Question

If \(y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}\) 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 any function of the form \(y = \sqrt{f(x) + \sqrt{f(x) + \dots}}\), the derivative follows the pattern \(\frac{dy}{dx} = \frac{f'(x)}{2y-1}\). Memorizing this general form can lead to an instant answer for such problems.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an infinitely nested radical function and asked to find its derivative.

Step 2: Key Formula or Approach:
The key to solving this is to recognize the repeating pattern. The entire expression under the first square root is simply the original function \(y\) itself. This allows us to write a simple implicit equation for \(y\), which can then be differentiated.

Step 3: Detailed Explanation:
The given function is:
\[ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}} \]
We can rewrite this as:
\[ y = \sqrt{\sin x + y} \]
To remove the square root, we square both sides of the equation:
\[ y^2 = \sin x + y \]
Now, we differentiate this equation implicitly with respect to \(x\):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(y) \]
\[ 2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \]
Now, we need to solve for \(\frac{dy}{dx}\). Group all the \(\frac{dy}{dx}\) terms on one side:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x \]
Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (2y - 1) = \cos x \]
Isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]
To match the given options, we can multiply the numerator and denominator by -1:
\[ \frac{dy}{dx} = \frac{-\cos x}{-(2y - 1)} = \frac{-\cos x}{1 - 2y} \]

Step 4: Final Answer:
The derivative \(\frac{dy}{dx}\) is \(-\frac{\cos x}{1 - 2y}\).