If e^y(x+1)=1,then frac{d^2 y}{d x^2} = . . | vedclass

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(D) Given the equation $e^y(x+1)=1$. Taking the natural logarithm on both sides: $\ln(e^y) + \ln(x+1) = \ln(1)$. $y + \ln(x+1) = 0$. Differentiating w

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$\left(\frac{d y}{d x}\right)^2$

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(D) Given the equation $e^y(x+1)=1$. Taking the natural logarithm on both sides: $\ln(e^y) + \ln(x+1) = \ln(1)$. $y + \ln(x+1) = 0$. Differentiating with respect to $x$: $\frac{d y}{d x} + \frac{1}{x+1} = 0$. So,$\frac{d y}{d x} = -\frac{1}{x+1} = -(x+1)^{-1}$. Differentiating again with respect to $x$: $\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$. Since $\frac{d y}{d x} = -\frac{1}{x+1}$,we have $\left(\frac{d y}{d x}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$. Therefore,$\frac{d^2 y}{d x^2} = \left(\frac{d y}{d x}\right)^2$.

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

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