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To solve for the derivative of the given function using the quotient rule, we follow this general rule:

\[
\text{If } y = \frac{f(x)}{g(x)}, \text{ then } \frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.
\]

Now, looking at your function:

\[
y = \frac{(3x^2 + 4)(2x + 2)}{5x - 6},
\]

we will treat the numerator as:

\[
f(x) = (3x^2 + 4)(2x + 2),
\]
and the denominator as:
\[
g(x) = 5x - 6.
\]

### Step 1: Differentiate \( f(x) \)
First, expand \( f(x) \):

\[
f(x) = (3x^2 + 4)(2x + 2) = 6x^3 + 6x^2 + 8x + 8.
\]

Now, take the derivative of \( f(x) \):

\[
f'(x) = 18x^2 + 12x + 8.
\]

### Step 2: Differentiate \( g(x) \)
Now, differentiate \( g(x) = 5x - 6 \):

\[
g'(x) = 5.
\]

### Step 3: Apply the Quotient Rule
Substitute everything into the quotient rule formula:

\[
\frac{dy}{dx} = \frac{(18x^2 + 12x + 8)(5x - 6) - (6x^3 + 6x^2 + 8x + 8)(5)}{(5x - 6)^2}.
\]

### Step 4: Simplify the expression
You can simplify the numerator by expanding both terms and then combining like terms. After completing these steps, the final expression will be:

\[
\frac{dy}{dx} = \frac{(first \: term) - (second \: term)}{(5x - 6)^2}.
\]

Do you need further details on the expansion and final simplification?

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Here are 5 related questions you might consider:

1. What happens when the numerator \( f(x) \) has a higher degree than the denominator?
2. Can you use the product rule within the quotient rule? How?
3. How does the quotient rule handle complex fractions with nested functions?
4. What are other applications of the quotient rule in physics or engineering?
5. How does the behavior of the derivative change near points where \( g(x) = 0 \)?

**Tip:** When applying the quotient rule, be careful with signs when subtracting the two terms. Errors in sign can lead to incorrect results.

Use the quotient rule to find the derivative of the following: y = ((3x^2 + 4)(2x + 2))/(5x - 6)

Find the derivative of the function y = ((3x^2 + 4)(2x + 2))/(5x - 6) using the quotient rule. This solution involves calculus concepts such as differentiation, the quotient rule, and the product rule, suitable for students in Grades 11-12 or college level.

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