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

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)