Question Number 202255 by SANOGO last updated on 23/Dec/23
$${calcul}\:{f}_{{n}} '\left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)={n}^{\alpha} {x}\left(\mathrm{1}−{x}\right)^{{n}\:} \: \\ $$
Answered by SANOGO last updated on 23/Dec/23
$${calculer}\:{la}\:{derivee}\:{de}\:{cette}\:{suite}\:{de}\:{fonction} \\ $$
Answered by elcogito last updated on 24/Dec/23
$${f}'_{{n}} \left({x}\right)={n}^{\alpha} \left(\left(\mathrm{1}−{x}\right)^{{n}} +{nx}\left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={n}^{\alpha} \left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} \left(\left({n}−\mathrm{1}\right){x}+\mathrm{1}\right) \\ $$$$ \\ $$