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

The correct answer is Option (1) → \( \displaystyle\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2 \)

Taking logarithms gives y explicitly:

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

    \( \displaystyle y=-\log(x+1) \)

Differentiating twice,

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

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

Since \( \left(\frac{dy}{dx}\right)^2=\frac{1}{(x+1)^2} \),

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

NCERT (2026-27 edition):
Second Order Derivative