Partial Derivatives of ln(e^x + e^y)

One logarithm, one frozen exponential

Set $u = e^{x}+e^{y}$ and use $\frac{\partial}{\partial x}\ln(u) = \frac{u_x}{u}$. When $x$ moves, $e^{y}$ is a constant with derivative $0$, so $u_x = e^{x}$ and only the first term survives in the numerator.

At $(0,0)$ both partials are $\frac{1}{2}$ and the height is $\ln 2 \approx 0.693$. At $(1,0)$ they are $\frac{e}{e+1} \approx 0.731$ and $\frac{1}{e+1} \approx 0.269$: the larger input takes the larger share of the slope.

The mistake: turning a log of a sum into a sum of logs

• $\ln(e^{x}+e^{y})$ is not $x + y$. That would be $\ln(e^{x}e^{y})$, a product rather than a sum, and it would make both partials equal $1$.
• Writing $\frac{\partial f}{\partial x} = \frac{e^{x}+e^{y}}{e^{x}+e^{y}} = 1$ differentiates $e^{y}$ as though $y$ moved too. Only $e^{x}$ belongs in the numerator of the $x$ partial.

There is a clean check on the diagonal. Since $e^{x}+e^{x} = 2e^{x}$, the function reduces to $f(x,x) = x + \ln 2$, a plane of slope $1$ along that line. The two partials there are $\frac{1}{2}$ each, and they add to the $1$ you see.

Why the two partials add to one

The sum is $1$ at every point, so the two slopes are weights that trade against each other. When one input dominates, its weight runs to $1$ and the other to $0$: at $(3,0)$ the $x$ partial is already about $0.953$.

That is the sense in which this log-sum-exp function is a smooth maximum. It stays between $\max(x,y)$ and $\max(x,y)+\ln 2$, and its partials switch smoothly between $0$ and $1$ instead of jumping the way the slopes of $\max(x,y)$ do along $y = x$. Neither partial is ever zero, so the surface has no critical point.