If $e^y(x+1)= 1$, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{d^2y}{dx^2}=(\frac{dy}{dx})^2$
Given: $e^y (x + 1) = 1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}[e^y (x+1)] = 0$
$e^y \frac{dy}{dx} (x+1) + e^y = 0$
$\frac{dy}{dx} (x+1) + 1 = 0 \Rightarrow \frac{dy}{dx} = -\frac{1}{x+1}$
Differentiate again to find second derivative:
$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(-\frac{1}{x+1}\right) = \frac{1}{(x+1)^2}$
Express in terms of $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{1}{x+1} \Rightarrow ( \frac{dy}{dx} )^2 = \frac{1}{(x+1)^2}$
So, $ \frac{d^2 y}{dx^2} = (\frac{dy}{dx})^2 $