1. **State the problem:** Find the derivative $f'(a)$ of the function $$f(x) = \frac{x^2}{x + 20}$$ at the point $a = 5$ using the definition of the derivative: $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$ 2. **Calculate $f(5)$:** $$f(5) = \frac{5^2}{5 + 20} = \frac{25}{25} = 1$$ 3. **Set up the difference quotient:** $$\frac{f(x) - f(5)}{x - 5} = \frac{\frac{x^2}{x + 20} - 1}{x - 5}$$ 4. **Simplify the numerator:** $$\frac{x^2}{x + 20} - 1 = \frac{x^2 - (x + 20)}{x + 20} = \frac{x^2 - x - 20}{x + 20}$$ 5. **Rewrite the difference quotient:** $$\frac{\frac{x^2 - x - 20}{x + 20}}{x - 5} = \frac{x^2 - x - 20}{(x + 20)(x - 5)}$$ 6. **Factor the numerator:** $$x^2 - x - 20 = (x - 5)(x + 4)$$ 7. **Substitute back and cancel common factors:** $$\frac{\cancel{(x - 5)}(x + 4)}{(x + 20)\cancel{(x - 5)}} = \frac{x + 4}{x + 20}$$ 8. **Take the limit as $x \to 5$:** $$f'(5) = \lim_{x \to 5} \frac{x + 4}{x + 20} = \frac{5 + 4}{5 + 20} = \frac{9}{25}$$ **Final answer:** $$f'(5) = \frac{9}{25}$$