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.