assignment 5: problem 3 (1 point) let $f(x)=\frac{3\sin(x)}{3 + \cos(x)}$. find the following: 1. $f(x)=$ 2. $f(3)=$ note: you can earn partial credit on this problem

Question

Accepted Answer

# 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 $u'=3\cos(x)$, and $v = 3+\cos(x)$, so $v'=-\sin(x)$.
$$
\begin{align*}
f'(x)&=\frac{(3\cos(x))(3 + \cos(x))-3\sin(x)(-\sin(x))}{(3+\cos(x))^{2}}\
&=\frac{9\cos(x)+3\cos^{2}(x)+ 3\sin^{2}(x)}{(3+\cos(x))^{2}}
\end{align*}
$$
## Step2: Use the trigonometric identity $\sin^{2}(x)+\cos^{2}(x) = 1$
Substitute $\sin^{2}(x)+\cos^{2}(x)=1$ into the numerator:
$$
\begin{align*}
f'(x)&=\frac{9\cos(x)+3(\cos^{2}(x)+\sin^{2}(x))}{(3+\cos(x))^{2}}\
&=\frac{9\cos(x)+3\times1}{(3+\cos(x))^{2}}\
&=\frac{9\cos(x)+ 3}{(3+\cos(x))^{2}}\
&=\frac{3(3\cos(x)+1)}{(3+\cos(x))^{2}}
\end{align*}
$$
## Step3: Evaluate $f'(3)$
Substitute $x = 3$ into $f'(x)$:
$$
f'(3)=\frac{3(3\cos(3)+1)}{(3+\cos(3))^{2}}
$$
Using a calculator (assuming $x$ is in radians), $\cos(3)\approx - 0.989992$
$$
\begin{align*}
f'(3)&=\frac{3(3\times(-0.989992)+1)}{(3+(-0.989992))^{2}}\
&=\frac{3(-2.969976 + 1)}{(1.910008)^{2}}\
&=\frac{3\times(-1.969976)}{3.64813}\
&=\frac{-5.909928}{3.64813}\approx - 1.62
\end{align*}
$$

# Answer:
1. $f'(x)=\frac{3(3\cos(x)+1)}{(3+\cos(x))^{2}}$
2. $f'(3)\approx - 1.62$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the quotient rule

Step2: Use the trigonometric identity \(\sin^{2}(x)+\cos^{2}(x) = 1\)

Step3: Evaluate \(f'(3)\)

Answer: