Question

If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \text{ to } \infty}}} \), then \( \frac{dy}{dx} = \):

• A. \( \frac{\sin x}{2y - 1} \)
• B. \( \frac{\cos x}{y - 1} \)
• C. \( \frac{\cos x}{2y - 1} \)
• D. \( \frac{1}{2y - 1} \)

Hint:

For infinite nested radicals, define the expression as a variable and solve using algebraic manipulation. Then apply implicit differentiation.

Answer 1

The Correct Option is C

We are given: \[ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}} \] Let: \[ y = \sqrt{\sin x + y} \] Now square both sides: \[ y^2 = \sin x + y \Rightarrow y^2 - y = \sin x \quad \cdots (1) \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\sin x) \Rightarrow 2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x \Rightarrow (2y - 1) \frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]