To differentiate \( x^2 \sqrt{x - 5} \) with respect to \( x \), we need to apply both the product rule and the chain rule.

The function can be written as:

\[
f(x) = x^2 \cdot \sqrt{x - 5}
\]

### Step-by-Step Solution

1. **Product Rule**: 
   The product rule states that:
   \[
   \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   Here, \( u(x) = x^2 \) and \( v(x) = \sqrt{x - 5} \).

2. **Differentiate \( u(x) = x^2 \)**:
   \[
   u'(x) = 2x
   \]

3. **Differentiate \( v(x) = \sqrt{x - 5} \)**:
   Rewriting \( \sqrt{x - 5} \) as \( (x - 5)^{1/2} \), we apply the chain rule:
   \[
   v'(x) = \frac{1}{2} (x - 5)^{-1/2} \cdot 1 = \frac{1}{2 \sqrt{x - 5}}
   \]

4. **Apply the product rule**:
   Now apply the product rule:
   \[
   \frac{d}{dx} [x^2 \cdot \sqrt{x - 5}] = 2x \cdot \sqrt{x - 5} + x^2 \cdot \frac{1}{2 \sqrt{x - 5}}
   \]

5. **Simplify**:
   The first term is:
   \[
   2x \cdot \sqrt{x - 5}
   \]
   The second term is:
   \[
   \frac{x^2}{2 \sqrt{x - 5}}
   \]

So, the final derivative is:

\[
\frac{d}{dx} \left( x^2 \sqrt{x - 5} \right) = 2x \sqrt{x - 5} + \frac{x^2}{2 \sqrt{x - 5}}
   \]

---

Would you like further details or explanations on any part?

### Related Questions:
1. How do you apply the product rule in other similar functions?
2. Can you differentiate functions with higher powers of \( x \) inside the square root?
3. How do you integrate \( x^2 \sqrt{x - 5} \)?
4. How can you solve for critical points of the function \( x^2 \sqrt{x - 5} \)?
5. What are the applications of chain rule and product rule in physics?

### Tip:
When using the product rule, always differentiate one function at a time and keep the other function unchanged, then add the results together!

Learn how to differentiate x^2 * sqrt(x - 5) using the product and chain rules. This detailed solution includes step-by-step application of the product rule and chain rule to solve this calculus problem. Suitable for students in Grades 11-12 studying advanced differentiation techniques.

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