If e^y(x+1)=1, then frac{d^2y}{dx^2} - left(f | vedclass

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(D) Given $e^y(x+1) = 1$.Taking the natural logarithm on both sides, we get: $y + \ln(x+1) = 0 \implies y = -\ln(x+1)$.Differentiating with respect to

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(D) Given $e^y(x+1) = 1$.Taking the natural logarithm on both sides, we get: $y + \ln(x+1) = 0 \implies y = -\ln(x+1)$.Differentiating with respect to $x$: $\frac{dy}{dx} = -\frac{1}{x+1} = -(x+1)^{-1}$.Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2} = \left(\frac{1}{x+1}\right)^2$.Now, calculating the expression: $\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{x+1}\right)^2 - \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2} - \frac{1}{(x+1)^2} = 0$.

If $e^y(x+1)=1$, then $\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 = $ . . . . . . .

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