Question: The expression $\frac{x^{-2}y^{\frac{1}{2}}}{x^{\frac{1}{3}}y^{-1}}$, where $x > 1$ and $y > 1$, is equivalent to which of the following?
1. **State the problem:** Simplify the expression $$\frac{x^{-2}y^{\frac{1}{2}}}{x^{\frac{1}{3}}y^{-1}}$$ and find which option it matches.
2. **Recall the rules for exponents:**
- When dividing like bases, subtract exponents: $$\frac{a^m}{a^n} = a^{m-n}$$
- Negative exponents mean reciprocal: $$a^{-m} = \frac{1}{a^m}$$
- Fractional exponents represent roots: $$a^{\frac{1}{2}} = \sqrt{a}, \quad a^{\frac{1}{3}} = \sqrt[3]{a}$$
3. **Apply the division rule to $x$ terms:**
$$x^{-2} \div x^{\frac{1}{3}} = x^{-2 - \frac{1}{3}} = x^{-\frac{6}{3} - \frac{1}{3}} = x^{-\frac{7}{3}}$$
4. **Apply the division rule to $y$ terms:**
$$y^{\frac{1}{2}} \div y^{-1} = y^{\frac{1}{2} - (-1)} = y^{\frac{1}{2} + 1} = y^{\frac{3}{2}}$$
5. **Rewrite the expression:**
$$\frac{x^{-2}y^{\frac{1}{2}}}{x^{\frac{1}{3}}y^{-1}} = x^{-\frac{7}{3}} y^{\frac{3}{2}}$$
6. **Rewrite negative exponent as reciprocal:**
$$x^{-\frac{7}{3}} = \frac{1}{x^{\frac{7}{3}}}$$
7. **Express fractional exponents as roots:**
$$x^{\frac{7}{3}} = x^{2 + \frac{1}{3}} = x^2 \cdot x^{\frac{1}{3}} = x^2 \sqrt[3]{x}$$
$$y^{\frac{3}{2}} = y^{1 + \frac{1}{2}} = y \cdot y^{\frac{1}{2}} = y \sqrt{y}$$
8. **Combine all parts:**
$$x^{-\frac{7}{3}} y^{\frac{3}{2}} = \frac{y \sqrt{y}}{x^2 \sqrt[3]{x}}$$
9. **Match with options:** This matches option D.
**Final answer:** D. $$\frac{y \sqrt{y}}{x^2 \sqrt[3]{x}}$$