To analyze the function $y = \frac{2x - 1}{2x + 6} - 4$ and compute the derivative $\frac{dy}{dx}$, let's proceed step by step.

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The function:

$y = \frac{2x - 1}{2x + 6} - 4$

Steps to find $\frac{dy}{dx}$:

1. Rewrite the function for simplicity: Let $u = \frac{2x - 1}{2x + 6}$, so $y = u - 4$. Thus, we only need to differentiate $u$.

2. Apply the quotient rule for $u = \frac{2x - 1}{2x + 6}$: $u' = \frac{(v \cdot u' - u \cdot v')}{v^2}$ where: $u = 2x - 1, \quad v = 2x + 6, \quad u' = 2, \quad v' = 2.$

3. Substituting: $u' = \frac{(2x + 6)(2) - (2x - 1)(2)}{(2x + 6)^2}.$

4. Simplify: $u' = \frac{4x + 12 - 4x + 2}{(2x + 6)^2} = \frac{14}{(2x + 6)^2}.$

5. Since $y = u - 4$, the derivative is: $\frac{dy}{dx} = u' = \frac{14}{(2x + 6)^2}.$

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Final result:

$\frac{dy}{dx} = \frac{14}{(2x + 6)^2}.$

Let me know if you'd like more clarification or additional details. 😊

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Relative Questions:

1. What is the behavior of $y$ as $x \to \infty$?
2. How does the quotient rule apply to rational functions?
3. What is the second derivative $\frac{d^2y}{dx^2}$?
4. How do vertical and horizontal asymptotes affect the function?
5. Can we rewrite $y$ to simplify the computation of its derivative further?

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Tip:

Always simplify the denominator first when applying the quotient rule—it can save time and reduce errors.