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

One variable sits only in the numerator

Look at where each variable lives. The variable $y$ appears only in the numerator, and only to the first power, so the $y$-partial is the reciprocal of the denominator with nothing else attached.

The variable $x$ appears in both places, so the $x$-partial is a real quotient rule problem. The numerator $x+y$ has $x$-derivative 1, and the denominator $1+e^{x}$ has $x$-derivative $e^{x}$.

You can factor the $e^{x}$ terms if you want a compact form: the numerator is $1 + e^{x}\left(1 - x - y\right)$. Both versions are correct, and the factored one makes the sign change easier to locate.

The mistake: differentiating the exponential in the wrong variable

Because $e^{x}$ has no $y$ in it, its $y$-derivative is zero. Students who run the full quotient rule for $\frac{\partial f}{\partial y}$ sometimes still write a $-\left(x+y\right)e^{x}$ term, which does not belong: nothing in the denominator responds to $y$.

• $\frac{\partial}{\partial y}\left(1+e^{x}\right) = 0$, so the second quotient rule term vanishes and one factor of $1+e^{x}$ cancels.
• $\frac{\partial}{\partial x}\left(x+y\right) = 1$, not $1+y$. The $y$ is a constant when $x$ moves.
• $1+e^{x}$ is never zero, since $e^{x}>0$. The function is smooth on the whole plane.

Notice also that $\frac{\partial f}{\partial y}$ is strictly positive everywhere. That says the surface always rises as $y$ increases, which you can see directly in the original formula: increasing $y$ increases the numerator while leaving the positive denominator alone.

Checking at the origin

At $(0,0)$ we have $e^{0}=1$, so the denominator is 2 and its square is 4. The $x$-partial numerator is $1 + 1 - 0 = 2$.

Equal components mean the tangent plane at the origin rises at the same rate in both coordinate directions, and the steepest ascent from there points along the diagonal $\langle 1, 1 \rangle$.