Critical Points of the Gaussian Bump e^(-x^2-y^2)

The gradient points straight at the origin

Both partials come from the chain rule on the exponent $u = -(x^{2}+y^{2})$, whose derivatives are $-2x$ and $-2y$.

Written as a vector, the gradient is $-2e^{-(x^{2}+y^{2})}\,(x, y)$, a negative multiple of the position vector, so at every point other than the origin it aims straight back at the origin. That is the picture of a surface which climbs toward the centre from every direction.

Since the exponential never vanishes, the gradient is zero only when $(x, y) = (0,0)$. One critical point, and the list is complete.

For the second partials, differentiate the products above. Each one needs the product rule, since the polynomial factor and the exponential both depend on the variable.

At the origin the exponential is 1, so $f_{xx} = f_{yy} = -2$ and $f_{xy} = 0$. Then $D = 4 > 0$ with a negative diagonal entry, a local maximum of value $f(0,0) = 1$.

Two independent confirmations

The discriminant is one route to the answer. Here is a second one that needs no derivatives at all. The exponent $-(x^{2}+y^{2})$ is at most 0, with equality only at the origin, and $t \mapsto e^{t}$ is strictly increasing. So $f \le e^{0} = 1$ everywhere, with equality only at the origin.

That is stronger than what the second derivative test gives. The test says local maximum; this argument says global maximum, and says the maximiser is unique. Whenever a function is a monotone transformation of something obviously extremal, that comparison is worth doing, because it settles global questions the Hessian cannot see.

A third confirmation comes from the quadratic approximation near the origin. Expanding the exponential gives $f \approx 1 - (x^{2}+y^{2})$, a downward paraboloid whose Hessian is $-2$ on the diagonal and 0 off it. That matches the second partials computed above, which is the check worth running whenever a Hessian looks surprising.

The mistake students make

This surface is constant along every circle centred at the origin, and that leads to a tempting wrong conclusion: since $f$ does not change as you walk around a circle of radius 1, every point on that circle must be critical. It is not. Being constant along a curve makes the directional derivative along the curve zero, which is one direction out of infinitely many.

The gradient measures all directions at once. At a point on that circle, the gradient is $-2e^{-1}(x, y)$, which has length $2e^{-1} \approx 0.736$ and points radially inward. The surface is flat around the circle and steep across it, so no point on the circle is critical.

• A level curve tells you where the directional derivative is zero along that curve.
• A critical point needs every directional derivative to be zero, which means the whole gradient vector is zero.
• On this surface only the origin qualifies, because the gradient is a nonzero multiple of $(x,y)$ everywhere else.

The other slip is a chain rule error: writing $f_x = -x^{2}e^{-(x^{2}+y^{2})}$ instead of $-2x\,e^{-(x^{2}+y^{2})}$, by multiplying by the exponent rather than by its derivative. Paired with the matching $f_y = -y^{2}e^{-(x^{2}+y^{2})}$, that system still vanishes only at the origin, so the critical point survives and nothing looks wrong yet. The Hessian is where it shows. Differentiating $-x^{2}e^{-(x^{2}+y^{2})}$ gives $\left(2x^{3}-2x\right)e^{-(x^{2}+y^{2})}$ in $x$ and $2x^{2}y\,e^{-(x^{2}+y^{2})}$ in $y$, and both are zero at the origin, as is the matching $f_{yy}$. Every entry vanishes, so $D = 0$ and the test reports degenerate on a surface with an obvious peak. Differentiate the exponent, then multiply, and check the power of $x$ that comes out.