Question:

<b>Second Derivative of Inverse — dy/dx vs dx/dy</b><br>$\frac{dy}{dx} = 1 + e^x$, so $\frac{dx}{dy} = \frac{1}{1+e^x}$.<br>$\frac{d^2x}{dy^2} = \frac{d}{dy}\\!\left(\frac{dx}{dy}\right) = \frac{d}{dx}\\!\left(\frac{1}{1+e^x}\right) \cdot \frac{dx}{dy}$<br>$= \frac{-e^x}{(1+e^x)^2} \cdot \frac{1}{1+e^x} = \frac{-e^x}{(1+e^x)^3}$<br>Wait, rechecking: $\frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{(dy/dx)^3} = -\frac{e^x}{(1+e^x)^3}$.<br>The answer key says B: $-\frac{e^x}{(1+e^x)^2}$...<br><b>Key concept:</b> $\frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{(dy/dx)^3}$.<br><b>Trap:</b> Confusing $\frac{d^2x}{dy^2}$ with $\frac{1}{d^2y/dx^2}$.