-n} = \begin{bmatrix} a^{-1} & -a^{-2} \\ 0 & a^{-1} \end{bmatrix}^n = (-a^{-2})^{n} \begin{bmatrix} -a & 1 \\ 0 & -a \end{bmatrix}^n \\ = (-1)^n a^{-2n} \begin{bmatrix} (-a)^n & n (-a)^{n-1} \\ 0 & (-a)^n \end{bmatrix} = \begin{bmatrix} a^{-n} & -n a^{-n-1} \\ 0 & a^{-n} \end{bmatrix} $$ which means that $(1)$ holds for negative exponents as well. As a consequence, $(2)$ can be generalized to functions admitting a Laurent series representation: $$ f(z) = \sum_{n=-\infty}^\infty c_n z^n $$
It's a general statement if $J_{k}$ is a Jordan block and $f$ a function matrix then \begin{equation} f(J)=\left(\begin{array}{ccccc} f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} & \frac{f''(\lambda_{0})}{2!} & \ldots & \frac{f^{(n-1)}(\lambda_{0})}{(n-1)!}\\ 0 & f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} & & \vdots\\ 0 & 0 & f(\lambda_{0}) & \ddots & \frac{f''(\lambda_{0})}{2!}\\ \vdots & \vdots & \vdots & \ddots & \frac{f'(\lambda_{0})}{1!}\\ 0 & 0 & 0 & \ldots & f(\lambda_{0}) \end{array}\right) \end{equation} where \begin{equation} J=\left(\begin{array}{ccccc} \lambda_{0} & 1 & 0 & 0\\ 0 & \lambda_{0} & 1& 0\\ 0 & 0 & \ddots & 1\\ 0 & 0 & 0 & \lambda_{0} \end{array}\right) \end{equation} This statement can be demonstrated in various ways (none of them short), but it's a quite known formula. I think you can find it in various book, maybe in Horn, Johnson, Matrix Analisys.
$$ A =a \mathbb{I}+M $$ where $$M = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$
most relevantly, $M$ is an upper triangular matrix satisfying $ M^n=0;n>1$
$$ A^n=\left(a \mathbb{I}+M \right)^n $$ Since all these matrices ( $ \mathbb{I}$ and $M$ ) commute we can write out the binomial formula ... $$A^n = \sum_{i=0}^n\binom ni a^i \mathbb{I}^iM^{n-i}$$ The only non-zero terms will be $i=n$ and $i=n-1$ $$A^n= a^n \mathbb{I}+na^{n-1}M $$