Question:medium

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

Show Hint

For infinite repeated radicals, replace the repeated part again by \(y\).
  • \(\frac{\cos x}{1-2y}\)
  • \(\frac{\sin x}{1-2y}\)
  • \(\frac{-\sin x}{1-2y}\)
  • \(\frac{-\cos x}{1-2y}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The given function is an infinite nested radical. The key to solving such problems is to recognize the repeating pattern and express the function in a simple implicit form, which can then be differentiated.
Step 2: Key Formula or Approach:
Since the expression is infinite, we can write: \[ y = \sqrt{\sin x + y} \] By squaring both sides, we get an implicit equation relating x and y. We then use implicit differentiation to find $\frac{dy}{dx}$.
Step 3: Detailed Explanation:
The given function is $y = \sqrt{\sin x + \sqrt{\sin x + \dots}}$. We can rewrite this as: \[ y = \sqrt{\sin x + y} \] Square both sides to remove the outer square root: \[ y^2 = \sin x + y \] Rearrange the terms to prepare for differentiation: \[ y^2 - y = \sin x \] Now, differentiate both sides with respect to x using implicit differentiation. \[ \frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\sin x) \] \[ 2y\frac{dy}{dx} - 1\frac{dy}{dx} = \cos x \] Factor out $\frac{dy}{dx}$ on the left side: \[ (2y - 1)\frac{dy}{dx} = \cos x \] Solve for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \] Now, we check the options. The denominators are in the form $1-2y$. Let's rewrite our result to match this form. \[ 2y - 1 = -(1 - 2y) \] So, \[ \frac{dy}{dx} = \frac{\cos x}{-(1 - 2y)} = \frac{-\cos x}{1 - 2y} \] Step 4: Final Answer:
The derivative $\frac{dy}{dx}$ is $\frac{-\cos x}{1 - 2y}$. Therefore, option (D) is correct.
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