f(x) = (3 sin x) / (1 + cos x)

1. Identify the function and the goal: We are given $f(x) = \frac{3 \sin x}{1 + \cos x}$. We need to find the derivative $f'(x)$ and then evaluate it at $x=1$.

1. Apply the quotient rule: The quotient rule states that for $f(x) = \frac{u(x)}{v(x)}$, the derivative is $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$. Here, $u(x) = 3 \sin x$ and $v(x) = 1 + \cos x$. Thus, $u'(x) = 3 \cos x$ and $v'(x) = -\sin x$.

3. Calculate the derivative: Substituting these into the formula:
$$f'(x) = \frac{(3 \cos x)(1 + \cos x) - (3 \sin x)(-\sin x)}{(1 + \cos x)^2}$$
$$f'(x) = \frac{3 \cos x + 3 \cos^2 x + 3 \sin^2 x}{(1 + \cos x)^2}$$
Using the identity $\sin^2 x + \cos^2 x = 1$, we have:
$$f'(x) = \frac{3 \cos x + 3(1)}{(1 + \cos x)^2} = \frac{3(\cos x + 1)}{(1 + \cos x)^2}$$
Simplifying by canceling $(1 + \cos x)$:
$$f'(x) = \frac{3}{1 + \cos x}$$

4. Evaluate at $x=1$: Now, substitute $x=1$ into the expression for $f'(x)$:
$$f'(1) = \frac{3}{1 + \cos(1)}$$
Since $\cos(1)$ is in radians, this is the final exact value.