The $n$th derivative of $f(e^x)$

Because it comes up every so often, I was wondering if there was a nice general expansion for the $n$th derivative of $f(e^x)$.

$\frac{\mathrm d}{\mathrm dx}f(e^x)=e^xf'(e^x)$

$\frac{\mathrm d^2}{\mathrm dx^2}f(e^x)=e^xf'(e^x)+e^{2x}f''(e^x)$

$\frac{\mathrm d^3}{\mathrm dx^3}f(e^x)=e^xf'(e^x)+3e^{2x}f''(e^x)+e^{3x}f'''(e^x)$

$\vdots$

One can see that we end up with the form

$$\frac{\mathrm d^n}{\mathrm dx^n}f(e^x)=\sum_{k=1}^nC(n,k)e^{kx}f^{(k)}(e^x)$$

for some constants $C(n,k)$. I've found that it reduces to solving the following recurrence relation:

$$C(n,k)=\begin{cases}0,&n<k\\1,&n>0\land k=1\\kC(n-1,k)+C(n-1,k-1),&n>0\land k>1\end{cases}$$

From which you can deduce things such as $C(n,2)=2^{n-1}-1$.

Is there a nice general solution?

Solution 1:

Short answer: No, there is no nice solution, at least in terms of elementary functions. But if you consider the Stirling numbers to be nice (which is a good practice), then the solution is very nice.

As mentioned in the comments, $C(n,k)$ are the Stirling numbers of the second kind, ${n \brace k}$. They have the following "closed form:" $$ {n\brace k}=\frac1{k!}\sum_j(-1)^j\binom{k}{j}(k-j)^n. $$ If you like, you can substitute that mess for $C(n,k)$, but I think it is better to leave it as ${n \brace k}$. This is as simple as it gets.