1. **State the problem:** We are given the function $$f(x) = -\frac{2 \sqrt{x}}{5} + \frac{2 \sqrt{x^3}}{3}$$ and need to find its derivative at $$x=4$$, i.e., $$f'(4)$$. 2. **Rewrite the function in terms of exponents:** Recall that $$\sqrt{x} = x^{1/2}$$ and $$\sqrt{x^3} = (x^3)^{1/2} = x^{3/2}$$. So, $$f(x) = -\frac{2}{5} x^{1/2} + \frac{2}{3} x^{3/2}$$. 3. **Differentiate term-by-term:** Using the power rule $$\frac{d}{dx} x^n = n x^{n-1}$$, \begin{align*} f'(x) &= -\frac{2}{5} \cdot \frac{1}{2} x^{1/2 - 1} + \frac{2}{3} \cdot \frac{3}{2} x^{3/2 - 1} \\ &= -\frac{2}{5} \cdot \frac{1}{2} x^{-1/2} + \frac{2}{3} \cdot \frac{3}{2} x^{1/2} \\ &= -\frac{1}{5} x^{-1/2} + 1 \cdot x^{1/2} \\ &= -\frac{1}{5} x^{-1/2} + x^{1/2} \end{align*} 4. **Evaluate at $$x=4$$:** \begin{align*} f'(4) &= -\frac{1}{5} \cdot 4^{-1/2} + 4^{1/2} \\ &= -\frac{1}{5} \cdot \frac{1}{\sqrt{4}} + \sqrt{4} \\ &= -\frac{1}{5} \cdot \frac{1}{2} + 2 \\ &= -\frac{1}{10} + 2 \\ &= \frac{-1 + 20}{10} = \frac{19}{10} \end{align*} **Final answer:** $$f'(4) = \frac{19}{10}$$.