1. **State the problem:** Find the derivative $\frac{dy}{dx}$ of the curve $y = 4x \sqrt{3x - 2}$ for $x \geq \frac{2}{3}$, and then find the equation of the tangent line at the point $(6, 96)$. 2. **Rewrite the function:** Express $y$ as $$y = 4x (3x - 2)^{\frac{1}{2}}$$ 3. **Use the product rule:** Since $y$ is a product of two functions, $u = 4x$ and $v = (3x - 2)^{\frac{1}{2}}$, the derivative is $$\frac{dy}{dx} = u'v + uv'$$ 4. **Calculate derivatives:** - $u' = \frac{d}{dx}(4x) = 4$ - For $v = (3x - 2)^{\frac{1}{2}}$, use the chain rule: $$v' = \frac{1}{2}(3x - 2)^{-\frac{1}{2}} \times 3 = \frac{3}{2}(3x - 2)^{-\frac{1}{2}}$$ 5. **Substitute back:** $$\frac{dy}{dx} = 4 (3x - 2)^{\frac{1}{2}} + 4x \times \frac{3}{2} (3x - 2)^{-\frac{1}{2}}$$ 6. **Simplify:** $$\frac{dy}{dx} = 4 (3x - 2)^{\frac{1}{2}} + 6x (3x - 2)^{-\frac{1}{2}}$$ 7. **Find the slope at $x=6$:** Calculate each term: - $(3 \times 6 - 2) = 18 - 2 = 16$ - $(3x - 2)^{\frac{1}{2}} = \sqrt{16} = 4$ - $(3x - 2)^{-\frac{1}{2}} = \frac{1}{4}$ So, $$\frac{dy}{dx}\bigg|_{x=6} = 4 \times 4 + 6 \times 6 \times \frac{1}{4} = 16 + 9 = 25$$ 8. **Equation of the tangent line:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ where $m=25$ and point $(x_1, y_1) = (6, 96)$. So, $$y - 96 = 25(x - 6)$$ $$y = 25x - 150 + 96 = 25x - 54$$ **Final answers:** $$\frac{dy}{dx} = 4 (3x - 2)^{\frac{1}{2}} + 6x (3x - 2)^{-\frac{1}{2}}$$ Equation of tangent at $(6, 96)$: $$y = 25x - 54$$