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:
Answer & explanation
Correct answer: option 1
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).