1. **State the problem:** We are given the function $$f(x) = \frac{1}{\sqrt{x}} + \frac{3\sqrt{x}}{2}$$ and need to find its derivative at $$x=6$$, i.e., $$f'(6)$$, expressed as a single fraction in simplest radical form. 2. **Rewrite the function using exponents:** Recall that $$\sqrt{x} = x^{\frac{1}{2}}$$, so $$f(x) = x^{-\frac{1}{2}} + \frac{3}{2} x^{\frac{1}{2}}$$ 3. **Differentiate term-by-term using the power rule:** - For $$x^{-\frac{1}{2}}$$, the derivative is $$-\frac{1}{2} x^{-\frac{3}{2}}$$ - For $$\frac{3}{2} x^{\frac{1}{2}}$$, the derivative is $$\frac{3}{2} \cdot \frac{1}{2} x^{-\frac{1}{2}} = \frac{3}{4} x^{-\frac{1}{2}}$$ So, $$f'(x) = -\frac{1}{2} x^{-\frac{3}{2}} + \frac{3}{4} x^{-\frac{1}{2}}$$ 4. **Evaluate the derivative at $$x=6$$:** $$f'(6) = -\frac{1}{2} \cdot 6^{-\frac{3}{2}} + \frac{3}{4} \cdot 6^{-\frac{1}{2}}$$ 5. **Simplify the powers:** - $$6^{-\frac{3}{2}} = \frac{1}{6^{\frac{3}{2}}} = \frac{1}{(\sqrt{6})^3} = \frac{1}{6 \sqrt{6}}$$ - $$6^{-\frac{1}{2}} = \frac{1}{\sqrt{6}}$$ 6. **Substitute back:** $$f'(6) = -\frac{1}{2} \cdot \frac{1}{6 \sqrt{6}} + \frac{3}{4} \cdot \frac{1}{\sqrt{6}} = -\frac{1}{12 \sqrt{6}} + \frac{3}{4 \sqrt{6}}$$ 7. **Find common denominator and combine:** The common denominator is $$12 \sqrt{6}$$. Rewrite $$\frac{3}{4 \sqrt{6}} = \frac{3 \cdot 3}{4 \cdot 3 \sqrt{6}} = \frac{9}{12 \sqrt{6}}$$. So, $$f'(6) = -\frac{1}{12 \sqrt{6}} + \frac{9}{12 \sqrt{6}} = \frac{8}{12 \sqrt{6}}$$ 8. **Simplify the fraction:** $$\frac{8}{12} = \frac{2}{3}$$, so $$f'(6) = \frac{2}{3 \sqrt{6}}$$ 9. **Rationalize the denominator:** Multiply numerator and denominator by $$\sqrt{6}$$: $$f'(6) = \frac{2 \sqrt{6}}{3 \sqrt{6} \cdot \sqrt{6}} = \frac{2 \sqrt{6}}{3 \cdot 6} = \frac{2 \sqrt{6}}{18} = \frac{\sqrt{6}}{9}$$ **Final answer:** $$\boxed{\frac{\sqrt{6}}{9}}$$