Question

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

• 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 radicals, replace the repeating radical by \(y\). Then square both sides and differentiate implicitly.

Answer 1

The Correct Option is D

We are given: \[ y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\cdots}}}. \] Since the expression under the first square root repeats infinitely, the repeated radical is again equal to \(y\). So we can write: \[ y=\sqrt{\sin x+y}. \] Now square both sides: \[ y^2=\sin x+y. \] Bring all terms to one side: \[ y^2-y=\sin x. \] Now differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(y^2-y)=\frac{d}{dx}(\sin x). \] Differentiate term by term: \[ 2y\frac{dy}{dx}-\frac{dy}{dx}=\cos x. \] Take \(\frac{dy}{dx}\) common: \[ (2y-1)\frac{dy}{dx}=\cos x. \] Therefore, \[ \frac{dy}{dx}=\frac{\cos x}{2y-1}. \] Now multiply numerator and denominator by \(-1\): \[ \frac{dy}{dx}=\frac{-\cos x}{1-2y}. \] Hence, \[ \frac{dy}{dx}=\frac{-\cos x}{1-2y}. \]