Question Number 177133 by Matica last updated on 01/Oct/22
$$\:\:{f}\left({x}\right)\:=\:{x}^{\mathrm{2022}} \:.\:{Find}\:\:{derivative}\:{f}^{\left({n}\right)} . \\ $$
Answered by JDamian last updated on 01/Oct/22
$${f}^{\left({n}\right)} =\begin{cases}{\frac{\mathrm{2022}!}{\left(\mathrm{2022}−{n}\right)!}\:{x}^{\mathrm{2022}−{n}} \:\:\:\:\:\forall{n}\leqslant\mathrm{2022}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\forall{n}>\mathrm{2022}}\end{cases} \\ $$
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](https://www.tinkutara.com/question/Q177141.png)
$${f}^{\left(\mathrm{1}\right)} =\mathrm{2022}{x}^{\mathrm{2021}} \\ $$$${f}^{\left(\mathrm{2}\right)} =\left(\mathrm{2022}×\mathrm{2021}\right){x}^{\mathrm{2020}} \\ $$$${f}^{\left(\mathrm{3}\right)} =\left(\mathrm{2022}×\mathrm{2021}×\mathrm{2020}\right){x}^{\mathrm{2019}} \\ $$$$ \\ $$$$……. \\ $$$${f}^{\left({n}\right)} =\left[\mathrm{2022}×\mathrm{2021}×\mathrm{2020}×…..×\left(\mathrm{2022}−{n}+\mathrm{1}\right)\right]{x}^{\mathrm{2022}−{n}} \\ $$$${f}^{\left({n}\right)} =\frac{\mathrm{2022}!}{\left(\mathrm{2022}−{n}\right)!}{x}^{\mathrm{2022}−{n}} \\ $$$$ \\ $$$$\left.{f}^{\left(\mathrm{2022}\right)} =\mathrm{2022}×\mathrm{2021}×\mathrm{2020}×…..×\mathrm{1}\right)=\mathrm{2022}! \\ $$$${f}^{\left(\mathrm{2023}\right)} =\left(\mathrm{2022}!\right)^{'} =\left({constante}\right)^{'} =\mathrm{0}\:\:\Rightarrow{f}^{\left(\mathrm{2023}+{p}\right)} =\mathrm{0} \\ $$$${donc}\:\:\:\:{n}\leqslant\mathrm{2022}\:\:\:{f}^{\left({n}\right)} =\frac{\mathrm{2022}!}{\left(\mathrm{2022}−{n}\right)!}{x}^{\mathrm{2022}−{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{n}\geqslant\mathrm{2023}\:\:\:\:{f}^{\left({n}\right)} =\mathrm{0} \\ $$$$ \\ $$