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

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

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

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

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

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