Question

quotient rule: problem 3 (1 point)
let $f(x)=\frac{-7x}{sin(x)+cos(x)}$. evaluate $f(x)$ at $x = 2pi$.
$f(2pi)=square$
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Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = - 7x$, so $u'=-7$, and $v=\sin(x)+\cos(x)$, so $v'=\cos(x)-\sin(x)$.

Step2: Apply quotient - rule

$f'(x)=\frac{-7(\sin(x)+\cos(x))-(-7x)(\cos(x)-\sin(x))}{(\sin(x)+\cos(x))^{2}}$.

Step3: Evaluate at $x = 2\pi$

When $x = 2\pi$, $\sin(2\pi)=0$ and $\cos(2\pi)=1$.
Substitute these values into $f'(x)$:
\[

$$\begin{align*} f'(2\pi)&=\frac{-7(0 + 1)-(-7\times2\pi)(1 - 0)}{(0 + 1)^{2}}\\ &=\frac{-7+14\pi}{1}\\ &=-7 + 14\pi \end{align*}$$

\]

Answer:

$-7 + 14\pi$