1. **State the problem:** We are given the function $$y(x) = \frac{12}{2\sqrt{x}} \left(2 - 8x + \sqrt{x} - 4\right)$$ and asked to find its derivative $$y'(x)$$. 2. **Simplify the function first:** Simplify the coefficient: $$\frac{12}{2\sqrt{x}} = \frac{12}{2} \cdot \frac{1}{\sqrt{x}} = 6 \cdot \frac{1}{\sqrt{x}} = \frac{6}{\sqrt{x}}$$ So, $$y(x) = \frac{6}{\sqrt{x}} (2 - 8x + \sqrt{x} - 4)$$ 3. **Simplify inside the parentheses:** $$2 - 8x + \sqrt{x} - 4 = (2 - 4) - 8x + \sqrt{x} = -2 - 8x + \sqrt{x}$$ Thus, $$y(x) = \frac{6}{\sqrt{x}} (-2 - 8x + \sqrt{x})$$ 4. **Distribute the term:** $$y(x) = 6 \cdot \frac{-2}{\sqrt{x}} + 6 \cdot \frac{-8x}{\sqrt{x}} + 6 \cdot \frac{\sqrt{x}}{\sqrt{x}}$$ Rewrite each term: - $$6 \cdot \frac{-2}{\sqrt{x}} = -\frac{12}{\sqrt{x}}$$ - $$6 \cdot \frac{-8x}{\sqrt{x}} = -48 \cdot \frac{x}{\sqrt{x}} = -48 \sqrt{x}$$ (since $$\frac{x}{\sqrt{x}} = \sqrt{x}$$) - $$6 \cdot \frac{\sqrt{x}}{\sqrt{x}} = 6 \cdot 1 = 6$$ So, $$y(x) = -\frac{12}{\sqrt{x}} - 48 \sqrt{x} + 6$$ 5. **Find the derivative term by term:** Recall: - $$\frac{d}{dx} x^{n} = n x^{n-1}$$ - $$\sqrt{x} = x^{1/2}$$ - $$\frac{1}{\sqrt{x}} = x^{-1/2}$$ Derivatives: - $$\frac{d}{dx} \left(-\frac{12}{\sqrt{x}}\right) = -12 \cdot \frac{d}{dx} x^{-1/2} = -12 \cdot \left(-\frac{1}{2} x^{-3/2}\right) = 6 x^{-3/2} = \frac{6}{x^{3/2}}$$ - $$\frac{d}{dx} (-48 \sqrt{x}) = -48 \cdot \frac{d}{dx} x^{1/2} = -48 \cdot \frac{1}{2} x^{-1/2} = -24 x^{-1/2} = -\frac{24}{\sqrt{x}}$$ - $$\frac{d}{dx} 6 = 0$$ 6. **Combine the derivatives:** $$y'(x) = \frac{6}{x^{3/2}} - \frac{24}{\sqrt{x}} + 0 = \frac{6}{x^{3/2}} - \frac{24}{\sqrt{x}}$$ 7. **Rewrite in radical form:** $$y'(x) = \frac{6}{x^{3/2}} - \frac{24}{x^{1/2}} = \frac{6}{(\sqrt{x})^{3}} - \frac{24}{\sqrt{x}} = \frac{6}{x \sqrt{x}} - \frac{24}{\sqrt{x}}$$ **Final answer:** $$y'(x) = \frac{6}{x \sqrt{x}} - \frac{24}{\sqrt{x}}$$