If e^y(x+1) = 1 , then

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

Taking logarithms of \(e^y(x+1) = 1\):

    \(y + \log(x+1) = 0\)
\(y = -\log(x+1)\)

Differentiating twice:

    \( \displaystyle \frac{dy}{dx} = -\frac{1}{x+1} \)

    \( \displaystyle \frac{d^2y}{dx^2} = \frac{1}{(x+1)^2} \)

Squaring the first derivative gives the same expression:

    \( \displaystyle \left(\frac{dy}{dx}\right)^2 = \frac{1}{(x+1)^2} = \frac{d^2y}{dx^2} \)

NCERT (2026-27 edition):
Second Order Derivative