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I'm trying to solve a problem here where I came across the following term

$\frac{\rm d^k}{\rm dx^k}\left(e^{f(x)}\right)$

Is there a summation formula for this kind of derivative? It would really help me.

Thank you very much.

Addendum

for me, $f(x)=-\frac{a\,x}{1-x}$.

So for k=7, for example, I have the coefficients

$\frac{k!}{1!}\,\,\,6\,\frac{k!}{2!}\,\,\,15\,\frac{k!}{3!}\,\,\,20\,\frac{k!}{4!}\,\,\,15\,\frac{k!}{5!}\,\,\,6\,\frac{k!}{6!}\,\,\,\frac{k!}{7!}$

where the number in the denominator is the order of the term, so $k!/7!$ is multiplied by $a^7$, for example.

There is an order in the way the coefficients are placed but I cannot express it in a simpler way.

Gabu
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1 Answers1

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It can be seen that: $D (e^{f(x)}) = f'(x) \, e^{f(x)}$ and $$D^2 (e^{f(x)}) = D(f'(x) \, e^{f(x)}) = [(f'(x))^2 + f''(x)] \, e^{f(x)}$$ which is an example of $$D^{n+1}(e^{f(x)}) = \sum_{k=0}^{n}\binom{n}{k} D^{k}(e^{f(x)}) \, D^{n-k}(f'(x)). $$

Leucippus
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