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

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

$${f}^{\left(\mathrm{1}\right)} =\mathrm{2022}{x}^{\mathrm{2021}} \\ $$$${f}^{\left(\mathrm{2}\right)} =\left(\mathrm{2022}×\mathrm{2021}\right){x}^{\mathrm{2020}} \\ $$$${f}^{\left(\mathrm{3}\right)} =\left(\mathrm{2022}×\mathrm{2021}×\mathrm{2020}\right){x}^{\mathrm{2019}} \\ $$$$ \\ $$$$……. \\ $$$${f}^{\left({n}\right)} =\left[\mathrm{2022}×\mathrm{2021}×\mathrm{2020}×…..×\left(\mathrm{2022}−{n}+\mathrm{1}\right)\right]{x}^{\mathrm{2022}−{n}} \\ $$$${f}^{\left({n}\right)} =\frac{\mathrm{2022}!}{\left(\mathrm{2022}−{n}\right)!}{x}^{\mathrm{2022}−{n}} \\ $$$$ \\ $$$$\left.{f}^{\left(\mathrm{2022}\right)} =\mathrm{2022}×\mathrm{2021}×\mathrm{2020}×…..×\mathrm{1}\right)=\mathrm{2022}! \\ $$$${f}^{\left(\mathrm{2023}\right)} =\left(\mathrm{2022}!\right)^{'} =\left({constante}\right)^{'} =\mathrm{0}\:\:\Rightarrow{f}^{\left(\mathrm{2023}+{p}\right)} =\mathrm{0} \\ $$$${donc}\:\:\:\:{n}\leqslant\mathrm{2022}\:\:\:{f}^{\left({n}\right)} =\frac{\mathrm{2022}!}{\left(\mathrm{2022}−{n}\right)!}{x}^{\mathrm{2022}−{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{n}\geqslant\mathrm{2023}\:\:\:\:{f}^{\left({n}\right)} =\mathrm{0} \\ $$$$ \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com