1. We are asked to find the derivative of the function $$h(x) = \frac{x^2 + 6x + 9}{x + 3}$$. 2. First, recognize that the numerator can be factored: $$x^2 + 6x + 9 = (x + 3)^2$$. 3. So, $$h(x) = \frac{(x + 3)^2}{x + 3}$$. 4. Simplify the function by canceling common factors: $$h(x) = \frac{\cancel{(x + 3)}(x + 3)}{\cancel{x + 3}} = x + 3$$. 5. Now, find the derivative of the simplified function: $$h'(x) = \frac{d}{dx}(x + 3) = 1$$. 6. Therefore, the derivative of the original function is $$h'(x) = 1$$. This simplification works for all $$x \neq -3$$ where the original function is defined.