If $f ( x ) = 6 \cos x + \sin ^ { 2 } x$, find $f ^ { \prime } ( x )$ and $f ^ { \prime \prime } ( x )$.
A) $f ^ { tt } ( x ) = - 6 \cos ( 2 x ) + 2 \cos ( x )$
B) $f ^ { t } ( x ) = - 6 \sin ( x ) + \sin ( 2 x )$
C) $f ^ { t } ( x ) = - 6 \sin ( 2 x ) + \sin ( x )$
D) $f ^ { tt } ( x ) = - 6 \cos ( x ) + 2 \cos ( 2 x )$
E) $f ^ { tt } ( x ) = - 2 \cos ( 2 x ) + 6 \cos ( x )$

Question

Quizplus · Accepted Answer

First derivative $f'(x) = -6\sin(x) + 2\sin(x)\cos(x)$, using the chain rule and the derivative of $\sin^2(x)$ which is $2\sin(x)\cos(x)$. The term $2\sin(x)\cos(x)$ can be simplified to $\sin(2x)$ using the double angle formula, making the correct first derivative $f'(x) = -6\sin(x) + \sin(2x)$.Second derivative $f''(x) = -6\cos(x) + 2\cos(2x)$, derived by taking the derivative of $f'(x)$, where the derivative of $\sin(x)$ is $\cos(x)$ and the derivative of $\sin(2x)$ is $2\cos(2x)$ using the chain rule.

If f(x)=6cos⁡x+sin⁡2xf ( x ) = 6 \cos x + \sin ^ { 2 } xf(x)=6cosx+sin2x

If $f ( x ) = 6 \cos x + \sin ^ { 2 } x$