let ( f(x)=\frac{3sin(x)}{3+cos(x)} ). find the following:\n1. ( f(x)=\frac{3(3cos(x)+1)}{(3+cos(x))^{2}}…

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y'=\frac{u'v - uv'}{v^{2}}). Here, (u = 3\sin(x)), so (u'=3\cos(x)); (v = 3+\cos(x)), so (v'=-\sin(x)). [ \begin{align*} f'(x)&=\frac{(3\cos(x))(3 + \cos(x))-3\sin(x)(-\sin(x))}{(3+\cos(x))^{2}}\ \end{align*} ]

Step2: Simplify the numerator

Expand the numerator: [ \begin{align*} (3\cos(x))(3 + \cos(x))-3\sin(x)(-\sin(x))&=9\cos(x)+3\cos^{2}(x)+ 3\sin^{2}(x)\ &=9\cos(x)+3(\cos^{2}(x)+\sin^{2}(x)) \end{align*} ] Since (\cos^{2}(x)+\sin^{2}(x) = 1), the numerator becomes (9\cos(x)+3=3(3\cos(x)+1))

Answer:

1. (f'(x)=\frac{3(3\cos(x)+1)}{(3 + \cos(x))^{2}})
2. To find (f'(3)), substitute (x = 3) into (f'(x)): [ f'(3)=\frac{3(3\cos(3)+1)}{(3+\cos(3))^{2}} ] (It seems there was an error in the original answer for (f'(3)) as (f'(x)) is a well - defined function for all real (x) and (f'(3)) can be calculated by substitution. If we assume the original problem had a typo and was perhaps asking for (f'(3\pi)) or something else, but based on the given function (y = f(x)=\frac{3\sin(x)}{3+\cos(x)}), (f'(x)) is defined for all (x) and (f'(3)) is (\frac{3(3\cos(3)+1)}{(3+\cos(3))^{2}}))