The function f(x)

Consider the function \(f(x) =\frac{4 \sin{(x)}-x\cdot \cos{(x)}-2x}{2+\cos{(x)} } \quad \text{ for }\quad 0\le x \le2\pi\)

1) Show that f'(x)=4cosx-cos^2x/(2+cosx)^2

 \(\begin{array}{rcll} f(x) = \frac{4 \sin{(x)}-x\cdot \cos{(x)}-2x}{2+\cos{(x)} } &=& \frac{4 \sin{(x)}-x\cdot [ 2+ \cos{(x)} ] }{2+\cos{(x)} }\\ &=& \frac{4 \sin{(x)} } {2+\cos{(x)} } - \frac{x\cdot [ 2+ \cos{(x)} ] } {2+\cos{(x)} }\\ &=& \frac{4 \sin{(x)} } {2+\cos{(x)} } - x\\\\ f(x) &=& \frac{u}{v} \qquad \Rightarrow \qquad f'(x) = \frac{u'v-uv'}{v^2} \\ u= 4 \sin{(x)} && u' = 4 \cos{(x)} \\ v= 2+\cos{(x)} && v' = - \sin{(x)} \\\\ f'(x) &=& \frac{ 4 \cos{(x)}(2+\cos{(x)}) - 4 \sin{(x)}(- \sin{(x)}) } {[2+\cos{(x)}]^2} - 1 \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 \cos^2{(x)} + 4 \sin^2{(x)} } {[2+\cos{(x)}]^2} - 1 \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 [\cos^2{(x)} + \sin^2{(x)}] } {[2+\cos{(x)}]^2} - 1 \qquad & \cos^2{(x)} + \sin^2{(x)} = 1\\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 } {[2+\cos{(x)}]^2} - 1 \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 - [2+\cos{(x)}]^2} {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 - [4 + 4\cos{(x)} + \cos^2{(x)}]} {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 - 4 - 4\cos{(x)} - \cos^2{(x)}]} {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 8 \cos{(x)} - 4\cos{(x)} - \cos^2{(x)} } {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 4 \cos{(x)} - \cos^2{(x)}} {[2+\cos{(x)}]^2} \\ \end{array}\)

2.