Q. $\frac{d}{dx}\ \lbrack\cos{x^{0}\rbrack}\$is equal to
A) $- \frac{\pi}{180}\sin x^{0}$
B) $- \sin x^{0}$
C) $\frac{\pi}{180}\sin x^{0}$
D) $- \frac{\pi x}{180}\sin x$
Explanation
(a)
$\frac{d}{dx}\lbrack\cos{x^{0}\rbrack = \frac{d}{dx}\left\lbrack \cos\frac{\pi x}{180} \right\rbrack}$ $\genfrac{[}{]}{0pt}{}{\because 1^{0} = \frac{\pi}{180}\ radians\ }{\therefore x^{0} = \frac{\pi x}{180}\ radians}$
$= - \sin\left( \frac{\pi x}{180} \right).\frac{\pi}{180} = - \frac{\pi}{180}\sin{(x^{0})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.