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

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)