Question

$\frac{d}{dx}\left(3\cos \left(\frac{\pi }{6}+x^{\circ }\right)-4{\cos }^3\left(\frac{\pi }{6}+x^{\circ }\right)\right)=$ _______

• A. $\cos \left(3x^{\circ }\right)$
• B. $\frac{\pi }{60}\sin \left(3x^{\circ }\right)$
• C. $\frac{\pi }{60}\cos \left(3x^{\circ }\right)$
• D. $-\frac{\pi }{60}\sin \left(3x^{\circ }\right)$

Answer 1

Answer: (C)

$\frac{d}{dx}\left(-\cos 3\left(\frac{\pi }{6}+x^{\circ }\right)\right)=\frac{d}{dx}\left(-\cos \left(\frac{\pi }{2}+3x^{\circ }\right)\right)=$
$\frac{d}{dx}\left(\sin 3x\frac{\pi }{180}\right)=\frac{3\pi }{180}\cos 3x^{\circ }$