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Question: Let $g(x) = e^{f(x)}$ where $f$ is infinitely differentiable function. Does there exist a formula for $n$-th derivative $g^{(n)}(x)$ where $n\geq 0?$

I obtain the following.

$$g'(x) = f'(x) e^{f(x)}$$ $$g''(x) = f''(x) e^{f(x)} + (f'(x))^2 e^{f(x)}$$ $$g'''(x) = f'''(x) e^{f(x)} + 3 f'(x) f''(x) e^{f(x)} + (f'(x))^3 e^{f(x)}.$$ But I could not obtain any pattern from above.

Idonknow
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    It looks like you would need another couple derivatives written out to start making generalizations. It looks like there should be a "biased binomial expansion" in the coefficients, but the exact pattern isn't clear yet. – abiessu Sep 19 '19 at 11:31
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    Massively related https://math.stackexchange.com/questions/441522/general-formula-or-a-pattern-for-the-nth-derivatives-of-efx/2810253 –  Sep 19 '19 at 11:32
  • Also: https://math.stackexchange.com/questions/397325/closed-form-for-nth-derivative-of-exponential-of-f – Martin R Sep 19 '19 at 11:38

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Hint: Work with $$\ln(y)=f(x)$$ to get the higher derivative.

Dr. Sonnhard Graubner
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  • +1 - I like this idea of spotting patterns via implicit differentiation –  Sep 19 '19 at 11:42
  • I would like to understand the downvotes since I am not sure I get the hint right. – nicomezi Sep 19 '19 at 12:15
  • Hi, there is a closed formula on this paper :"On a Closed Formula for the Derivatives of ef (x) and Related Financial Applications" written by Konstantinos Drakakis. And i have invented a method to calculate it, if you want i can explain it to you. – martín canullán Sep 25 '22 at 21:43