g(x)=2cos^2 x+sin(2x). g′(x)= ..........?


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Question Number 80519 by mathocean1 last updated on 03/Feb/20

$g (x) = 2 {c o s}^{2} x + s i n (2 x) .$ $g^{'} (x) = . . . . . . . . . . ?$
Commented by mr W last updated on 03/Feb/20

$g^{'} (x) = 2 \times 2 \cos x (- \sin x) + \cos (2 x) 2$ $= - 4 \cos x \sin x + 2 \cos (2 x)$ $= - 2 \sin (2 x) + 2 \cos (2 x)$ $= 2 [\cos (2 x) - \sin (2 x)]$ $= 2 \sqrt{2} [\frac{1}{\sqrt{2}} \cos (2 x) - \frac{1}{\sqrt{2}} \sin (2 x)]$ $= 2 \sqrt{2} [\cos \frac{π}{4} \cos (2 x) - \sin \frac{π}{4} \sin (2 x)]$ $= 2 \sqrt{2} \cos (2 x + \frac{π}{4})$
Commented by mathocean1 last updated on 03/Feb/20

$thank you sirs$
	Answered by Rio Michael last updated on 03/Feb/20

	$g^{'} (x) = - 4 \sin x + 2 \cos 2 x$
	Answered by Henri Boucatchou last updated on 03/Feb/20

	$g (x) = c o s 2 x + s i n 2 x + 1$ $g^{'} (x) = 2 (c o s 2 x - s i n 2 x)$ $= 2 \sqrt{2} c o s (2 x + \frac{π}{4})$
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