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

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

Given

$e^y=\log x$

Differentiate with respect to $x$

$e^y\frac{dy}{dx}=\frac{1}{x}$

Substitute $e^y=\log x$

$\frac{dy}{dx}=\frac{1}{x\log x}$

Differentiate again

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x\log x}\right)$

$=-\frac{1}{x^2\log x}-\frac{1}{x^2(\log x)^2}$

Now evaluate

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

$=x\left(-\frac{1}{x^2\log x}-\frac{1}{x^2(\log x)^2}\right) +x\left(\frac{1}{x^2(\log x)^2}\right) +\frac{1}{x\log x}$

$=-\frac{1}{x\log x}-\frac{1}{x(\log x)^2} +\frac{1}{x(\log x)^2} +\frac{1}{x\log x}$

$=0$

The correct relation is $x\frac{d^2y}{dx^2}+x\left(\frac{dy}{dx}\right)^2+\frac{dy}{dx}=0$.