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f-x-x-2022-Find-derivative-f-n-

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Question Number 177133 by Matica last updated on 01/Oct/22
f(x) = x^(2022) . Find derivative f^((n)) .
f(x)=x2022.Findderivativef(n).
Answered by JDamian last updated on 01/Oct/22
f^((n)) = { ((((2022!)/((2022−n)!)) x^(2022−n) ∀n≤2022)),(( 0 ∀n>2022)) :}
f(n)={2022!(2022−n)!x2022−n∀n⩽20220∀n>2022
Answered by a.lgnaoui last updated on 01/Oct/22
f^((1)) =2022x^(2021) f^((2)) =(2022×2021)x^(2020) f^((3)) =(2022×2021×2020)x^(2019) ....... f^((n)) =[2022×2021×2020×.....×(2022−n+1)]x^(2022−n) f^((n)) =((2022!)/((2022−n)!))x^(2022−n) f^((2022)) =2022×2021×2020×.....×1)=2022! f^((2023)) =(2022!)^′ =(constante)^′ =0 ⇒f^((2023+p)) =0 donc n≤2022 f^((n)) =((2022!)/((2022−n)!))x^(2022−n) n≥2023 f^((n)) =0
f(1)=2022x2021f(2)=(2022×2021)x2020f(3)=(2022×2021×2020)x2019…….f(n)=[2022×2021×2020×…..×(2022−n+1)]x2022−nf(n)=2022!(2022−n)!x2022−nf(2022)=2022×2021×2020×…..×1)=2022!f(2023)=(2022!)′=(constante)′=0⇒f(2023+p)=0doncn⩽2022f(n)=2022!(2022−n)!x2022−nn⩾2023f(n)=0

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