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The given function is \( y = x \sqrt{x} \). To find the derivative, we'll use the product rule along with the rules of differentiation.

### Step-by-Step Solution

1. **Rewrite the function**:
   \[
   y = x \cdot x^{1/2}
   \]

2. **Apply the product rule**:
   The product rule states that:
   \[
   \frac{d}{dx}(u \cdot v) = u'v + uv'
   \]
   Here, \( u = x \) and \( v = x^{1/2} \).

3. **Differentiate each term**:
   - \( u = x \) gives \( u' = 1 \),
   - \( v = x^{1/2} \) gives \( v' = \frac{1}{2}x^{-1/2} \) (using the power rule).

4. **Substitute into the product rule**:
   \[
   \frac{dy}{dx} = u'v + uv' = (1)(x^{1/2}) + (x)\left(\frac{1}{2}x^{-1/2}\right)
   \]

5. **Simplify the expression**:
   - First term: \( x^{1/2} \),
   - Second term: \( x \cdot \frac{1}{2}x^{-1/2} = \frac{1}{2}x^{1/2} \).

   Add the terms:
   \[
   \frac{dy}{dx} = x^{1/2} + \frac{1}{2}x^{1/2}
   \]

6. **Combine like terms**:
   \[
   \frac{dy}{dx} = \frac{2}{2}x^{1/2} + \frac{1}{2}x^{1/2} = \frac{3}{2}x^{1/2}
   \]

### Final Answer:
\[
\frac{dy}{dx} = \frac{3}{2}\sqrt{x}
\]

Would you like further details or examples on this type of differentiation?

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**Questions to explore further:**

1. How does the product rule work for more complex functions?
2. Can this derivative be applied to real-world problems involving roots?
3. What happens if \( x < 0 \) in this context (domain issues)?
4. How does this derivative change when combined with other operations?
5. What are other scenarios where roots and powers are involved in differentiation?

**Tip:** Always simplify roots and exponents to their fractional forms for easier differentiation.

Learn how to find the derivative of y = x√x step-by-step using the product rule. This guide covers differentiation techniques for functions involving roots and powers. Suitable for Grades 11-12 calculus students.

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