frac{d}{dx} left( sqrt{3} sin left(2x + frac{ | vedclass

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(B) Let $y = \sqrt{3} \sin \left(2x + \frac{\pi}{3} \right) + \cos \left(2x + \frac{\pi}{3} \right)$.We can rewrite this expression by multiplying and

Vedclass · Accepted Answer

$-4 \sin 2x$

Vedclass · Answer

(B) Let $y = \sqrt{3} \sin \left(2x + \frac{\pi}{3} ight) + \cos \left(2x + \frac{\pi}{3} ight)$.We can rewrite this expression by multiplying and dividing by $2$:$y = 2 \left( \frac{\sqrt{3}}{2} \sin \left(2x + \frac{\pi}{3} ight) + \frac{1}{2} \cos \left(2x + \frac{\pi}{3} ight) ight)$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,where $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$:$y = 2 \left( \sin \left(2x + \frac{\pi}{3} ight) \cos \left(\frac{\pi}{6} ight) + \cos \left(2x + \frac{\pi}{3} ight) \sin \left(\frac{\pi}{6} ight) ight)$.$y = 2 \sin \left(2x + \frac{\pi}{3} + \frac{\pi}{6} ight) = 2 \sin \left(2x + \frac{\pi}{2} ight)$.Since $\sin(	heta + \frac{\pi}{2}) = \cos 	heta$,we have $y = 2 \cos(2x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (2 \cos 2x) = 2 	imes (- \sin 2x) 	imes 2 = -4 \sin 2x$.Thus,the correct option is $B$.

$\frac{d}{dx} \left( \sqrt{3} \sin \left(2x + \frac{\pi}{3} \right) + \cos \left(2x + \frac{\pi}{3} \right) \right) = $ . . . . . .

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