Question

let ( f(x)=\frac{3sin(x)}{3+cos(x)} ). find the following:

1. ( f(x)=\frac{3(3cos(x)+1)}{(3+cos(x))^{2}} )
2. ( f(3)=dne )

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 \(v'=-\sin(x)\).

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Step2: Simplify the numerator

Expand the numerator:

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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 \(\frac{3(3\cos(3)+1)}{(3+\cos(3))^{2}}\))