While differentation the function $f(x) = (1-e^x)^{5}$, I realize one can use the Binomial Theorem to find formulas for nth derivative: The idea is as follows. Let $f(x) = (1-e^x)^n$. We have

$$ f(x) = \sum_{i=0}^n {n \choose i} (-1)^i e^{ix} $$

So, evidently, the kth derivative is

$$ f^{(k)}(x) = \sum i^k (-1)^i {n \choose i} e^{ix} $$

In particular, one has

$$ f^{(k)}(0) = \sum i^k (-1)^k {n \choose i} $$

Which seems to be a pretty neat formula. My question is whether anyone has more literature into this identities. In other words, can someone pinpoint me to references that discuss these ideas into more detail. Looking to explore things as $f(x) = (1-g(x))^n$

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    $\begingroup$ I don't have a specific reference, but one such tricks are common in the literature of generating functions. $\endgroup$
    – awkward
    Commented Apr 7 at 13:51
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    $\begingroup$ Stirling S2 gives the derivative at $x=0$ for $f(x)$. Also, one can use it to convert between $(h(e^x))^{(n)}$ and $\sum h^{(n)}(e^x)$ like in the “function from exponent” section here. Are these formulas wanted or what type of identities, besides binomial theorem, are wanted for $(1-g(x))^n$? $\endgroup$ Commented Apr 14 at 20:25

2 Answers 2


I see that your expression for the k-th derivative does not match what I obtained and tested. I will show my work on this problem below:

Given $$ f(x) = (1-e^{x})^{n} = (-1)^{n}(e^{x}-1)^{n}$$

Using Binomial theorem we can re-write $f(x)$ as:

$$f(x) = (-1)^{n} \sum_{i=0}^n {n \choose i} (-1)^{n-i} e^{ix}$$

$$f^{(k)}(x)= (-1)^{n} \sum_{i=0}^n {n \choose i} (-1)^{n-i}\frac{d^{k}}{dx^{k}}(e^{ix}) $$

$$f^{(k)}(x)= (-1)^{n} \sum_{i=0}^n {n \choose i} (-1)^{n-i} (i)^{k}( e^{ix}) $$

(2) The case for the expression $ t(x) = (1-g(x))^{n} = (-1)^{n}(g(x)-1) $ $$t(x) =(-1)^{n} \sum_{i=0}^n {n \choose i} (-1)^{n-i} g^{ix}$$

$$t^{(k)}(x)= (-1)^{n} \sum_{i=0}^n {n \choose i} (-1)^{n-i} \frac{d^{k}}{dx^{k}}( g(x)^{i}) $$

$$t^{(k)}(x)= \sum_{i=0}^n {n \choose i} (-1)^{-i} \frac{d^{k}}{dx^{k}}( g(x)^{i}) $$

The expression $\frac{d^{k}}{dx^{k}}( g(x)^{i}$ is not easy to simplify further. Yes, it can be written in terms of different expansions such as General Leibniz Rule, Faà di Bruno's formula or Taylor's series expansion. However, using any of these will only make the expression more complex and will probably add no more insight (at least I think so :) ).


There are probably many sources on the internet. Your question isn't very specific and many such formulas occur as exercises. Here is one source: https://mathweb.ucsd.edu/~gptesler/184a/slides/184a_ch4slides_17-handout.pdf which I found by a Google search using the key 'formulas + binomial coefficients'. Here is a source with applications: https://www.whitman.edu/mathematics/cgt_online/book/section05.03.html

Another promising search key would be 'combinatorics + pdf' which usually produces lecture notes. Hundreds of universities around the globe publish their lecture notes on their servers. They are high-quality and cheap sources for further studies. The '+pdf' part makes sure to end up with notes instead of webpages.

  • $\begingroup$ Your second link is probably off-topic? $\endgroup$
    – NoChance
    Commented Apr 14 at 23:59
  • $\begingroup$ @NoChance It has been meant to provide examples and intended to answer "My question is whether anyone has more literature into this identities." where I read "this" as "such". A parallel section of the same website was whitman.edu/mathematics/cgt_online/book/section01.03.html which directly deals with binomial coefficients. I just wanted to add examples since the original question is quite broad. I mean, $g(x)$ could be anything. Combinatorics and Graph Theory are both potential search keys for formulas wth binomial coefficients. $\endgroup$ Commented Apr 15 at 0:47

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