Let $e^y (x + 1) = 1$. Then which of the following are TRUE?

(A) $\frac{d^2y}{dx^2}=-\frac{1}{(x + 1)^2}$
(B) $\frac{d^2y}{dx^2}=(\frac{dy}{dx})^2$
(C) $\left. \frac{d^2y}{dx^2}\right|_{x=0}=-1$
(D) $\left. \frac{d^2y}{dx^2}\right|_{x=0}=1$
(E) $\left. \frac{d^2y}{dx^2}\right|_{x=1}=\frac{1}{4}$

Choose the correct answer from the options given below:

(B), (D) and (E) only

The correct answer is Option (1) → (B), (D) and (E) only

Given

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

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

$y=\ln\left(\frac{1}{x+1}\right)=-\ln(x+1)$

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

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

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

False

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

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

True

(C) $\left.\frac{d^2y}{dx^2}\right|_{x=0}=-1$

$\left.\frac{d^2y}{dx^2}\right|_{x=0}=1$

False

(D) $\left.\frac{d^2y}{dx^2}\right|_{x=0}=1$

True

(E) $\left.\frac{d^2y}{dx^2}\right|_{x=1}=\frac{1}{4}$

True

The correct options are (B) , (D) ana (E).