Partial Derivatives of ln(1 + e^(xy))

Peel the layers from the outside in

There are three layers: a logarithm, then $1 + e^{xy}$, then the exponent $xy$. Work outward to inward. With $u = 1+e^{xy}$, the logarithm gives $\frac{u_x}{u}$, and $u_x$ needs the chain rule once more, on $e^{xy}$ with $y$ held fixed.

At $(0,2)$ the exponent is $0$, so $e^{xy} = 1$ and the $x$ partial is $\frac{2}{2} = 1$ while the $y$ partial is $\frac{0}{2} = 0$. At $(1,2)$ the $x$ partial is $\frac{2e^{2}}{1+e^{2}} \approx 1.762$.

The mistake: stopping after one chain rule

• Writing $\frac{y}{1+e^{xy}}$ keeps the inside derivative of the exponent but forgets that $e^{xy}$ itself belongs in the numerator, since $\frac{\partial}{\partial x}\left(1+e^{xy}\right) = y\,e^{xy}$.
• Writing $\frac{e^{xy}}{1+e^{xy}}$ stops one layer too early and never differentiates the exponent $xy$, so the factor $y$ is missing.
• Writing $\frac{1}{1+e^{xy}}$ applies only the logarithm rule and skips both inner steps.

Pick a point where the wrong versions separate from the right one. At $(1,2)$ the correct $x$ partial is about $1.762$, while the version that skips the exponent gives about $0.881$, exactly half. A single well-chosen test point catches this.

The softplus and its logistic slope

Write $t = xy$. Then $f = \ln(1+e^{t})$, the softplus function, and its one-variable derivative is the logistic function, because dividing the numerator and denominator by $e^{t}$ turns $\frac{e^{t}}{1+e^{t}}$ into $\frac{1}{1+e^{-t}}$.

That factor is always between $0$ and $1$, so the size of each partial is controlled by the frozen variable out front. The gradient vanishes only at the origin, where both $x$ and $y$ are zero, and the argument $1+e^{xy}$ is always greater than $1$, so nothing here is ever undefined.