Second Partial Derivatives of xy e^(x+y)

Product rule twice, then factor

Write $E = e^{x+y}$, so that $\frac{\partial E}{\partial x} = \frac{\partial E}{\partial y} = E$. For the $x$ partial, treat $y$ as constant and apply the product rule to $x \cdot E$.

Differentiate that again in $x$. The bracket $y(1+x)$ contributes $yE$ and the exponential contributes $y(1+x)E$.

For the mixed partial, differentiate $y(1 + x)E$ in $y$. Here $(1 + x)$ is constant and the product $y \cdot E$ needs the rule again, which factors into a tidy symmetric form.

Since swapping $x$ and $y$ leaves $f$ unchanged, $f_{yy} = x(2 + y)E$ follows without further work, and the mixed partial is already symmetric under the swap, which is the Clairaut consistency you should expect.

Classifying both critical points

Because $E$ is never zero, the critical point equations reduce to $y(1 + x) = 0$ and $x(1 + y) = 0$. The solutions are $(0,0)$ and $(-1,-1)$.

At the origin, $f_{xx} = 0$, $f_{yy} = 0$, and $f_{xy} = 1$, so $D = -1 < 0$ and the origin is a saddle point. That matches the surface: $f$ is positive in the first and third quadrants and negative in the second and fourth.

At $(-1,-1)$ the mixed partial vanishes because of the factor $(1 + x)$, and both pure partials are negative.

So $D = e^{-4} > 0$ with $f_{xx} < 0$, giving a local maximum of value $f(-1,-1) = e^{-2} \approx 0.135$.

The mistake students make

The frequent error is differentiating the two factors separately and multiplying the results, which turns $f_x$ into $yE$ instead of $y(1 + x)E$. The product rule adds two terms, it does not multiply two derivatives, and the term that gets dropped is the one that grows: the correct $f_{xx}$ is $y(2 + x)E$, while the mistaken route just repeats $yE$ at every stage.

A useful pattern for this family: differentiating $p(x)E$ in $x$ gives $\left(p + p'\right)E$, so $xE$ goes to $(1 + x)E$ and $(1 + x)E$ goes to $(2 + x)E$. The constant climbs by 1 with each derivative while the $x$ term stays put.

The second slip is classifying the origin as a maximum or minimum because both pure second partials are zero there. Zero pure curvature with a nonzero mixed partial means $D < 0$, which is a saddle. The twist term alone decides it.