\[
y = \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x))} + \dots \infty}}
\]
\[ y = \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x))} + \dots \infty}} \]
Show Hint
- \(\frac{\cos(\log(2x))}{2x(2y-1)}\)
- \(\frac{\cos(\log(2x))}{(2y-1)}\)
- \(\frac{\cos(\log(2x))}{x(2y-1)}\)
\(\frac{\sin(\log(2x))}{(2y-1)}\)
The Correct Option is C
Solution and Explanation
- 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: \[ \frac{dy}{dx} = \frac{\cos(\log(2x))}{x(2y-1)} \]
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