1. **State the problem:** We want to find the derivative of the function $f(x) = \frac{1}{x-2}$ using the first principle (definition of derivative). 2. **Recall the definition of the derivative:** $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ 3. **Substitute $f(x)$ into the formula:** $$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)-2} - \frac{1}{x-2}}{h} $$ 4. **Simplify the numerator:** Find a common denominator: $$ \frac{1}{x+h-2} - \frac{1}{x-2} = \frac{x-2 - (x+h-2)}{(x+h-2)(x-2)} = \frac{x-2 - x - h + 2}{(x+h-2)(x-2)} = \frac{-h}{(x+h-2)(x-2)} $$ 5. **Rewrite the expression for the limit:** $$ f'(x) = \lim_{h \to 0} \frac{-h}{h \cdot (x+h-2)(x-2)} = \lim_{h \to 0} \frac{-1}{(x+h-2)(x-2)} $$ 6. **Take the limit as $h \to 0$:** $$ f'(x) = \frac{-1}{(x-2)^2} $$ **Final answer:** $$ \frac{d}{dx}\left( \frac{1}{x-2} \right) = -\frac{1}{(x-2)^2} $$