Question

Derivative of \[ y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\cdots\infty}}} \] is:

• A.
• B.
• C.
• D.

Hint:

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

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This is a derivative of an infinite nested radical function. We can replace the repeating part with \( y \).
Step 3: Detailed Explanation:
The function is:
\[ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots}}} \]
Since it is an infinite series, the part inside the first square root is also equal to \( y \):
\[ y = \sqrt{\sin x + y} \]
Squaring both sides:
\[ y^2 = \sin x + y \]
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \]
Grouping \( \frac{dy}{dx} \) terms:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x \]
\[ \frac{dy}{dx} (2y - 1) = \cos x \]
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]
Step 4: Final Answer:
The derivative is \( \frac{\cos x}{2y - 1} \).