Multivariable calculus

Critical Points of e^(xy)

Q: Can I always classify a composition by looking at the exponent alone?

Only when the outer function is strictly increasing, as e^{t} is. Then \nabla f = g'(u)\nabla u with g' > 0, so the critical points match and the local ordering of values is preserved, which means the classification transfers exactly. If the outer function is not monotone, say g(t) = t^{2}, then g'(u) = 0 can create critical points that the exponent does not have, and maxima can turn into minima.

f(x,y) = e^(xy) has exactly one critical point, the origin, and it is a saddle with D = -1. Both partials carry a factor of e^(xy), which is never zero, so f_x = y e^(xy) vanishes only when y = 0 and f_y = x e^(xy) only when x = 0. At the origin f_xx = f_yy = 0 and f_xy = 1.

f(x,y) = e^{xy}

$\left(0, 0\right)$ saddle pointdiscriminant D = -1

A factor that can never be zero

Differentiating $e^{xy}$ is a chain rule with inside function $u = xy$ , so each partial is the exponential times the corresponding derivative of the exponent.

\frac{\partial f}{\partial x} = y\,e^{xy}, \qquad \frac{\partial f}{\partial y} = x\,e^{xy}

The exponential is strictly positive for every real input, so it can never be the reason a product is zero. Divide it out of both equations and what is left is $y = 0$ and $x = 0$ . The origin is the only critical point, and this time the list really is complete.

The second partials need the product rule only in the mixed case, since $f_x = y e^{xy}$ has a loose factor of $y$ that the $y$ derivative sees.

f_{xx} = y^{2}e^{xy}, \qquad f_{yy} = x^{2}e^{xy}, \qquad f_{xy} = (1 + xy)\,e^{xy}

At the origin the exponential equals 1, so those become $0$ , $0$ and $1$ , and the discriminant is $D = -1$ . A negative discriminant is a saddle.

Reading the answer off the exponent

There is a shortcut here that is worth understanding rather than just using. The outer function $t \mapsto e^{t}$ is strictly increasing, so $f = e^{u}$ is large exactly where $u = xy$ is large and small exactly where $u$ is small.

That has two consequences. First, by the chain rule $\nabla f = e^{u}\nabla u$ , and since $e^{u} > 0$ , the gradient of $f$ vanishes precisely where the gradient of $u$ does. The two surfaces have the same critical points. Second, a strictly increasing outer function preserves the local ordering of values, so a point that is a local maximum of $u$ is a local maximum of $f$ , and likewise for minima and saddles.

\nabla f = e^{xy}\,\nabla(xy) = e^{xy}\,(y, x)

The exponent $u = xy$ is the standard saddle: it is positive in the first and third quadrants and negative in the second and fourth, so it takes both signs arbitrarily close to the origin. Its discriminant is $u_{xx}u_{yy} - u_{xy}^{2} = -1$ , the same value the full computation gave for $f$ . The exponential wraps the surface but does not change its shape near the origin.

The mistake students make

The costly error is computing the mixed partial as $f_{xy} = xy\,e^{xy}$ , which drops the term the product rule contributes. Differentiating $y\,e^{xy}$ with respect to $y$ means differentiating a product of $y$ and $e^{xy}$ , and the loose $y$ has a derivative of its own.

\frac{\partial}{\partial y}\left(y\,e^{xy}\right) = 1 \cdot e^{xy} + y \cdot x\,e^{xy} = (1 + xy)\,e^{xy}

The missing term is exactly the one that matters. At the origin the correct mixed partial is 1, while the truncated version gives 0. That turns $D = -1$ into $D = 0$ , and a clear saddle gets reported as a degenerate point where the test fails. The tell is that a page of exponential algebra ends with the test giving up; if that happens, check the product rule first.

A second slip is trying to solve $y\,e^{xy} = 0$ by setting the exponential to zero. It has no solutions: $e^{t} > 0$ for every real $t$ . Recognising that early is what makes the critical point search on exponential surfaces a one line job.

Frequently asked questions

Can I always classify a composition by looking at the exponent alone?

Only when the outer function is strictly increasing, as $e^{t}$ is. Then $\nabla f = g'(u)\nabla u$ with $g' > 0$ , so the critical points match and the local ordering of values is preserved, which means the classification transfers exactly. If the outer function is not monotone, say $g(t) = t^{2}$ , then $g'(u) = 0$ can create critical points that the exponent does not have, and maxima can turn into minima.

Is the origin really the only critical point?

Yes, and the argument is short. Both partials are a positive quantity times a single variable: $f_x = y e^{xy}$ and $f_y = x e^{xy}$ . Since $e^{xy}$ is never zero, $f_x = 0$ requires $y = 0$ and $f_y = 0$ requires $x = 0$ . The only point meeting both conditions is $(0,0)$ , so the list on this page is genuinely complete.