To find the derivative \(\frac{dy}{dx}\) of the function \(y = (3x + 1)^4(x - 3)\), we need to apply the product rule and the chain rule. 

### Step 1: Identify the parts for the product rule.
Let:
\[ u = (3x + 1)^4 \quad \text{and} \quad v = (x - 3) \]

The product rule states:
\[
\frac{dy}{dx} = u'v + uv'
\]

### Step 2: Find the derivatives \(u'\) and \(v'\).

- **For \(u\):**
  \[
  u = (3x + 1)^4
  \]
  Using the chain rule, differentiate \(u\) with respect to \(x\):
  \[
  u' = 4(3x + 1)^3 \cdot \frac{d}{dx}(3x + 1) = 4(3x + 1)^3 \cdot 3 = 12(3x + 1)^3
  \]

- **For \(v\):**
  \[
  v = (x - 3)
  \]
  The derivative of \(v\) with respect to \(x\) is:
  \[
  v' = 1
  \]

### Step 3: Apply the product rule.
\[
\frac{dy}{dx} = u'v + uv' = \left[12(3x + 1)^3\right](x - 3) + \left[(3x + 1)^4\right](1)
\]

### Step 4: Simplify the expression.
\[
\frac{dy}{dx} = 12(3x + 1)^3(x - 3) + (3x + 1)^4
\]

This is the derivative of the given function.

Would you like further details or have any questions? Here are five related questions to explore:

1. How would the derivative change if the power in the function were different?
2. Can you explain how the chain rule is applied in this context?
3. What happens if we consider higher-order derivatives of the function?
4. How can this derivative be used to find critical points of the function?
5. What if the function were a product of three terms instead of two?

**Tip:** When differentiating functions that are products of two or more functions, always consider using the product rule first, and then apply the chain rule as needed for more complex terms.

Learn how to differentiate the function y = (3x + 1)^4(x - 3) by applying the product rule and chain rule. This problem is suitable for students in Grades 11-12 or college-level calculus courses. Detailed steps are provided for understanding the solution.

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