let $f(x)=-3x(sin(x)+cos(x))$. find the following:
1. $f(x)=$
2. $f(\frac{pi}{6})=$
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Accepted Answer

# Explanation:
## Step1: Apply the product rule
The product rule states that if $y = uv$, then $y'=u'v + uv'$. Let $u=-3x$ and $v = \sin(x)+\cos(x)$. Then $u'=-3$ and $v'=\cos(x)-\sin(x)$.
$$
\begin{align*}
f'(x)&=(-3)(\sin(x)+\cos(x))+(-3x)(\cos(x)-\sin(x))\
&=-3\sin(x)-3\cos(x)-3x\cos(x)+3x\sin(x)
\end{align*}
$$
## Step2: Evaluate $f'(\frac{\pi}{6})$
Substitute $x = \frac{\pi}{6}$ into $f'(x)$. We know that $\sin(\frac{\pi}{6})=\frac{1}{2}$, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$.
$$
\begin{align*}
f'(\frac{\pi}{6})&=-3\times\frac{1}{2}-3\times\frac{\sqrt{3}}{2}-3\times\frac{\pi}{6}\times\frac{\sqrt{3}}{2}+3\times\frac{\pi}{6}\times\frac{1}{2}\
&=-\frac{3}{2}-\frac{3\sqrt{3}}{2}-\frac{\pi\sqrt{3}}{4}+\frac{\pi}{4}\
&=\frac{-6 - 6\sqrt{3}-\pi\sqrt{3}+\pi}{4}
\end{align*}
$$

# Answer:
1. $f'(x)=-3\sin(x)-3\cos(x)-3x\cos(x)+3x\sin(x)$
2. $f'(\frac{\pi}{6})=\frac{-6 - 6\sqrt{3}-\pi\sqrt{3}+\pi}{4}$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the product rule

Step2: Evaluate \(f'(\frac{\pi}{6})\)

Answer: