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

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(B) Given $x+1 = e^{-y}$.Taking the natural logarithm on both sides,we get $\ln(x+1) = -y$,which implies $y = -\ln(x+1)$.Differentiating with respect

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

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(B) Given $x+1 = e^{-y}$.Taking the natural logarithm on both sides,we get $\ln(x+1) = -y$,which implies $y = -\ln(x+1)$.Differentiating with respect to $x$,we get $\frac{d y}{d x} = -\frac{1}{x+1}$.Differentiating again with respect to $x$,we get $\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$.Since $\frac{d y}{d x} = -\frac{1}{x+1}$,we have $\left(\frac{d y}{d x}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$.Thus,$\frac{d^2 y}{d x^2} = \left(\frac{d y}{d x}\right)^2$.

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

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