Question

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

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

Hint:

For infinite repeated radicals, replace the repeated part again by \(y\).

Answer 1

The Correct Option is D

Step 1: Given: \[ y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\cdots}}} \]

Step 2: Since the repeated radical is again \(y\), write: \[ y=\sqrt{\sin x+y} \]

Step 3: Squaring both sides: \[ y^2=\sin x+y \]

Step 4: Differentiate: \[ 2y\frac{dy}{dx}=\cos x+\frac{dy}{dx} \]

Step 5: Bring derivative terms together: \[ (2y-1)\frac{dy}{dx}=\cos x \] \[ \frac{dy}{dx}=\frac{\cos x}{2y-1} \] \[ \frac{dy}{dx}=\frac{-\cos x}{1-2y} \] \[ \boxed{\frac{-\cos x}{1-2y}} \]