To find the derivative of the function \((3x-5)(x^2 - 2)\), we'll apply the **Product Rule**. The Product Rule states:

\[
\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Let \( u(x) = 3x - 5 \) and \( v(x) = x^2 - 2 \). 

1. **Find the derivative of** \( u(x) \):
   \[
   u'(x) = \frac{d}{dx}[3x - 5] = 3
   \]

2. **Find the derivative of** \( v(x) \):
   \[
   v'(x) = \frac{d}{dx}[x^2 - 2] = 2x
   \]

3. **Apply the Product Rule**:
   \[
   \frac{d}{dx}[(3x-5)(x^2-2)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   \[
   = 3 \cdot (x^2 - 2) + (3x - 5) \cdot 2x
   \]

4. **Simplify the expression**:
   \[
   = 3x^2 - 6 + 6x^2 - 10x
   \]
   \[
   = 9x^2 - 10x - 6
   \]

So, the derivative of \((3x-5)(x^2 - 2)\) is:

\[
\boxed{9x^2 - 10x - 6}
\]

Would you like any more details or explanations? 

Here are five related questions to consider:
1. What is the derivative of \( (2x + 1)(x^2 + 3x + 2) \)?
2. How would you apply the Product Rule to \( (x^2 + x + 1)(x^3 - 4x + 5) \)?
3. What is the second derivative of the function \( (3x-5)(x^2-2) \)?
4. Can you find the critical points of the function \( f(x) = (3x-5)(x^2-2) \)?
5. How does the Chain Rule differ from the Product Rule when taking derivatives?

**Tip:** Always simplify your function as much as possible before applying differentiation rules; it can often make the process easier!

Learn how to find the derivative of the function (3x-5)(x^2 - 2) using the Product Rule. This problem involves differentiation of polynomials and application of the Product Rule formula. Suitable for advanced high school students and college calculus learners.

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