Question Number 143274 by Aditya9886 last updated on 12/Jun/21
Answered by Olaf_Thorendsen last updated on 12/Jun/21
$${y}\:=\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{cos}{x}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\left(−\mathrm{sin}{x}\right)×\mathrm{2cos}\left(\mathrm{cos}{x}\right)×\mathrm{sin}\left(\mathrm{cos}{x}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:−\mathrm{sin}{x}.\mathrm{sin}\left(\mathrm{2cos}{x}\right) \\ $$$$ \\ $$$$\mathrm{Nota}\:: \\ $$$${y}\:=\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\:=\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2cos}{x}\right)}{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}\:=\:−\frac{\left(−\mathrm{sin}{x}\right)\left(−\mathrm{2sin}\left(\mathrm{2cos}{x}\right)\right)}{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}\:=\:−\mathrm{sin}{x}.\mathrm{sin}\left(\mathrm{2cos}{x}\right) \\ $$
Answered by justtry last updated on 12/Jun/21
$${u}={cosx}\Rightarrow\frac{{du}}{{dx}}=−{sinx} \\ $$$${y}={sin}^{\mathrm{2}} {u}\Rightarrow\frac{{dy}}{{du}}=\mathrm{2}{sinu}.{cosu}={sin}\mathrm{2}{u}={sin}\mathrm{2}\left({cosx}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}.\frac{{du}}{{dx}}={sin}\mathrm{2}\left({cosx}\right).\left(−{sinx}\right)=−{sinx}.{sin}\mathrm{2}\left({cosx}\right) \\ $$
Answered by mathmax by abdo last updated on 12/Jun/21
$$\mathrm{we}\:\mathrm{have}\:\mathrm{y}=\mathrm{sin}^{\mathrm{2}} \left(\mathrm{cosx}\right)\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2cosx}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)\left(−\mathrm{2sinx}\right)\mathrm{sin}\left(\mathrm{2cosx}\right)\:\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{sinx}\:\mathrm{sin}\left(\mathrm{2cosx}\right) \\ $$