1. **Problem statement:** Given the function $y = (\sqrt{4x+1})^3$, find the derivative $\frac{dy}{dx}$. Then, use this derivative to find the approximate increase in $y$ as $x$ increases from 6 to $6 + p$, where $p$ is small. 2. **Formula and rules:** To find $\frac{dy}{dx}$, we use the chain rule. If $y = [u(x)]^3$ where $u(x) = \sqrt{4x+1}$, then $$\frac{dy}{dx} = 3[u(x)]^2 \cdot \frac{du}{dx}$$ Also, $\sqrt{4x+1} = (4x+1)^{1/2}$, so $$\frac{du}{dx} = \frac{1}{2}(4x+1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x+1}}$$ 3. **Calculate $\frac{dy}{dx}$:** $$\frac{dy}{dx} = 3(\sqrt{4x+1})^2 \cdot \frac{2}{\sqrt{4x+1}} = 3(4x+1) \cdot \frac{2}{\sqrt{4x+1}} = \frac{6(4x+1)}{\sqrt{4x+1}} = 6\sqrt{4x+1}$$ 4. **Approximate increase in $y$ from $x=6$ to $x=6+p$:** For small $p$, the increase in $y$ is approximately $$\Delta y \approx \frac{dy}{dx} \bigg|_{x=6} \cdot p = 6\sqrt{4(6)+1} \cdot p = 6\sqrt{25} \cdot p = 6 \times 5 \times p = 30p$$ **Final answers:** $$\frac{dy}{dx} = 6\sqrt{4x+1}$$ Approximate increase in $y$ when $x$ changes from 6 to $6+p$ is $$\Delta y \approx 30p$$