CUET UG 2025 — Mathematics Calculus

Given $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots + \infty}}}$y=sinx+sinx+sinx+⋯+∞​​​

The expression inside the first square root contains the same infinite pattern as the original expression:

$$y = \sqrt{\sin x + \underbrace{\sqrt{\sin x + \sqrt{\sin x + \cdots}}}_{\text{equals } y}}$$y=sinx+equals ysinx+sinx+⋯​​​​​

Therefore:

$$y = \sqrt{\sin x + y}$$y=sinx+y​

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$$y = \sqrt{\sin x + y}$$y=sinx+y​

$$y^2 = \sin x + y$$y2=sinx+y

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Differentiating both sides with respect to $x$x:

$$\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$$dxd​(y2)=dxd​(sinx+y)

$$2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$$2ydxdy​=cosx+dxdy​

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$$2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$$2ydxdy​=cosx+dxdy​

$$2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x$$2ydxdy​−dxdy​=cosx

$$\frac{dy}{dx}(2y - 1) = \cos x$$dxdy​(2y−1)=cosx

$$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$$dxdy​=2y−1cosx​