1. We are given the function $$y = \frac{x^2 + 4}{x - 3}$$ and asked to find its derivative $$\frac{dy}{dx}$$. 2. To differentiate a quotient, we use the Quotient Rule: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where $$u = x^2 + 4$$ and $$v = x - 3$$. 3. Compute the derivatives of numerator and denominator: $$\frac{du}{dx} = 2x$$ $$\frac{dv}{dx} = 1$$ 4. Apply the Quotient Rule: $$\frac{dy}{dx} = \frac{(x - 3)(2x) - (x^2 + 4)(1)}{(x - 3)^2}$$ 5. Expand the numerator: $$= \frac{2x^2 - 6x - x^2 - 4}{(x - 3)^2}$$ 6. Simplify the numerator: $$= \frac{(2x^2 - x^2) - 6x - 4}{(x - 3)^2} = \frac{x^2 - 6x - 4}{(x - 3)^2}$$ 7. The derivative is: $$\boxed{\frac{dy}{dx} = \frac{x^2 - 6x - 4}{(x - 3)^2}}$$ This matches the structure of the derivative given but with correct simplification and signs.