1. **State the problem:** We are given the function $$f(x) = - \frac{4}{3\sqrt{x}} - 2\sqrt{x}$$ and need to find its derivative at $$x=2$$, expressed as a single fraction in simplest radical form. 2. **Rewrite the function using exponents:** Recall that $$\sqrt{x} = x^{\frac{1}{2}}$$ and $$\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$$. So, $$f(x) = - \frac{4}{3} x^{-\frac{1}{2}} - 2 x^{\frac{1}{2}}$$ 3. **Find the derivative $$f'(x)$$ using the power rule:** $$f'(x) = - \frac{4}{3} \cdot \left(-\frac{1}{2}\right) x^{-\frac{3}{2}} - 2 \cdot \frac{1}{2} x^{-\frac{1}{2}}$$ Simplify coefficients: $$f'(x) = \frac{4}{3} \cdot \frac{1}{2} x^{-\frac{3}{2}} - x^{-\frac{1}{2}} = \frac{2}{3} x^{-\frac{3}{2}} - x^{-\frac{1}{2}}$$ 4. **Evaluate $$f'(2)$$:** $$f'(2) = \frac{2}{3} \cdot 2^{-\frac{3}{2}} - 2^{-\frac{1}{2}}$$ Recall that $$2^{-\frac{3}{2}} = \frac{1}{2^{3/2}} = \frac{1}{2^{1} \cdot \sqrt{2}} = \frac{1}{2 \sqrt{2}}$$ and $$2^{-\frac{1}{2}} = \frac{1}{\sqrt{2}}$$. So, $$f'(2) = \frac{2}{3} \cdot \frac{1}{2 \sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{2}{3} \cdot \frac{1}{2 \sqrt{2}} - \frac{1}{\sqrt{2}}$$ Simplify the first term: $$\frac{2}{3} \cdot \frac{1}{2 \sqrt{2}} = \frac{1}{3 \sqrt{2}}$$ Therefore, $$f'(2) = \frac{1}{3 \sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{1 - 3}{3 \sqrt{2}} = -\frac{2}{3 \sqrt{2}}$$ 5. **Rationalize the denominator:** $$f'(2) = -\frac{2}{3 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{2 \sqrt{2}}{3 \cdot 2} = -\frac{\sqrt{2}}{3}$$ **Final answer:** $$f'(2) = - \frac{\sqrt{2}}{3}$$