Question Number 190583 by 073 last updated on 06/Apr/23
Answered by mr W last updated on 06/Apr/23
$${f}\left(\mathrm{2}\right)=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${f}'\left(\mathrm{2}\right)=\mathrm{tan}\:\mathrm{60}°=\sqrt{\mathrm{3}} \\ $$$${g}'\left({x}\right)=\mathrm{2}{xf}\left({x}\right)+{x}^{\mathrm{2}} {f}'\left({x}\right) \\ $$$${g}'\left(\mathrm{2}\right)=\mathrm{2}×\mathrm{2}{f}\left(\mathrm{2}\right)+\mathrm{2}^{\mathrm{2}} {f}'\left(\mathrm{2}\right)=\mathrm{8}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{12}\sqrt{\mathrm{3}}\:\checkmark \\ $$