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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.

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Interesting question. I assume that your function is built out of elementary functions using products, sums, and compositions? Here's a silly observation: I'd rather take a 50th derivative of a degree 49 polynomial than a 13th. A slightly less silly one: I'd rather take a 49th derivative of a degree 49 polynomial than a 13th. – user29743 May 3 '12 at 1:34
If one has a good sight, one may conjecture a formula after some derivations and prove it by induction on $n$. Else, I can't think of any other way right now. – Pedro Tamaroff May 3 '12 at 1:43
Maybe you could estimate it? For $x^n$, the nth derivative is $n!x$, and if there is a term $x^m$ where $m < n - 2$, that term will be 0. – Joel Cornett May 3 '12 at 1:44
up vote 17 down vote accepted

It really depends on the function. Some functions admit compact general expressions for arbitrary order derivatives:

$$\begin{align*} \frac{\mathrm d^k}{\mathrm dx^k}x^n&=k!\binom{n}{k}x^{n-k}\\ \frac{\mathrm d^k}{\mathrm dx^k}\sin\,x&=\sin\left(x+\frac{k\pi}{2}\right)\\ \frac{\mathrm d^k}{\mathrm dx^k}\cos\,x&=\cos\left(x+\frac{k\pi}{2}\right)\\ \end{align*}$$

and as already mentioned, if your function satisfies a nice differential equation, you can use that differential equation to derive a general expression for your derivatives.

(As a bonus, the formula for the derivative of the power function can be generalized to complex $k$, barring exceptional values of $n$ and $k$; this is the realm of the fractional calculus. The formulae for sine and cosine are no longer as simple in the general complex case, though.)

In general, however, there isn't always a nice expression. This page for instance displays a number of representations for the derivative of the tangent function. None of them look particularly nice. Sometimes, functions won't even allow for an explicit expression for derivatives, as in the case of the Lambert function. (Note that the last formula in that page requires an auxiliary recursive definition for the polynomials that turn up in the differentiation.)

Relatedly: formulae like the Leibniz rule and the Faà di Bruno formula are helpful when determining general expressions for derivatives of more complicated functions. There are also a number of formulae listed here.

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The following references for "formulae like the Leibniz rule" that J.M. mentioned in his last paragraph may be of interest. In what follows, AMM is American Mathematical Monthly and MG is Mathematical Gazette. [A] Closed form expressions for the $n$th derivative of the quotient of two functions [AMM 55 (1948), p. 491; MG 64 (1980), p. 52]. [B] Closed form expressions for the $n$th derivative of the inverse of a function [AMM 69 (1962), pp. 904-907]. (continued) – Dave L. Renfro May 3 '12 at 15:08
(continued) [C] Closed form expressions for the $n$th derivative of a composition of two functions [AMM 50 (1943), pp. 9-12 & 356; AMM 68 (1961), pp. 69-70; AMM 69 (1962), pp. 912-914; MG 54 (1970), p. 389; MG 55 (1971), pp. 395-397; AMM 87 (1980), pp. 805-809; Mathematical Spectrum 30 #3 (1997/98), pp. 54-57; AMM 109 (2002), pp. 217-234]. (continued) – Dave L. Renfro May 3 '12 at 15:08
(continued) [D] Closed form expressions for the $n$th iterate of integration by parts [Philosophical Magazine (3) 33 (1848), pp. 335-337; AMM 45 (1938), pp. 36-38; MG 35 (1951), pp. 122-123; AMM 67 (1960), p. 372; Nieuw Archief voor Wiskunde (4) 7 (1989), pp. 95-105]. – Dave L. Renfro May 3 '12 at 15:09
Thanks @Dave! I'll edit my answer later to incorporate your references. – J. M. May 3 '12 at 15:29

Yes, for the broad-class of functions admitting series expansions such that the coefficients satisfy a constant-coefficient difference equation (recurrence). In this case the coefficients (hence derivatives) can be computed quickly in polynomial time by converting to system form and repeatedly squaring the shift-matrix. This can be comprehended by examining the special case for Fibonacci numbers - see this prior question. Once you understand this 2-dimensional case it should be clear how it generalizes to higher-order constant coefficient recurrences.

Look up D-finite series and holonomic functions to learn more.

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Another way of putting it is that if $f$ satisfies a differential equation with constant coefficients, then you can reduce higher-order derivatives to linear combinations of the first few. Canonical example: $sin''=-sin$. – lhf May 3 '12 at 1:48
More generally: if you know that your function satisfies a differential equation of order $n$, it is always possible to write the $k$-th derivative ($k \geq n$) in terms of the functions and the first $(n-1)$-th derivatives. – J. M. May 3 '12 at 4:19

It is possible to write a single limit for the second derivative: $$f''(x) = \lim_{h \to 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$ and probably much more complicated ones for higher derivatives, but I'm not sure that's what you're looking for.

Source: Wikipedia.

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I assume you are using finite difference coefficients? I would think that the limits would be hard to evaluate with this method. I think this method is better for a high-accuracy numerical approximation of the nth derivative rather then getting the function of the nth derivative. – Argon May 3 '12 at 1:40
It's definitely closely related to finite difference coefficients, though it's a limit, not an approximation. – lhf May 3 '12 at 1:44
I see, it is a bit different because it has $h$ in the numerator. I think it would be a bit painful to evaluate this limit, especially in higher derivatives! – Argon May 3 '12 at 1:49
The generalization can be stated using the forward difference operator $\Delta_h$, $$\frac{d^n}{dx^n}=\lim_{h\to0}\frac{1}{h^n}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}\Delta_h^k.$$ – anon May 3 '12 at 8:21
In addition to the formula @anon mentioned, there are versions of the formula that use central or backward difference operators instead of forward ones. Books on the difference calculus will list such formulae. – J. M. May 3 '12 at 8:43

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.

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I have been working on the problem of finding formulas for the nth derivative and the nth anti derivative of elementary and special functions for many years. I have got many results which have been already published. I am referring you to my website, where you can find an answer to your question. Here is the link to my website here.

For example, In this paper, a complete solution the problem of finding the nth derivative and the nth anti derivative of the power exponential class has been given. Just read section 6 in the paper which answers your question for this class of functions.

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