Question

If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots}}} \), then \( \frac{dy}{dx} \) is:

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

Hint:

For $y = \sqrt{f(x) + \sqrt{f(x) + \dots}}$, the derivative is always $f'(x)/(2y-1)$.

Answer 1

The Correct Option is A

Step 1: Concept
Write the function as $y = \sqrt{\sin x + y}$ and square both sides.
Step 2: Analysis
$y^{2} = \sin x + y$. Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$.
Step 3: Conclusion
$2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx}(2y - 1) = \cos x \Rightarrow \frac{dy}{dx} = \frac{\cos x}{2y-1}$.
Final Answer: (A)