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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?
calculus sequences-and-series combinatorics derivatives recurrence-relations
Sri-Amirthan Theivendran
Simply Beautiful ArtSimply Beautiful Art
$\begingroup$ Your $C(n,k)$ look a lot like the Stirling numbers of the second kind $\endgroup$ – Sri-Amirthan Theivendran Mar 15 '19 at 15:50
$\begingroup$ @FoobazJohn look at me getting rusty $\endgroup$ – Simply Beautiful Art Mar 15 '19 at 15:51
$\begingroup$ If you calculate the fourth derivative as well and enter the coefficients 1, 1, 1, 1, 3, 1, 1, 7, 6, 1 at oeis.org then you'll find oeis.org/A008277. $\endgroup$ – Martin R Mar 15 '19 at 16:01
$\begingroup$ Ah, bad to use $C(n,k)$, since that is sometimes used for $\binom n k$. I almost missed the definition of $C.$ $\endgroup$ – Thomas Andrews Mar 15 '19 at 16:02
$\begingroup$ You have seen Faa Di Bruno's rule? $\endgroup$ – Teresa Lisbon Mar 15 '19 at 16:07
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.
Mike EarnestMike Earnest
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