1. The problem is to find the derivative of the function $$f(x) = \frac{3 e^x - 1}{e^x + 1}$$. 2. We use the quotient rule for derivatives, which states: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $u = 3 e^x - 1$ and $v = e^x + 1$. 3. Compute the derivatives of $u$ and $v$: $$u' = \frac{d}{dx}(3 e^x - 1) = 3 e^x$$ $$v' = \frac{d}{dx}(e^x + 1) = e^x$$ 4. Apply the quotient rule: $$f'(x) = \frac{3 e^x (e^x + 1) - (3 e^x - 1) e^x}{(e^x + 1)^2}$$ 5. Simplify the numerator: $$3 e^x (e^x + 1) - (3 e^x - 1) e^x = 3 e^{2x} + 3 e^x - 3 e^{2x} + e^x = 3 e^x + e^x = 4 e^x$$ 6. So the derivative is: $$f'(x) = \frac{4 e^x}{(e^x + 1)^2}$$ 7. This derivative shows the rate of change of the original function with respect to $x$. Final answer: $$\boxed{f'(x) = \frac{4 e^x}{(e^x + 1)^2}}$$