Question Number 80519 by mathocean1 last updated on 03/Feb/20
$$\mathrm{g}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} {x}+{sin}\left(\mathrm{2}{x}\right). \\ $$$${g}'\left({x}\right)=\:……….? \\ $$
Commented by mr W last updated on 03/Feb/20
![g′(x)=2×2 cos x (−sin x)+cos (2x) 2 =−4 cos x sin x+2 cos (2x) =−2 sin (2x)+2 cos (2x) =2 [cos (2x)−sin (2x)] =2(√2)[(1/( (√2))) cos (2x)−(1/( (√2))) sin (2x)] =2(√2)[cos (π/4) cos (2x)−sin (π/4) sin (2x)] =2(√2)cos (2x+(π/4))](https://www.tinkutara.com/question/Q80527.png)
$${g}'\left({x}\right)=\mathrm{2}×\mathrm{2}\:\mathrm{cos}\:{x}\:\left(−\mathrm{sin}\:{x}\right)+\mathrm{cos}\:\left(\mathrm{2}{x}\right)\:\mathrm{2} \\ $$$$=−\mathrm{4}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}+\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right) \\ $$$$=−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)+\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right) \\ $$$$=\mathrm{2}\:\left[\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\left[\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right) \\ $$
Commented by mathocean1 last updated on 03/Feb/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sirs} \\ $$
Answered by Rio Michael last updated on 03/Feb/20
$${g}'\left({x}\right)\:=\:−\mathrm{4sin}\:{x}\:+\:\mathrm{2cos}\:\mathrm{2}{x} \\ $$
Answered by Henri Boucatchou last updated on 03/Feb/20
$${g}\left({x}\right)={cos}\mathrm{2}{x}+{sin}\mathrm{2}{x}+\mathrm{1} \\ $$$${g}'\left({x}\right)=\mathrm{2}\left({cos}\mathrm{2}{x}−{sin}\mathrm{2}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{2}}{cos}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right) \\ $$