1. **State the problem:** Find the constant $A$ in the derivative $$\frac{dy}{dx} = \frac{Ax^2 + 12}{x^4 (x^2 - 4)^{1/2}}$$ for the function $$y = \frac{(x^2 - 4)^{1/2}}{x^3}$$ where $x > 2$. 2. **Recall the quotient rule:** For $$y = \frac{u}{v}$$, the derivative is $$y' = \frac{u'v - uv'}{v^2}$$. 3. **Identify $u$ and $v$:** $$u = (x^2 - 4)^{1/2}, \quad v = x^3$$ 4. **Compute $u'$:** Using the chain rule, $$u' = \frac{1}{2}(x^2 - 4)^{-1/2} \cdot 2x = \frac{x}{(x^2 - 4)^{1/2}}$$ 5. **Compute $v'$:** $$v' = 3x^2$$ 6. **Apply the quotient rule:** $$y' = \frac{u'v - uv'}{v^2} = \frac{\frac{x}{(x^2 - 4)^{1/2}} \cdot x^3 - (x^2 - 4)^{1/2} \cdot 3x^2}{x^6}$$ 7. **Simplify numerator:** $$= \frac{x^4}{(x^2 - 4)^{1/2}} - 3x^2 (x^2 - 4)^{1/2}$$ 8. **Put over common denominator $(x^2 - 4)^{1/2}$:** $$= \frac{x^4 - 3x^2 (x^2 - 4)}{(x^2 - 4)^{1/2}} = \frac{x^4 - 3x^2 x^2 + 12 x^2}{(x^2 - 4)^{1/2}} = \frac{x^4 - 3x^4 + 12 x^2}{(x^2 - 4)^{1/2}} = \frac{-2x^4 + 12 x^2}{(x^2 - 4)^{1/2}}$$ 9. **Divide by $v^2 = x^6$:** $$y' = \frac{-2x^4 + 12 x^2}{x^6 (x^2 - 4)^{1/2}} = \frac{-2x^4 + 12 x^2}{x^6 (x^2 - 4)^{1/2}}$$ 10. **Factor numerator:** $$= \frac{x^2(-2x^2 + 12)}{x^6 (x^2 - 4)^{1/2}} = \frac{-2x^4 + 12 x^2}{x^6 (x^2 - 4)^{1/2}}$$ 11. **Simplify denominator:** $$= \frac{-2x^4 + 12 x^2}{x^6 (x^2 - 4)^{1/2}} = \frac{-2x^4 + 12 x^2}{x^6 (x^2 - 4)^{1/2}}$$ 12. **Rewrite denominator as $x^4 \cdot x^2 (x^2 - 4)^{1/2}$:** $$y' = \frac{-2x^4 + 12 x^2}{x^6 (x^2 - 4)^{1/2}} = \frac{-2x^4 + 12 x^2}{x^4 \cdot x^2 (x^2 - 4)^{1/2}}$$ 13. **Divide numerator and denominator by $x^2$:** $$y' = \frac{-2x^2 + 12}{x^4 (x^2 - 4)^{1/2}}$$ 14. **Compare with given form:** $$\frac{dy}{dx} = \frac{Ax^2 + 12}{x^4 (x^2 - 4)^{1/2}}$$ 15. **Identify $A$:** $$A = -2$$ **Final answer:** $$\boxed{A = -2}$$