Question Number 149291 by mathdanisur last updated on 04/Aug/21
$${f}\left({x}\right)\:=\:\mathrm{2}{sinx}^{\mathrm{2}} −{cos}^{\mathrm{2}} {x}−\mathrm{2} \\ $$$${f}\:^{'} \left({x}\right)\:=\:? \\ $$
Answered by nimnim last updated on 04/Aug/21
$${f}\:'\left({x}\right)=\mathrm{4}{x}.{cosx}^{\mathrm{2}} +\mathrm{2}{cosx}.{sinx}−\mathrm{0}=\mathrm{4}{x}.{cos}\left({x}^{\mathrm{2}} \right)+{sin}\left(\mathrm{2}{x}\right) \\ $$
Answered by Ar Brandon last updated on 04/Aug/21
$${f}\left({x}\right)=\mathrm{2sin}\left({x}^{\mathrm{2}} \right)−\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{2} \\ $$$${f}\:'\left({x}\right)=\mathrm{4}{x}\mathrm{cos}\left({x}^{\mathrm{2}} \right)+\mathrm{2sin}{x}\mathrm{cos}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{x}\mathrm{cos}\left({x}^{\mathrm{2}} \right)+\mathrm{sin2}{x} \\ $$
Commented by Ar Brandon last updated on 04/Aug/21
$$\mathrm{Haha}\:!\:\mathrm{J}'\mathrm{avais}\:\mathrm{commis}\:\mathrm{cette}\:\mathrm{m}\hat {\mathrm{e}me}\:\mathrm{erreur}\: \\ $$$$\mathrm{avant}\:\mathrm{de}\:\mathrm{m}'\mathrm{en}\:\mathrm{rendre}\:\mathrm{compte},\:\mathrm{apr}\grave {\mathrm{e}s}\:\mathrm{avoir} \\ $$$$\mathrm{vu}\:\mathrm{les}\:\mathrm{diff}\acute {\mathrm{e}rentes}\:\mathrm{solutions}\:\mathrm{propos}\acute {\mathrm{e}es}. \\ $$
Commented by Ar Brandon last updated on 04/Aug/21
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Commented by mathdanisur last updated on 04/Aug/21
$${Thank}\:{You}\:{Ser} \\ $$