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

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