Question Number 194386 by SANOGO last updated on 05/Jul/23
$${calcul}\:{the}\:{the}\:{derive}\:{of} \\ $$$${cos}^{\mathrm{3}} \left(\frac{\theta}{\mathrm{3}}\right)' \\ $$
Answered by Frix last updated on 05/Jul/23
$${f}\left({t}\right)={t}^{\mathrm{3}} \\ $$$${g}\left({t}\right)=\mathrm{cos}\:{t} \\ $$$${h}\left({t}\right)=\frac{{t}}{\mathrm{3}} \\ $$$$\left({f}\left({g}\left({h}\left({t}\right)\right)\right)\right)'={h}'\left({t}\right)×{g}'\left({h}\left({t}\right)\right)×{f}'\left({g}\left({h}\left({t}\right)\right)\right) \\ $$$$\mathrm{W}{e}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\left(−\mathrm{sin}\:\frac{\theta}{\mathrm{3}}\right)×\mathrm{3cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{3}}=−\mathrm{sin}\:\frac{\theta}{\mathrm{3}}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{3}} \\ $$