If y = x + e^x,then at x = 1,frac{d^2x}{dy^2} | 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,$\frac{dx}{dy} = \frac{1}{1 + e^x}$.Now,differe

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

Vedclass · Answer

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

If $y = x + e^x$,then at $x = 1$,$\frac{d^2x}{dy^2}$ is equal to

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