If \(y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots + \infty}}}\) then \(\displaystyle\frac{dy}{dx}\) equals to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\cos x}{2y-1}\)
The infinite root repeats itself: after the first \(\sin x\), the expression under the outer root is y again. So
\( \displaystyle y = \sqrt{\sin x + y} \)
\( y^2 = \sin x + y \)
Differentiating with respect to x:
\( \displaystyle 2y\frac{dy}{dx} = \cos x + \frac{dy}{dx} \)
\( \displaystyle (2y - 1)\frac{dy}{dx} = \cos x \)
\( \displaystyle \frac{dy}{dx} = \frac{\cos x}{2y - 1} \)