Question:

<b>Second Derivative of Inverse Function</b><br> $y = x+e^x$, so $\dfrac{dy}{dx} = 1+e^x$, meaning $\dfrac{dx}{dy} = \dfrac{1}{1+e^x}$.<br> For the second derivative: $\dfrac{d^2x}{dy^2} = \dfrac{d}{dy}\!\left(\dfrac{dx}{dy}\right) = \dfrac{d}{dy}\!\left(\dfrac{1}{1+e^x}\right)$.<br> $= \dfrac{d}{dx}\!\left(\dfrac{1}{1+e^x}\right)\cdot\dfrac{dx}{dy} = \dfrac{-e^x}{(1+e^x)^2}\cdot\dfrac{1}{1+e^x} = -\dfrac{e^x}{(1+e^x)^3}$.<br> <b>Key concept:</b> $\dfrac{d^2x}{dy^2} = -\dfrac{d^2y/dx^2}{(dy/dx)^3}$. Here $d^2y/dx^2 = e^x$ and $(dy/dx)^3=(1+e^x)^3$, giving $-e^x/(1+e^x)^3$.<br> <b>Trap:</b> Writing $d^2x/dy^2 = 1/(d^2y/dx^2)$ — completely wrong; use the formula above.