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

(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:

Given: $e^y(x+1) = 1$

Dividing by $(x+1)$:

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

Taking natural logarithm:

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

$y = -\ln(x+1)$

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Differentiating with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}[-\ln(x+1)]$

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

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Differentiating again with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{1}{x+1}\right]$

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

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

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

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

The second derivative is $\frac{1}{(x+1)^2}$, which is positive.

Option (A) is FALSE.

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

$\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2$

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

This equals $\frac{d^2y}{dx^2}$.

Option (B) is TRUE.

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

Substituting $x=0$:

$\frac{1}{(0+1)^2} = \frac{1}{1} = 1$

Option (C) is FALSE.

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

Substituting $x=0$:

$\frac{1}{(0+1)^2} = 1$

Option (D) is TRUE.

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

Substituting $x=1$:

$\frac{1}{(1+1)^2} = \frac{1}{4}$

Option (E) is TRUE.

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Therefore, statements (B), (D), and (E) are TRUE.