what is the derivative of (2x-7)^3 * (x^2-3x-2)^3

To differentiate \((2x - 7)^3 \cdot (x^2 - 3x - 2)^3\), we will apply the **product rule** and **chain rule**.

### Step-by-Step Solution:

Let:
\[
u = (2x - 7)^3 \quad \text{and} \quad v = (x^2 - 3x - 2)^3
\]

The product rule for derivatives is:
\[
\frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v'
\]
Thus, we need to find \(u'\) and \(v'\).

### 1. Differentiate \(u = (2x - 7)^3\):
Using the chain rule:
\[
u' = 3(2x - 7)^2 \cdot \frac{d}{dx}(2x - 7) = 3(2x - 7)^2 \cdot 2 = 6(2x - 7)^2
\]

### 2. Differentiate \(v = (x^2 - 3x - 2)^3\):
Again using the chain rule:
\[
v' = 3(x^2 - 3x - 2)^2 \cdot \frac{d}{dx}(x^2 - 3x - 2) = 3(x^2 - 3x - 2)^2 \cdot (2x - 3)
\]

### 3. Apply the product rule:
Now, substitute these results into the product rule:
\[
\frac{d}{dx} \left[ (2x - 7)^3 \cdot (x^2 - 3x - 2)^3 \right] = u'v + uv'
\]
This becomes:
\[
= 6(2x - 7)^2 \cdot (x^2 - 3x - 2)^3 + (2x - 7)^3 \cdot 3(x^2 - 3x - 2)^2 \cdot (2x - 3)
\]

### Final Answer:
\[
\frac{d}{dx} \left[ (2x - 7)^3 \cdot (x^2 - 3x - 2)^3 \right] = 6(2x - 7)^2 (x^2 - 3x - 2)^3 + 3(2x - 7)^3 (x^2 - 3x - 2)^2 (2x - 3)
\]

Would you like me to simplify the expression further, or do you have any questions about this solution?

### Relative Questions:
1. How does the chain rule work in differentiation?
2. What is the product rule, and how is it applied?
3. Can you explain the difference between \(u'\) and \(v'\) in the product rule?
4. How do you simplify a derivative expression?
5. What happens if one of the functions in the product rule is constant?

**Tip**: In any derivative involving products or compositions of functions, it’s essential to carefully apply the product rule and chain rule to avoid errors.

Find the derivative of the function (2x-7)^3 * (x^2-3x-2)^3 by applying the product and chain rule. This step-by-step solution breaks down the use of differentiation techniques for polynomial functions. Suitable for students in Grades 11-12 or early college Calculus.

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