1. **State the problem:** We are given the function $$f(x) = \frac{2}{x} - \frac{1}{2x^2}$$ and need to find the derivative at $$x=2$$, i.e., $$f'(2)$$. 2. **Rewrite the function:** Express the function with negative exponents for easier differentiation: $$f(x) = 2x^{-1} - \frac{1}{2}x^{-2}$$ 3. **Differentiate the function:** Using the power rule $$\frac{d}{dx} x^n = nx^{n-1}$$, \begin{align*} f'(x) &= 2 \cdot (-1) x^{-2} - \frac{1}{2} \cdot (-2) x^{-3} \\ &= -2x^{-2} + x^{-3} \end{align*} 4. **Simplify the derivative:** $$f'(x) = -\frac{2}{x^2} + \frac{1}{x^3}$$ 5. **Evaluate at $$x=2$$:** $$f'(2) = -\frac{2}{2^2} + \frac{1}{2^3} = -\frac{2}{4} + \frac{1}{8} = -\frac{1}{2} + \frac{1}{8}$$ 6. **Combine into a single fraction:** $$-\frac{1}{2} + \frac{1}{8} = -\frac{4}{8} + \frac{1}{8} = -\frac{3}{8}$$ **Final answer:** $$f'(2) = -\frac{3}{8}$$