Question

\[ y = \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x))} + \dots \infty}} \]

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

Hint:

When handling infinite sums in calculus, first examine if the sum converges and apply appropriate rules like geometric series sum or differentiation techniques.

Answer 1

The Correct Option is C

- First, recognize \( y \) as an infinite series: \[ y = \sqrt{\sin(\log(2x))} + \sqrt{\sin(\log(2x))} + \sqrt{\sin(\log(2x))} + \ldots \] which is a geometric series with the first term \( \sqrt{\sin(\log(2x))} \) and common ratio 1. - Therefore, the sum of the infinite series is: \[ y = \frac{\sqrt{\sin(\log(2x))}}{1 - 1} = \infty \] This gives \(y = \infty\). Now, we calculate \( \frac{dy}{dx} \): - Applying differentiation, you get the answer as: \[ \frac{dy}{dx} = \frac{\cos(\log(2x))}{x(2y-1)} \]