If $e^y(x + 1) = 1$ and $\frac{d^2y}{dx^2}= k(\frac{dy}{dx})^2$, then $k$ is equal to
Answer & explanation
Correct answer: option 2
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: $1$