Question Number 90630 by jagoll last updated on 25/Apr/20
$${f}\left({x}\right)\:=\:{xe}^{−{x}} \\ $$$${f}^{\left(\mathrm{2020}\right)} \left({x}\right)\:=\: \\ $$
Commented by john santu last updated on 25/Apr/20
$${f}^{\left({n}\right)} \left({x}\right)\:=\:\begin{cases}{\left({n}−{x}\right).{e}^{−{x}} \:,\:{n}\:{odd}}\\{\left({x}−{n}\right).{e}^{−{x}} ,\:{n}\:{even}}\end{cases} \\ $$$${f}^{\left(\mathrm{2020}\right)} \left({x}\right)=\:\left({x}−\mathrm{2020}\right).{e}^{−{x}} \\ $$
Answered by MWSuSon last updated on 25/Apr/20
![I′m assuming you mean 2020th derivative using Leibnitz theorem f^((n)) (x)=[(−1)^n xe^(−x) +n(−1)^(n−1) e^(−x) ] f^((2020)) =[(−1)^(2020) xe^(−x) +2020(−1)^(2020−1) e^(−x) ] f^((2020)) =[xe^(−x) −2020e^(−x) ]=e^(−x) [x−2020]](https://www.tinkutara.com/question/Q90690.png)
$${I}'{m}\:{assuming}\:{you}\:{mean}\:\mathrm{2020}{th}\:{derivative} \\ $$$${using}\:{Leibnitz}\:{theorem}\:{f}^{\left({n}\right)} \left({x}\right)=\left[\left(−\mathrm{1}\right)^{{n}} {xe}^{−{x}} +{n}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {e}^{−{x}} \right] \\ $$$${f}^{\left(\mathrm{2020}\right)} =\left[\left(−\mathrm{1}\right)^{\mathrm{2020}} {xe}^{−{x}} +\mathrm{2020}\left(−\mathrm{1}\right)^{\mathrm{2020}−\mathrm{1}} {e}^{−{x}} \right] \\ $$$${f}^{\left(\mathrm{2020}\right)} =\left[{xe}^{−{x}} −\mathrm{2020}{e}^{−{x}} \right]={e}^{−{x}} \left[{x}−\mathrm{2020}\right] \\ $$