assignment 5: problem 6
(1 point)
let ( f(x)=\frac{-8 x}{sin (x)+cos (x)} ). evaluate ( f^{prime}(x) ) at ( x=-pi ).
( f^{prime}(-pi)= )

Question

Accepted Answer

# Explanation:
## Step1: Apply the quotient rule
The quotient rule is $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Let $u = - 8x$, then $u^\prime=-8$. Let $v=\sin(x)+\cos(x)$, then $v^\prime=\cos(x)-\sin(x)$.
$$
\begin{align*}
f^\prime(x)&=\frac{-8(\sin(x)+\cos(x))-(-8x)(\cos(x)-\sin(x))}{(\sin(x)+\cos(x))^{2}}\
&=\frac{-8\sin(x)-8\cos(x)+8x\cos(x)-8x\sin(x)}{(\sin(x)+\cos(x))^{2}}
\end{align*}
$$
## Step2: Substitute $x =-\pi$
We know that $\sin(-\pi)=0$ and $\cos(-\pi)=-1$.
$$
\begin{align*}
f^\prime(-\pi)&=\frac{-8\times0 - 8\times(-1)+8\times(-\pi)\times(-1)-8\times(-\pi)\times0}{(0 + (-1))^{2}}\
&=\frac{0 + 8+8\pi-0}{1}\
&=8 + 8\pi
\end{align*}
$$

# Answer:
$8 + 8\pi$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the quotient rule

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

Answer: