Let \(e^y (x + 1) = 1\). Then which of the following are TRUE?
(A) \(\displaystyle\frac{d^2y}{dx^2}=-\frac{1}{(x + 1)^2}\)
(B) \(\displaystyle\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2\)
(C) \(\displaystyle\left. \frac{d^2y}{dx^2}\right|_{x=0}=-1\)
(D) \(\displaystyle\left. \frac{d^2y}{dx^2}\right|_{x=0}=1\)
(E) \(\displaystyle\left. \frac{d^2y}{dx^2}\right|_{x=1}=\frac{1}{4}\)
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
Taking logarithms, y can be written explicitly:
\(y = -\log_e(x + 1)\)
\( \displaystyle \frac{dy}{dx} = -\frac{1}{x + 1} \)
\( \displaystyle \frac{d^2y}{dx^2} = \frac{1}{(x + 1)^2} \)
So \(\frac{dy}{dx} = -\frac{1}{x + 1}\) and \(\frac{d^2y}{dx^2} = \frac{1}{(x + 1)^2}\).
Checking the statements
(A) \(\frac{d^2y}{dx^2}=-\frac{1}{(x + 1)^2}\) — Incorrect. The second derivative is \(+\frac{1}{(x + 1)^2}\); the sign is wrong.
(B) \(\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2\) — Correct. \(\left(\frac{dy}{dx}\right)^2 = \frac{1}{(x + 1)^2}\), which equals \(\frac{d^2y}{dx^2}\).
(C) \(\left. \frac{d^2y}{dx^2}\right|_{x=0}=-1\) — Incorrect. At x = 0 the value is \(\frac{1}{1^2} = 1\).
(D) \(\left. \frac{d^2y}{dx^2}\right|_{x=0}=1\) — Correct. At x = 0 the value is \(\frac{1}{1^2} = 1\).
(E) \(\left. \frac{d^2y}{dx^2}\right|_{x=1}=\frac{1}{4}\) — Correct. At x = 1 the value is \(\frac{1}{2^2} = \frac{1}{4}\).