1. **State the problem:** We are given the function $$f(x) = \frac{5\sqrt{x}}{3} + 2\sqrt{x^3}$$ and need to find its derivative at $$x=1$$, i.e., $$f'(1)$$. We will express the answer as a single simplified fraction. 2. **Rewrite the function using exponents:** Recall that $$\sqrt{x} = x^{\frac{1}{2}}$$ and $$\sqrt{x^3} = (x^3)^{\frac{1}{2}} = x^{\frac{3}{2}}$$. So, $$f(x) = \frac{5}{3} x^{\frac{1}{2}} + 2 x^{\frac{3}{2}}$$ 3. **Differentiate term-by-term:** Using the power rule $$\frac{d}{dx} x^n = n x^{n-1}$$, \begin{align*} f'(x) &= \frac{5}{3} \cdot \frac{1}{2} x^{\frac{1}{2} - 1} + 2 \cdot \frac{3}{2} x^{\frac{3}{2} - 1} \\ &= \frac{5}{6} x^{-\frac{1}{2}} + 3 x^{\frac{1}{2}} \end{align*} 4. **Evaluate the derivative at $$x=1$$:** $$f'(1) = \frac{5}{6} (1)^{-\frac{1}{2}} + 3 (1)^{\frac{1}{2}} = \frac{5}{6} + 3 = \frac{5}{6} + \frac{18}{6} = \frac{23}{6}$$ 5. **Final answer:** $$f'(1) = \frac{23}{6}$$