Question

let ( f(x)=\frac{sin(x)+cos(x)}{7x}). evaluate ( f^{prime}(x)) at ( x = -pi). ( f^{prime}(-pi)=square)

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sin(x)+\cos(x)$ and $v = 7x$. So, $u'=\cos(x)-\sin(x)$ and $v' = 7$. Then $f'(x)=\frac{(\cos(x)-\sin(x))\cdot7x-7(\sin(x)+\cos(x))}{(7x)^{2}}=\frac{7x\cos(x)-7x\sin(x)-7\sin(x)-7\cos(x)}{49x^{2}}$.

Step2: Substitute $x =-\pi$

Substitute $x =-\pi$ into $f'(x)$.
\[

$$\begin{align*} f'(-\pi)&=\frac{7(-\pi)\cos(-\pi)-7(-\pi)\sin(-\pi)-7\sin(-\pi)-7\cos(-\pi)}{49(-\pi)^{2}}\\ &=\frac{- 7\pi(-1)+7\pi(0)-7(0)-7(-1)}{49\pi^{2}}\\ &=\frac{7\pi + 7}{49\pi^{2}}\\ &=\frac{\pi + 1}{7\pi^{2}} \end{align*}$$

\]

Answer:

$\frac{\pi + 1}{7\pi^{2}}$