(v.) If \( y=\sqrt{x+\sqrt{x+\ldots}} t_{0} \propto \), find \( \frac{d y}{d x} \)

To solve the problem, we start with the equation given for \( y \): \[ y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} \] This expression can be simplified by recognizing that the nested square root converges to a limit. We can set: \[ y = \sqrt{x + y} \] Next, we will square both sides to eliminate the square root: \[ y^2 = x + y \] Rearranging this equation gives us: \[ y^2 - y - x = 0 \] This is a quadratic equation in \( y \). We can use the quadratic formula to solve for \( y \): \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -1 \), and \( c = -x \). Plugging in these values, we have: \[ y = \frac{1 \pm \sqrt{1 + 4x}}{2} \] Since \( y \) must be non-negative (as it represents a square root), we take the positive root: \[ y = \frac{1 + \sqrt{1 + 4x}}{2} \] Now, we need to find \( \frac{dy}{dx} \). We will differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1 + \sqrt{1 + 4x}}{2} \right) \] Using the chain rule, we differentiate the square root: \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 + 4x}} \cdot \frac{d}{dx}(1 + 4x) \] Calculating the derivative of \( 1 + 4x \): \[ \frac{d}{dx}(1 + 4x) = 4 \] Substituting this back into our expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{4}{\sqrt{1 + 4x}} = \frac{2}{\sqrt{1 + 4x}} \] Thus, the final result is: \[ \frac{dy}{dx} = \frac{2}{\sqrt{1 + 4x}} \]