Partial Derivatives of e^x/(1 + e^y)

Only one variable sits in the denominator

The variable $x$ appears only in the numerator, so when $x$ moves the denominator is a fixed number and you divide its derivative straight through. No quotient rule is needed.

For $y$, rewrite the function as $e^{x}\left(1+e^{y}\right)^{-1}$ and use the power rule with the chain rule. The outside gives $-\left(1+e^{y}\right)^{-2}$ and the inside gives $e^{y}$.

At the origin the height is $\frac{1}{2}$, the $x$ slope is $\frac{1}{2}$, and the $y$ slope is $-\frac{1}{4}$.

The mistake: quotient rule where there is nothing to differentiate

• Running the full quotient rule for the $x$ partial adds a term containing $\frac{\partial}{\partial x}\left(1+e^{y}\right)$. That derivative is $0$, so the extra term is not wrong so much as wasted, and it is where sign errors creep in.
• Writing $-\frac{e^{x}}{\left(1+e^{y}\right)^{2}}$ for the $y$ partial forgets the inside derivative $e^{y}$ from the chain rule.
• Writing $\frac{e^{x}e^{y}}{\left(1+e^{y}\right)^{2}}$ loses the minus sign that the power $-1$ produces.

The sign is checkable without any algebra. Raising $y$ makes the denominator larger and the fraction smaller, so $\frac{\partial f}{\partial y}$ must be negative at every point. The correct formula is a negative number over a positive square, so it is.

A logistic curve in disguise

The factor $\frac{1}{1+e^{y}}$ is $1-\sigma(y)$, where $\sigma$ is the logistic function, so this surface is an exponential in $x$ scaled by a logistic decay in $y$. Its $y$ derivative is minus $e^{x}$ times the logistic derivative $\sigma'(y) = \frac{e^{y}}{\left(1+e^{y}\right)^{2}}$.

Because $\sigma'$ peaks at $y = 0$ and decays in both directions, the surface is most sensitive to $y$ near the $x$-axis and nearly flat in $y$ far from it, while it keeps growing without bound in $x$. The function is positive everywhere and has no critical points.