Question

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

• A. $4\cos 2x$
• B. $-4\sin 2x$
• C. $4\sin 2x$
• D. $-4\cos 2x$

Answer 1

Answer: (B)

$-4\sin 2x$

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

$=\frac{d}{dx}\left(2\sin (2x+\frac{\pi }{3}+\frac{\pi }{6})\right)$

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

$=\frac{d}{dx}\left(2\cos 2x\right)$

$=-4\sin 2x$.