let $f(x)=\frac{-3x}{sin(x)+cos(x)}$. evaluate $f(x)$ at $x = 2pi$. $f(2pi)=$

Question

Accepted Answer

# Explanation:
## Step1: Apply the quotient rule
The quotient rule states that if $y = \frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u=-3x$, so $u'=-3$; $v = \sin(x)+\cos(x)$, and $v'=\cos(x)-\sin(x)$.
$$
f'(x)=\frac{-3(\sin(x)+\cos(x))-(-3x)(\cos(x)-\sin(x))}{(\sin(x)+\cos(x))^{2}}
$$
## Step2: Substitute $x = 2\pi$
We know that $\sin(2\pi)=0$ and $\cos(2\pi)=1$.
Substitute into $f'(x)$:
$$
\begin{align*}
f'(2\pi)&=\frac{-3(0 + 1)-(-3\times2\pi)(1 - 0)}{(0 + 1)^{2}}\
&=\frac{-3+6\pi}{1}
\end{align*}
$$

# Answer:
$-3 + 6\pi$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the quotient rule

Step2: Substitute \(x = 2\pi\)

Answer: