This is a follow up question to one I've asked previously.
How can I take the derivative of this exponential mapping / what is the solution?
Let
$$A (x) := \begin{bmatrix}0&0\\4x&1\end{bmatrix}$$
and let $x$ be a scalar. Then what is the solution to $\frac{\partial}{\partial x} ( e^{A(x)})$?
If there is an analytical solution, that is preferable, but numerical approximation are will work too. Thank you.
On
Since $A=A^2$, we can use the formula for idempotent matrices from the Matrix Cookbook, which is valid for any analytic function of $A$ $$\eqalign{ f(A) &= A\,f(1) + (I-A)\,f(0) \cr &= e\,A + (I-A) \cr\cr df &= e\,dA - dA \cr &= (e-1)\,dA\cr\cr \frac{\partial f}{\partial x} &= (e-1)\,\frac{\partial A}{\partial x} \cr }$$
$A^n=A$ for all $n\geq 1$ is easy to check.
Therefore, $$e^A=\sum_{n=0}^\infty \frac{1}{n!} A^n=I+A \sum_{n=1}^\infty \frac{1}{n!}.$$ Also, $$\frac{\partial}{\partial x}A = \begin{bmatrix}0&0\\4 &0\end{bmatrix},$$ which gives $$\frac{\partial}{\partial x} e^A=\frac{\partial}{\partial x}A \sum_{n=1}^\infty \frac{1}{n!}=(e-1)\begin{bmatrix}0&0\\4 &0\end{bmatrix}$$