If \(y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \ldots}}}}\), then \(\displaystyle\frac{dy}{dx}\) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{x(2y-1)}\)
The expression under the first root, after \(\log x\), is the same infinite expression y again. So
\(y = \sqrt{\log x + y}\)
\(y^{2} = \log x + y\)
Differentiating both sides with respect to x (implicit differentiation):
\( \displaystyle 2y\frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} \)
\( \displaystyle (2y - 1)\frac{dy}{dx} = \frac{1}{x} \)
\( \displaystyle \frac{dy}{dx} = \frac{1}{x(2y-1)} \)