Question:

Step 1: Recognize the self-similar structure of the infinite nested radical. Since the expression $y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\cdots}}}}}}}$ is infinite and self-similar, the denominator inside the first square root is also equal to $y$. Therefore, we can write: $$y = \sqrt{\frac{x}{y}}$$ Step 2: Eliminate the square root by squaring both sides. Squaring both sides of the equation: $$y^2 = \frac{x}{y}$$ Multiplying both sides by $y$: $$y^3 = x$$ Step 3: Express $y$ as a power function of $x$. From the relation $y^3 = x$, we can solve for $y$: $$y = x^{1/3}$$ Step 4: Differentiate with respect to $x$. Using the power rule for differentiation: $$\frac{dy}{dx} = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$$ Step 5: Evaluate the derivative at $x = 8$. Substituting $x = 8$: $$\frac{dy}{dx}\bigg|_{x=8} = \frac{1}{3}(8)^{-2/3}$$ Since $8 = 2^3$, we have $(8)^{-2/3} = (2^3)^{-2/3} = 2^{-2} = \frac{1}{4}$ Therefore: $$\frac{dy}{dx}\bigg|_{x=8} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$$ **Final Answer:** The value of $\frac{dy}{dx}$ at $x = 8$ is $\boxed{\frac{1}{12}}$, which corresponds to **Option 2**.