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To differentiate \( y = (x^2 - 9) \sqrt{x} \), let's proceed by applying the **product rule** and **chain rule**.

### Step-by-Step Solution:

Given:
\[
y = (x^2 - 9) \cdot \sqrt{x}
\]
Rewrite \( \sqrt{x} \) as \( x^{\frac{1}{2}} \).

Thus, we have:
\[
y = (x^2 - 9) \cdot x^{\frac{1}{2}}
\]

Now, use the **product rule**:
\[
y' = \left( \frac{d}{dx}(x^2 - 9) \cdot x^{\frac{1}{2}} \right) + \left( (x^2 - 9) \cdot \frac{d}{dx}(x^{\frac{1}{2}}) \right)
\]

1. **Differentiate \( x^2 - 9 \):**
   \[
   \frac{d}{dx}(x^2 - 9) = 2x
   \]

2. **Differentiate \( x^{\frac{1}{2}} \):**
   \[
   \frac{d}{dx}(x^{\frac{1}{2}}) = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}
   \]

Now, substitute these derivatives back into the product rule expression:
\[
y' = \left( 2x \cdot x^{\frac{1}{2}} \right) + \left( (x^2 - 9) \cdot \frac{1}{2\sqrt{x}} \right)
\]

3. **Simplify each term:**

   - For the first term:
     \[
     2x \cdot x^{\frac{1}{2}} = 2x^{1 + \frac{1}{2}} = 2x^{\frac{3}{2}}
     \]

   - For the second term:
     \[
     \frac{x^2 - 9}{2\sqrt{x}} = \frac{x^2 - 9}{2x^{\frac{1}{2}}} = \frac{1}{2} x^{\frac{3}{2}} - \frac{9}{2} x^{-\frac{1}{2}}
     \]

Combine these results:
\[
y' = 2x^{\frac{3}{2}} + \frac{x^{\frac{3}{2}}}{2} - \frac{9}{2}x^{-\frac{1}{2}}
\]

Thus, the derivative is:
\[
y' = \frac{5x^{\frac{3}{2}}}{2} - \frac{9}{2} x^{-\frac{1}{2}}
\]

---

Would you like further details on any step?

### Additional Questions
1. How would the result change if the expression included another function, like \( e^x \)?
2. What does the product rule state, and why is it used here?
3. Can we simplify the answer further if \( x \) has a specific value?
4. How would you handle the derivative if a different power of \( x \) was used, like \( x^{3/2} \)?
5. What is the practical application of finding the derivative in this form?

### Tip
When differentiating products of functions, always check if one term can be simplified to make the process easier.

Find the derivative of y = (x^2 - 9)√x with a detailed solution applying the product rule and chain rule. This problem explores advanced differentiation techniques, suitable for high school calculus students in Grades 11-12.

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