1. **State the problem:** We need to find the derivative of the function $$f(x) = \sqrt{1 + \sqrt{1 + x}}$$ and then evaluate it at $$x=8$$. 2. **Recall the formula:** The derivative of $$\sqrt{u}$$ with respect to $$x$$ is $$\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$$ by the chain rule. 3. **Apply the chain rule step-by-step:** Let $$u = 1 + \sqrt{1 + x}$$, so $$f(x) = \sqrt{u}$$. Then, $$f'(x) = \frac{1}{2\sqrt{u}} \cdot u'$$. Next, find $$u'$$: Since $$u = 1 + \sqrt{1 + x}$$, $$u' = 0 + \frac{1}{2\sqrt{1 + x}} \cdot 1 = \frac{1}{2\sqrt{1 + x}}$$. 4. **Combine the derivatives:** $$f'(x) = \frac{1}{2\sqrt{1 + \sqrt{1 + x}}} \cdot \frac{1}{2\sqrt{1 + x}} = \frac{1}{4 \sqrt{1 + \sqrt{1 + x}} \cdot \sqrt{1 + x}}$$. 5. **Evaluate at $$x=8$$:** Calculate inside the radicals: $$\sqrt{1 + 8} = \sqrt{9} = 3$$, so $$\sqrt{1 + \sqrt{1 + 8}} = \sqrt{1 + 3} = \sqrt{4} = 2$$. Therefore, $$f'(8) = \frac{1}{4 \times 2 \times 3} = \frac{1}{24}$$. **Final answer:** $$f'(8) = \frac{1}{24}$$.