Multivariable calculus

Second Partials of the Gaussian e^(-x^2 - y^2)

Q: Where is f_xx zero for the Gaussian bump?

Where 4x^{2} = 2, that is x = \pm\frac{1}{\sqrt{2}} \approx \pm 0.707, for any y. Those lines separate the concave down cap of the bump from the concave up tails.

Q: Why is the mixed partial zero on the axes?

Because f_{xy} = 4xyE contains both x and y as factors, so it vanishes whenever x = 0 or y = 0. The bump is symmetric about both axes, so there is no twist along them.

Write E = e^(-x^2 - y^2). For the Gaussian bump f(x,y) = E the second partials are f_xx = (4x^2 - 2)E, f_yy = (4y^2 - 2)E, and the mixed partial f_xy = 4xyE. At the only critical point, the origin, the Hessian determinant is 4 and f_xx = -2, so the origin is a local maximum with value 1.

f(x,y) = e^{-\left(x^{2} + y^{2}\right)}

\frac{\partial f}{\partial x} = -2x\,e^{-\left(x^{2} + y^{2}\right)}

\frac{\partial f}{\partial y} = -2y\,e^{-\left(x^{2} + y^{2}\right)}

Second and mixed partials

\frac{\partial^{2} f}{\partial x^{2}} = \left(4x^{2} - 2\right)e^{-\left(x^{2} + y^{2}\right)}

\frac{\partial^{2} f}{\partial y\,\partial x} = 4xy\,e^{-\left(x^{2} + y^{2}\right)}

\frac{\partial^{2} f}{\partial y^{2}} = \left(4y^{2} - 2\right)e^{-\left(x^{2} + y^{2}\right)}

Differentiating the bump twice

Abbreviate $E = e^{-(x^{2} + y^{2})}$ , and note that $\frac{\partial E}{\partial x} = -2xE$ . That one fact drives every computation on this page.

\frac{\partial f}{\partial x} = -2xE, \qquad \frac{\partial f}{\partial y} = -2yE

For $f_{xx}$ , differentiate the product $-2x \cdot E$ in $x$ . The first factor gives $-2E$ and the second contributes $(-2x)(-2xE)$ .

\frac{\partial^{2} f}{\partial x^{2}} = -2E + 4x^{2}E = \left(4x^{2} - 2\right)E

For the mixed partial, differentiate $-2xE$ in $y$ . The factor $-2x$ is now constant, so only $E$ moves and no extra term appears.

\frac{\partial^{2} f}{\partial y\,\partial x} = -2x \cdot (-2yE) = 4xyE

The second derivative test at the peak

Since $E$ is never zero, the equations $-2xE = 0$ and $-2yE = 0$ force $x = y = 0$ . The origin is the only critical point.

Evaluate the three second partials there. With $E(0,0) = 1$ you get $f_{xx} = -2$ , $f_{yy} = -2$ , and $f_{xy} = 0$ .

D = f_{xx}f_{yy} - \left(f_{xy}\right)^{2} = (-2)(-2) - 0 = 4 > 0

Positive $D$ with negative $f_{xx}$ means a local maximum, and here it is the global maximum, $f(0,0) = 1$ . Away from the origin $f_{xx}$ changes sign at $|x| = \frac{1}{\sqrt{2}}$ , which is where the profile switches from concave down at the peak to concave up in the tails. Those are the inflection points of the bell curve.

The mistake students make

The usual error is $f_{xx} = 4x^{2}E$ , which comes from chaining through the exponential twice but forgetting the product rule on the factor $-2x$ that the first derivative produced. The missing $-2E$ is exactly the term that makes the origin a maximum, and without it $D = 0$ there and the test says nothing.

A second slip is a sign error on the mixed partial. Two negatives multiply, so $f_{xy} = 4xyE$ is positive in the first quadrant, not negative.

Clairaut symmetry is easy to verify here: differentiating $f_y = -2yE$ in $x$ gives $-2y(-2xE) = 4xyE$ , the same expression. Every partial of this function is a polynomial times $E$ , so all of them are continuous everywhere and the two orders can never disagree.

Frequently asked questions

Where is f_xx zero for the Gaussian bump?

Where $4x^{2} = 2$ , that is $x = \pm\frac{1}{\sqrt{2}} \approx \pm 0.707$ , for any $y$ . Those lines separate the concave down cap of the bump from the concave up tails.

Why is the mixed partial zero on the axes?

Because $f_{xy} = 4xyE$ contains both $x$ and $y$ as factors, so it vanishes whenever $x = 0$ or $y = 0$ . The bump is symmetric about both axes, so there is no twist along them.