Question Number 166813 by mathocean1 last updated on 28/Feb/22
$${f}\left({x}\right)={x}^{\mathrm{2}{n}} \\ $$$${Determinate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$
Answered by TheSupreme last updated on 28/Feb/22
$${f}'\left({x}\right)=\alpha{x}^{\alpha−\mathrm{1}} \\ $$$${f}^{\left({k}\right)} \left({x}\right)=\frac{\mathrm{2}{n}!}{\left(\mathrm{2}{n}−{k}\right)!}{x}^{\mathrm{2}{n}−{k}} \\ $$
Answered by Mathspace last updated on 28/Feb/22
$${f}^{\left(\mathrm{1}\right)} \left({x}\right)=\mathrm{2}{nx}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{\mathrm{2}{n}−\mathrm{2}} \\ $$$${by}\:{recurrence}\:{f}^{\left({p}\right)} \left({x}\right)=\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)…\left(\mathrm{2}{n}−{p}+\mathrm{1}\right){x}^{\mathrm{2}{n}−{p}} \\ $$$${p}={n}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)….\left({n}+\mathrm{1}\right){x}^{{n}} \\ $$$$=\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)…\left({n}+\mathrm{1}\right){n}!}{{n}!}{x}^{{n}} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{{n}!}{x}^{{n}} \\ $$$$ \\ $$