If $e^y(x + 1) = 1$ and $\frac{d^2y}{dx^2}= k(\frac{dy}{dx})^2$, then $k$ is equal to

1

The correct answer is Option (2) → 1 **

Given:

$e^{y}(x+1)=1$

Rewrite:

$e^{y}=\frac{1}{x+1}$

Take log:

$y = -\ln(x+1)$

Differentiate:

$\frac{dy}{dx} = -\frac{1}{x+1}$

Differentiate again:

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

Compute $\left(\frac{dy}{dx}\right)^{2}$:

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

Given:

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

So:

$\frac{1}{(x+1)^{2}} = k \cdot \frac{1}{(x+1)^{2}}$

Thus:

$k = 1$

Final Answer: $1$