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?