AP DSC School Assistant Math Model Test 2

Concept:Infinite nested radicals can often be simplified by recognizing a self-repeating pattern. Here, $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}$ implies that squaring both sides yields $y^2 = x + y$, which then can be differentiated to find $\frac{dy}{dx}$.Explanation:Given $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}$. Squaring both sides gives:$y^{2} = x + \sqrt{x + \sqrt{x + \cdots}}$The infinite expression under the square root is exactly the same as the original $y$, so:$y^{2} = x + y \quad \text{...(1)}$Now differentiate equation (1) with respect to $x$:$\frac{d}{dx}(y^{2}) = \frac{d}{dx}(x) + \frac{d}{dx}(y)$$2y \frac{dy}{dx} = 1 + \frac{dy}{dx}$Bring terms with $\frac{dy}{dx}$ to one side:$2y \frac{dy}{dx} - \frac{dy}{dx} = 1$$(2y - 1) \frac{dy}{dx} = 1$Hence,$\frac{dy}{dx} = \frac{1}{2y - 1}$Answer:$\frac{dy}{dx} = \frac{1}{2y - 1}$   (Option D)