If e^y = log x , then which of the following is true?

The correct answer is Option (4) → \(x\displaystyle\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0\)

Let \(y_1 = \frac{dy}{dx}\) and \(y_2 = \frac{d^2y}{dx^2}\). Differentiating \(e^y = \log x\):

    \( \displaystyle e^y\, y_1 = \frac{1}{x} \)

    \(x\, e^y\, y_1 = 1\)

Differentiating again by the product rule (three factors x, \(e^y\), \(y_1\)):

    \(e^y y_1 + x\, e^y y_1 \cdot y_1 + x\, e^y y_2 = 0\)

Dividing by \(e^y\) (never zero):

    \(y_1 + x\, y_1^2 + x\, y_2 = 0\)

    \( \displaystyle x\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0 \)