Question Number 69979 by Maclaurin Stickker last updated on 29/Sep/19
$${Using}\:{the}\:{definition}\:{calculate}\:{the} \\ $$$${derivative}\:{at}\:{point}\:{x}=\mathrm{2} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 29/Sep/19
$${lim}_{{x}\rightarrow\mathrm{2}} \frac{{f}\left({x}\right)−{f}\left(\mathrm{2}\right)}{{x}−\mathrm{2}}={lim}_{{x}\rightarrow\mathrm{2}} \frac{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{16}}{{x}−\mathrm{2}}= \\ $$$${lim}_{{x}\rightarrow\mathrm{2}} \frac{\mathrm{2}\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{{x}−\mathrm{2}}={lim}_{{x}\rightarrow\mathrm{2}} \mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)= \\ $$$$\mathrm{2}\left(\mathrm{4}+\mathrm{4}+\mathrm{4}\right)=\mathrm{24} \\ $$
Commented by Rio Michael last updated on 29/Sep/19
$${f}\:^{'} \left({x}\right)\:=\:\mathrm{6}{x}^{\mathrm{2}} \:\Rightarrow\:{f}\:^{'} \left(\mathrm{2}\right)\:=\:\mathrm{24} \\ $$