How to compute the $n_{th}$ derivative of a composition: ${\left( {f \circ g} \right)^{(n)}}=?$

Question

I know that there is a general formula to compute the $n_{th}$ derivative of a product which is as follows

$${\left( {fg} \right)^{(n)}} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n\\ k \end{array}} \right){f^{(n - k)}}{g^{(k)}}}$$

I am wondering that if there exists such a general formula for the $n_{th}$ derivative of a composition. Specifically, how to find a general formula for the following:

$${\left( {f \circ g} \right)^{(n)}}=?$$

Well, there's Faá di Bruno's formula. It ain't pretty. – Milo Brandt Oct 29 '15 at 16:55 — Milo Brandt, Oct 29 '15 at 16:55
It is really ugly! :D – Hosein Rahnama Oct 29 '15 at 16:58 — Hosein Rahnama, Oct 29 '15 at 16:58
it sounds like there's a pattern there. – Hedwig Oct 29 '15 at 17:02 — Hedwig, Oct 29 '15 at 17:02

2'5 9'2 · Answer 1 · 2015-12-10T22:06:22.513

A nicer way to write Faa di Bruno's formula (even though there is no change with execution of it) starts by switching to divided derivatives: $$f^{[n]}(x)=\frac{1}{n!}f^{(n)}(x)$$ Then $$\left(f\circ g\right)^{[n]}(x)=\sum_{P\in\mathcal{P}_n}f^{[\left\lvert P\right\rvert]}(g(x))g^{[P]}(x)$$ where $\mathcal{P}_n$ is the ordered partitions of $n$, and for one partition $P={}$"$n_1+n_2+\cdots+n_k$", $\lvert P\rvert=k$, and $g^{[P]}(x)$ is the product $g^{[n_1]}(x)g^{[n_2]}(x)\cdots g^{[n_k]}(x)$.

Writing it even more nicely as an equation of functions rather than as an equation of outputs: $$\left(f\circ g\right)^{[n]}=\sum_{P\in\mathcal{P}_n}\left(f^{[\left\lvert P\right\rvert]}\circ g\right)\cdot g^{[P]}$$

For example, for $n=4$, the ordered partitions are $$1+1+1+1,\quad 1+1+2,\quad 1+2+1,\quad 2+1+1,\quad 1+3,\quad 3+1,\quad 2+2,\quad 4$$

So $$\left(f\circ g\right)^{[4]}=\left(f^{[4]}\circ g\right)\left(g^{[1]}\right)^4+3\left(f^{[3]}\circ g\right)\left(g^{[1]}\right)^2g^{[2]}+2\left(f^{[2]}\circ g\right)g^{[1]}g^{[3]}+\left(f^{[2]}\circ g\right)\left(g^{[2]}\right)^2+\left(f^{[1]}\circ g\right)g^{[4]}$$

Then if you want to mutliply through by $4!$, you have $$\left(f\circ g\right)^{(4)}=\left(f^{(4)}\circ g\right)\left(g^{(1)}\right)^4+6\left(f^{(3)}\circ g\right)\left(g^{(1)}\right)^2g^{(2)}+4\left(f^{(2)}\circ g\right)g^{(1)}g^{(3)}+3\left(f^{(2)}\circ g\right)\left(g^{(2)}\right)^2+\left(f^{(1)}\circ g\right)g^{(4)}$$

Thanks for further clarification. :) – Hosein Rahnama Dec 10 '15 at 21:59 — Hosein Rahnama, Dec 10 '15 at 21:59

score 1 · Accepted Answer · edited Dec 10 '15 at 21:26

1

Faà di Bruno's formula may interest you:

$$ {\left( {f \circ g} \right)^{(n)}}=\sum_{m_1+2m_2+\dots+nm_n=n} \frac{n!}{\prod_{i=1}^n m_i! (i!)^{m_i}} \left(\prod_{j=1}^n(g^{(i)})^{m_i} \right){\left(f^{(m_1+\ldots+m_n)}\circ g\right)} $$

edited Dec 10 '15 at 21:26

Hosein Rahnama

14,882

answered Oct 29 '15 at 17:02

Olivier Oloa

120,989

1

Yeah! It was mentioned in the comment above by @Milo Brandt. Thanks for writing it down. :) – Hosein Rahnama Oct 29 '15 at 17:06
I found that you are an expert in solving integrals. Will you please take a look at this one. – Hosein Rahnama Oct 29 '15 at 17:14
@H.R. You did a good job there! To simplify it more, one needs to spend a great deal of time working on it. :) – Olivier Oloa Oct 29 '15 at 17:58
@Oliver Oloa: Thanks for your comment! :) – Hosein Rahnama Oct 29 '15 at 18:09
Holy cow, that formula is horrible! – YoTengoUnLCD Dec 10 '15 at 22:04

How to compute the $n_{th}$ derivative of a composition: ${\left( {f \circ g} \right)^{(n)}}=?$

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