TITLE: Are there ways of finding the $n$-th derivative of a function without computing the $(n-1)$-th derivative?
QUESTION [31 upvotes]: Say we have a function $f(x)$ that is infinitely differentiable at some point.
Is it possible to find $f^{(n)}(x)$ without having to find first $f^{(n-1)}(x)$?  If so, does it take less effort than computing preceding derivatives (i.e. $f'(x), f''(x), \cdots, f^{(n-1)}(x)$)?
I often find it very tedious to find multiple derivatives so I was wondering if someone knows the answer to this question.

REPLY [4 votes]: Yes, there are many explicit formula for this.  One useful one is the Cauchy integral formula which allows a simple contour integral to give this info.
But basically any formula that gives you coefficients in the series expansion works.  Simpson multisection, for instance, gives you the series terms of index $m \pmod n$.  You can compose these operators to get a formula for the mph coefficient only, which can be expressed in terms of the function at roots of unity times the variable, using the standard expansion.  Or you can pull off the coefficient directly with a limit:
$$\frac{D^m f(x)}{(1)_m}  =
\lim_{x \to 0} \frac{\frac{1}{n}\left( \sum_{j=0}^{n-1} w_n^{-jm} f\left( w_n^j x\right)\right)}{x^m}$$
Where $w_n$ is an nth root of unity and $(1)_m$ is the pocchhammer.