1. **State the problem:** We want to find the derivative of the function $$f(x) = \frac{2}{x}$$ using the definition of the derivative, which is $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$$ 2. **Write the difference quotient:** Substitute the function into the definition: $$f'(x) = \lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}.$$ 3. **Find a common denominator inside the numerator:** $$\frac{2}{x+h} - \frac{2}{x} = \frac{2x - 2(x+h)}{x(x+h)} = \frac{2x - 2x - 2h}{x(x+h)} = \frac{-2h}{x(x+h)}.$$ 4. **Substitute back into the difference quotient:** $$f'(x) = \lim_{h \to 0} \frac{\frac{-2h}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-2h}{x(x+h)} \cdot \frac{1}{h}.$$ 5. **Cancel the common factor $h$:** $$f'(x) = \lim_{h \to 0} \frac{-2\cancel{h}}{x(x+h)} \cdot \frac{1}{\cancel{h}} = \lim_{h \to 0} \frac{-2}{x(x+h)}.$$ 6. **Evaluate the limit as $h \to 0$:** $$f'(x) = \frac{-2}{x \cdot x} = \frac{-2}{x^2}.$$ 7. **Conclusion:** The derivative of $$f(x) = \frac{2}{x}$$ is $$f'(x) = -\frac{2}{x^2}.$$ **Answer:** $$f'(x) = -\frac{2}{x^2}.$$