Just explain how one would go about doing this:
What's the rule to this?
to solve that you would use th quotient rules as well as the sin and cos rules.
$\displaystyle
\frac{dy}{dx}\sin{x}=\cos{x}
$
$\displaystyle
\frac{dy}{dx}\cos{x}=-\sin{x}
$
so using the quotient rule, $\displaystyle \frac{dy}{dx}\frac{u}{v}=\frac{vu'-uv'}{v^2}$
$\displaystyle
u=4\sin{x}
$
$\displaystyle
v=2+\cos{x}
$
$\displaystyle
u'=4\cos{x}
$
$\displaystyle
v'=-\sin{x}
$
fill it all in and solve
I would actually have done:
$\displaystyle \frac{8\cos{x} + 4\cos^2{x} + 4\sin^2{x}}{(2 + \cos{x})^2}$
$\displaystyle = \frac{8\cos{x} + 4(\cos^2{x} + \sin^2{x})}{(2 + \cos{x})^2}$
$\displaystyle = \frac{8\cos{x} + 4}{(2 + \cos{x})^2}$
$\displaystyle = \frac{4(2\cos{x} + 1)}{(2 + \cos{x})^2}$.