If y = x + e^x,then frac{d^2x}{dy^2} is: | vedclass

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(B) Given $y = x + e^x$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 1 + e^x$. Therefore,$\frac{dx}{dy} = \frac{1}{1 + e^x} = (1 + e^

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$-\frac{e^x}{(1 + e^x)^3}$

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(B) Given $y = x + e^x$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 1 + e^x$. Therefore,$\frac{dx}{dy} = \frac{1}{1 + e^x} = (1 + e^x)^{-1}$. Differentiating with respect to $y$ using the chain rule: $\frac{d^2x}{dy^2} = \frac{d}{dy} \left( (1 + e^x)^{-1} \right) = -(1 + e^x)^{-2} \cdot \frac{d}{dy}(1 + e^x)$. $\frac{d^2x}{dy^2} = -(1 + e^x)^{-2} \cdot e^x \cdot \frac{dx}{dy}$. Substituting $\frac{dx}{dy} = \frac{1}{1 + e^x}$: $\frac{d^2x}{dy^2} = -\frac{e^x}{(1 + e^x)^2} \cdot \frac{1}{1 + e^x} = -\frac{e^x}{(1 + e^x)^3}$.

If $y = x + e^x$,then $\frac{d^2x}{dy^2}$ is:

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