1. **State the problem:** Find the derivative of the function $$f(x) = \frac{x^2 - x - 2}{x^2 - 6x + 9}$$. 2. **Recall the formula:** For a function $$f(x) = \frac{u(x)}{v(x)}$$, the derivative is given by the quotient rule: $$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$ where $$u(x) = x^2 - x - 2$$ and $$v(x) = x^2 - 6x + 9$$. 3. **Find the derivatives of numerator and denominator:** $$ u'(x) = \frac{d}{dx}(x^2 - x - 2) = 2x - 1$$ $$v'(x) = \frac{d}{dx}(x^2 - 6x + 9) = 2x - 6$$ 4. **Apply the quotient rule:** $$ f'(x) = \frac{(2x - 1)(x^2 - 6x + 9) - (x^2 - x - 2)(2x - 6)}{(x^2 - 6x + 9)^2} $$ 5. **Expand the terms in the numerator:** - First term: $$ (2x - 1)(x^2 - 6x + 9) = 2x^3 - 12x^2 + 18x - x^2 + 6x - 9 = 2x^3 - 13x^2 + 24x - 9 $$ - Second term: $$ (x^2 - x - 2)(2x - 6) = 2x^3 - 6x^2 - 2x^2 + 6x - 4x + 12 = 2x^3 - 8x^2 + 2x + 12 $$ 6. **Subtract the second term from the first:** $$ (2x^3 - 13x^2 + 24x - 9) - (2x^3 - 8x^2 + 2x + 12) = 2x^3 - 13x^2 + 24x - 9 - 2x^3 + 8x^2 - 2x - 12 $$ $$ = (2x^3 - 2x^3) + (-13x^2 + 8x^2) + (24x - 2x) + (-9 - 12) = -5x^2 + 22x - 21 $$ 7. **Write the derivative:** $$ f'(x) = \frac{-5x^2 + 22x - 21}{(x^2 - 6x + 9)^2} $$ 8. **Simplify denominator if possible:** Note that $$x^2 - 6x + 9 = (x - 3)^2$$, so denominator is $$((x - 3)^2)^2 = (x - 3)^4$$. **Final answer:** $$ f'(x) = \frac{-5x^2 + 22x - 21}{(x - 3)^4} $$