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The problem asks to find the derivative of the function \( f(x) = -4x^2 \cdot \ln(-2x^3 - 4x^2) \).

We can approach this by using the **product rule** and the **chain rule** since the function is a product of two terms, \( -4x^2 \) and \( \ln(-2x^3 - 4x^2) \).

### Step-by-step solution:

1. **Recall the product rule** for differentiation:
   \[
   \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   where \( u(x) = -4x^2 \) and \( v(x) = \ln(-2x^3 - 4x^2) \).

2. **Differentiate \( u(x) = -4x^2 \):**
   \[
   u'(x) = \frac{d}{dx}(-4x^2) = -8x
   \]

3. **Differentiate \( v(x) = \ln(-2x^3 - 4x^2) \):**

   To differentiate the logarithmic function, apply the **chain rule**. The derivative of \( \ln(g(x)) \) is \( \frac{1}{g(x)} \cdot g'(x) \), where \( g(x) = -2x^3 - 4x^2 \).

   - First, differentiate \( g(x) = -2x^3 - 4x^2 \):
     \[
     g'(x) = \frac{d}{dx}(-2x^3 - 4x^2) = -6x^2 - 8x
     \]
   - Now, apply the chain rule:
     \[
     v'(x) = \frac{1}{-2x^3 - 4x^2} \cdot (-6x^2 - 8x)
     \]

4. **Now, apply the product rule:**
   Using the results from steps 2 and 3, apply the product rule:
   \[
   f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   Substituting the derivatives and expressions for \( u(x) \) and \( v(x) \):
   \[
   f'(x) = (-8x) \cdot \ln(-2x^3 - 4x^2) + (-4x^2) \cdot \frac{-6x^2 - 8x}{-2x^3 - 4x^2}
   \]

5. **Simplify the expression**:
   - The first term remains \( -8x \cdot \ln(-2x^3 - 4x^2) \).
   - Simplify the second term:
     \[
     \frac{-6x^2 - 8x}{-2x^3 - 4x^2} = \frac{-2x(3x + 4)}{-2x^2( x + 2)} = \frac{3x + 4}{x + 2}
     \]
   Thus, the second term becomes:
     \[
     -4x^2 \cdot \frac{3x + 4}{x + 2}
     \]

6. **Final expression for the derivative:**
   \[
   f'(x) = -8x \ln(-2x^3 - 4x^2) - 4x^2 \cdot \frac{3x + 4}{x + 2}
   \]

This is the derivative of the given function.

Let me know if you'd like more details or have any further questions!

### Related questions:
1. How do you apply the chain rule for more complex functions?
2. Can the product rule be used when there are more than two factors in a product?
3. What are the conditions for using the logarithmic differentiation method?
4. How can we simplify expressions with common factors in derivatives?
5. How do logarithmic functions behave near zero or infinity?

### Tip:
When differentiating products of functions, always break down the terms using the product rule and simplify step by step to avoid errors.

Find the derivative of f(x) = -4x^2 · ln(-2x^3 - 4x^2).

Learn how to differentiate the function f(x) = -4x^2 · ln(-2x^3 - 4x^2) using the product and chain rule. This calculus problem explores logarithmic and polynomial functions and is ideal for college-level students studying calculus.

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