I'm interested in tidying up the calculation of arbitrarily high order derivatives of a function containing an exponential. Although any function can have it's derivative expressed as
$$\frac{\mathrm{d}y}{\mathrm{d}x} = y' = ay $$
for the functions I'm working with $a$ is simpler than the function as a whole. If we keep differentiating:
$$y'' = ay' + ya' = y (a^{2} + a') $$
$$y''' = y (a^{3} + 3aa' + a'') $$
We always end up with the form $y^{(n)} = y f(a)$. We could also define the coefficient on $y$ recursively:
$$ y^{(n)} = b_{n}y$$ $$b_{0} = 1$$ $$b_{1} = a$$ $$b_{n+1} = b_{n} a + b'_{n}$$
Does a closed expression exist for $b_{n} = f(a)$?
I've evaluated the $b_{n}$ by hand up to $b_{5}$ by which point it starts getting messy and I can't see an obvious pattern.
Not an answer, but here is a possible way to attack this. From Leibnitz formula on $y' = ay$ we get
$$y^{(n+1)} = \sum_{k=0}^n{n\choose k} y^{(k)}a^{(n-k)}$$
giving us the following recursion formula
$$b_{n+1} = \sum_{k=0}^n{n\choose k} b_ka^{(n-k)}$$
for $b_n$ which avoids the derivative of $b_n$ in the recursion.
We can also guess (and then confirm using the formula above) a few of the terms in $b_n$:
$$b_n = a^n + {n\choose 2} a^{n-2}a' + \ldots+ a^{(n)}$$
Getting a general formula for the $\ldots$ in terms of all the possible $a^{(k)}$ combinations seems like a hard problem since $b_n$ will contain all possible terms on the form $a^{k_0} [a^{(k_1)}]^{r_1}[a^{(k_2)}]^{r_2}\ldots [a^{(k_i)}]^{r_i}$ where $k_0 + \sum (k_j+1)r_j = n$.