1. **State the problem:** Differentiate $y = (\sqrt{x} + 4)(\sqrt{x} - 4)$ using the Product Rule and show that $\frac{dy}{dx} = 1$. 2. **Recall the Product Rule:** For two functions $u(x)$ and $v(x)$, the derivative of their product is $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$ 3. **Identify functions:** Let $$u = \sqrt{x} + 4 = x^{\frac{1}{2}} + 4,$$ $$v = \sqrt{x} - 4 = x^{\frac{1}{2}} - 4.$$ 4. **Find derivatives:** $$u' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}},$$ $$v' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.$$ 5. **Apply Product Rule:** $$\frac{dy}{dx} = u'v + uv' = \frac{1}{2\sqrt{x}}(\sqrt{x} - 4) + (\sqrt{x} + 4)\frac{1}{2\sqrt{x}}.$$ 6. **Combine terms:** $$\frac{dy}{dx} = \frac{\sqrt{x} - 4}{2\sqrt{x}} + \frac{\sqrt{x} + 4}{2\sqrt{x}} = \frac{(\sqrt{x} - 4) + (\sqrt{x} + 4)}{2\sqrt{x}}.$$ 7. **Simplify numerator:** $$(\sqrt{x} - 4) + (\sqrt{x} + 4) = \sqrt{x} - 4 + \sqrt{x} + 4 = 2\sqrt{x}.$$ 8. **Substitute back:** $$\frac{dy}{dx} = \frac{2\sqrt{x}}{2\sqrt{x}}.$$ 9. **Cancel common factors:** $$\frac{dy}{dx} = \frac{\cancel{2\sqrt{x}}}{\cancel{2\sqrt{x}}} = 1.$$ **Final answer:** $$\boxed{\frac{dy}{dx} = 1}.$$