If \( y=\sqrt{\sin x+\sqrt{\sin x}+\ldots \text { to }} \alpha \), find \( \frac{d y}{d x} \)

To solve the problem, we first need to understand the expression given for \( y \): \[ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}} \] This expression can be interpreted as an infinite nested radical. We can denote the entire expression by \( y \) itself: \[ y = \sqrt{\sin x + y} \] Next, we will square both sides to eliminate the square root: \[ y^2 = \sin x + y \] Rearranging this equation gives us: \[ y^2 - y - \sin x = 0 \] This is a quadratic equation in \( y \). We can apply the quadratic formula to solve for \( y \): \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -1 \), and \( c = -\sin x \). Plugging in these values, we have: \[ y = \frac{1 \pm \sqrt{1 + 4\sin x}}{2} \] Since \( y \) represents a length (as it is derived from a square root), we take the positive root: \[ y = \frac{1 + \sqrt{1 + 4\sin x}}{2} \] Now, we need to find \( \frac{dy}{dx} \). We will differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{d}{dx} \left( 1 + \sqrt{1 + 4\sin x} \right) \] Using the chain rule, we differentiate \( \sqrt{1 + 4\sin x} \): \[ \frac{d}{dx} \left( \sqrt{1 + 4\sin x} \right) = \frac{1}{2\sqrt{1 + 4\sin x}} \cdot \frac{d}{dx}(1 + 4\sin x) \] Now, we differentiate \( 1 + 4\sin x \): \[ \frac{d}{dx}(1 + 4\sin x) = 4\cos x \] Putting it all together, we have: \[ \frac{d}{dx} \left( \sqrt{1 + 4\sin x} \right) = \frac{4\cos x}{2\sqrt{1 + 4\sin x}} = \frac{2\cos x}{\sqrt{1 + 4\sin x}} \] Now substituting this back into our expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{2\cos x}{\sqrt{1 + 4\sin x}} = \frac{\cos x}{\sqrt{1 + 4\sin x}} \] Thus, the final result is: \[ \boxed{\frac{\cos x}{\sqrt{1 + 4\sin x}}} \]