Question

If \(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+}}}}.....+\) then \(\frac{dy}{dx}\) is :

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

Answer 1

The Correct Option is C

To solve for \(\frac{dy}{dx}\), we start by understanding the recursive relationship of the equation given: \(y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + \ldots}}}\). This expression suggests that it is a nested square root function. Let this be represented as:

\(y = \sqrt{\log x + y}\)

To find \(\frac{dy}{dx}\), we must differentiate both sides with respect to \(x\). Begin by squaring both sides to eliminate the square root:

\(y^2 = \log x + y\)

Rearranging gives us:

\(y^2 - y = \log x\)

Differentiating both sides with respect to \(x\) results in:

\(\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\log x)\)

Using the chain rule on the left side:

\(2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x}\)

Factor out \(\frac{dy}{dx}\):

\((2y - 1) \frac{dy}{dx} = \frac{1}{x}\)

Finally, solving for \(\frac{dy}{dx}\) gives:

\(\frac{dy}{dx} = \frac{1}{x(2y - 1)}\)

Therefore, the answer is:

\(\frac{1}{x(2y-1)}\)