To find the derivative of $f(x) = (2x - 7)^3 \cdot (x^2 - 3x - 2)^3$, we will need to apply the product rule and chain rule of differentiation.

Product Rule:

If $f(x) = u(x) \cdot v(x)$, then $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ where $u(x) = (2x - 7)^3$ and $v(x) = (x^2 - 3x - 2)^3$.

Chain Rule:

To differentiate $u(x)$ and $v(x)$, we will use the chain rule. For $u(x) = (g(x))^n$, the chain rule states: $u'(x) = n \cdot (g(x))^{n-1} \cdot g'(x)$

Let's compute the derivatives step by step.

Step 1: Differentiate $u(x) = (2x - 7)^3$

Using the chain rule: $u'(x) = 3 \cdot (2x - 7)^2 \cdot \frac{d}{dx}(2x - 7) = 3 \cdot (2x - 7)^2 \cdot 2 = 6 \cdot (2x - 7)^2$

Step 2: Differentiate $v(x) = (x^2 - 3x - 2)^3$

Again, using the chain rule: $v'(x) = 3 \cdot (x^2 - 3x - 2)^2 \cdot \frac{d}{dx}(x^2 - 3x - 2)$ Now, find the derivative of $x^2 - 3x - 2$: $\frac{d}{dx}(x^2 - 3x - 2) = 2x - 3$ Thus, $v'(x) = 3 \cdot (x^2 - 3x - 2)^2 \cdot (2x - 3)$

Step 3: Apply the Product Rule

Now, we apply the product rule to differentiate the entire function: $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ Substitute $u'(x)$, $u(x)$, $v'(x)$, and $v(x)$: $f'(x) = \left( 6 \cdot (2x - 7)^2 \right) \cdot (x^2 - 3x - 2)^3 + (2x - 7)^3 \cdot \left( 3 \cdot (x^2 - 3x - 2)^2 \cdot (2x - 3) \right)$

This is the derivative of the function. You can simplify further if needed, but this is the result in its expanded form.

Would you like any clarifications or details on the steps?

Here are 5 related questions:

1. What is the general formula for the product rule?
2. How do you apply the chain rule to composite functions?
3. What is the derivative of a product of three functions?
4. How do you simplify the derivative of a product of powers?
5. What role does the chain rule play in multivariable calculus?

Tip: Practice recognizing when to apply the product and chain rules together, as they frequently appear in more complex differentiation problems.