Question Number 207127 by hardmath last updated on 07/May/24
$$\mathrm{If}\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{sin}^{\mathrm{4}} \:\mathrm{2x} \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{f}\:^{'} \:\left(\frac{\pi}{\mathrm{12}}\right)\:=\:? \\ $$
Answered by Skabetix last updated on 07/May/24
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{\mathrm{4}} \left(\mathrm{2x}\right) \\ $$$$\Leftrightarrow\:{f}'\left({x}\right)=\mathrm{8}{sin}^{\mathrm{3}} \left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{f}\:^{'} \:\left(\frac{\pi}{\mathrm{12}}\right)\:=\mathrm{8}{sin}^{\mathrm{3}} \:\left(\frac{\pi}{\mathrm{6}}\right)\:{cos}\:\left(\frac{\pi}{\mathrm{6}}\right)\: \\ $$$$=\mathrm{8}×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} ×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{1}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by hardmath last updated on 07/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{ser} \\ $$
Answered by mathzup last updated on 08/May/24
$${f}^{'} \left({x}\right)=\mathrm{4}\:×\mathrm{2}{cos}\left(\mathrm{2}{x}\right){sin}^{\mathrm{3}} \left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${f}^{'} \left(\frac{\pi}{\mathrm{12}}\right)=\mathrm{8}{cos}\left(\frac{\pi}{\mathrm{6}}\right)×{sin}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\mathrm{8}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$