I am trying to derive an expression for the derivative of an exponential. Here is the task phrased as a problem
For the function $g(x) = \exp(f(x))$ where $f(x)$ is real valued and continuously differentiable for all derivatives, what is a general expression for the $N$th derivative of $g(x)$ in terms of $f(x)$?
Also, assuming $f(x)|_{x=0}$, what is an expression for $g^{(N)}(x)|_{x=0}$?
I think this should be a standard calculus problem, but I haven't found a clean solution. Any help would be much appreciated.
First approach: Using the principle of induction, one can prove that the $k$th derivative of $g(x)$ satisfies
\begin{equation} g^{(k)}(x) = H_{k}(x) \exp(f(x)), \qquad \qquad (1) \end{equation}
where $H_k(x)$ is defined recursively through
\begin{equation} H_{k+1}(x) = H'_{k}(x) + f'(x) H_{k}(x), \qquad \qquad (2) \end{equation} for $k\geq 0$ and where $H_0(x) = 1$.
Attempting a generating function approach, we can define $F(x, y) \equiv \sum_{\ell=0}^{\infty} y^{\ell} H_{\ell}(x)$. (2) can then be written as the PDE
\begin{equation} \big(1- y f'(x)\big)F(x, y)- y\frac{\partial}{\partial x} F(x, y) = 1, \end{equation}
which is not really helpful since the boundary conditions are not fully specified. That is, although we have $F(x, 0) = H_0(x) =1$ computing $F(0, y)$ requires that we know $H_k(0)$ which is what we are trying to find.
Second approach: Iterating (2), one can show that for $k\geq0$, $H_{k+1}(x)$ satisfies
\begin{equation} H_{k+1}(x) = \sum_{\ell=0}^{k} \left(\frac{d}{dx}\right)^{\ell}\left(f'(x) H_{k-\ell}(x)\right), \end{equation}
for but it is not clear what this approach buys us.
It seems you will get a sum of products of derivatives of $f$ with the number of total derivatives in each product is $n$, but I don't know if there is a general formula for the coefficients ...
So something like $$ g^{(n)}(x) = g(x) \sum_{\substack{(k_1,\dots,k_n)\in\mathbb{N}^n\\k_1+\dots+k_n = n}} c_{k_1,\dots,k_n} \prod_{j=1}^n f^{[k_j]}(x) $$ where $f^{[0]} = 1$ and $f^{[k]} = f^{(k)}$ if $k>0$. You can verify this formula by your recursive formula for example.