Second derivative test — Hessian determinant: Foundations

This page assumes you know nothing. Before you can read Second derivative test — Hessian determinant, you need a small toolkit of symbols and pictures. We build each one from zero, in an order where every new idea leans only on the ones before it.

1. What is $f (x, y)$ ? — a landscape

Think of a map of a hilly region. Each dot on the map has an east–west coordinate $x$ and a north–south coordinate $y$ . The function $f$ tells you the altitude there. Stack all those altitudes up and you get a surface floating above the flat $x y$ -plane.

We give that vertical altitude its own name. Let $z$ stand for how high above the map floor you are — the up–down coordinate, at right angles to both $x$ and $y$ . Then the surface is simply the collection of all points where the height equals what $f$ returns:

$z = f (x, y)$

Figure — Second derivative test — Hessian determinant

The flat floor is the $x y$ -plane — the map, seen from above.
The vertical coordinate $z$ measures altitude; the equation $z = f (x, y)$ lifts each map-point up to the surface (red).
Why we need this: everything else — slopes, curving, critical points — is a statement about this surface.

2. What is a partial derivative $f_{x}$ ? — slope in ONE direction

On a hill you can walk in many directions and the steepness differs. A partial derivative answers: if I walk in exactly one compass direction, how steep is it?

The little subscript is the direction you froze everything else against. To compute $f_{x}$ you differentiate treating $y$ as a plain constant, and vice versa.

$f_{x} > 0$ : ground rises as you step east (red arrow points uphill).
$f_{x} < 0$ : ground falls as you step east.
$f_{x} = 0$ : ground is momentarily flat in the east–west direction.

Why the topic needs it: the very first step of the test is finding where the ground is flat in every direction — and "every direction" turns out to be captured by just $f_{x} = 0$ and $f_{y} = 0$ together.

3. What is the gradient $\nabla f$ , and a critical point?

If even one slope were nonzero, you could still walk somewhere uphill or downhill, so you couldn't be at a peak, pit, or pass. Only when both vanish does the tangent plane go horizontal.

Why the topic needs it: the whole test is only valid at critical points. See Critical points and gradient for how to solve $\nabla f = 0$ .

4. Second derivatives — CURVING, not slope

At a flat spot the slope is zero in every direction, yet the spot could be a pit, a peak, or a pass. What distinguishes them is not slope but how the slope is changing — the curving.

Read the subscripts left to right as "differentiate by this, then by that." Because slope was zero at the critical point, the curving is the first thing that carries real information — that is exactly why the test is called the second derivative test.

5. Clairaut's theorem — the two mixed partials agree

Differentiating "east then north" gives the same answer as "north then east." So the four second derivatives are really only three distinct numbers: $f_{xx}$ , $f_{y y}$ , and the shared cross-term $f_{x y}$ . This is why the matrix in the next step is symmetric (a mirror across its diagonal). See Clairaut's theorem.

6. Packaging into the Hessian matrix

Why bundle them? Because the shape of the surface is controlled by all three curvatures at once, and a matrix is the natural container for "curving that depends on direction." See Hessian matrix.

That last number $D$ is the star of the parent note. The subtracted $(f_{x y})^{2}$ is precisely the tilt term from step 4 — it is not optional.

7. Quadratic form $Q (u, v)$ — the local shape

Near a critical point, if we let $u = x - a$ and $v = y - b$ be small steps away, the surface's local shape is captured by a single expression:

The tool that produces this expression — dropping the constant and the (zero) linear terms and keeping the second-order part — is the multivariable Taylor expansion. See Taylor series multivariable. And the study of when such an expression stays one sign is exactly Quadratic forms and definiteness.

Why we need it here: this is the bridge. Slopes told us where to look (critical point); the quadratic form tells us what shape is there, and $D$ reads off that shape.

Prerequisite map

Equipment checklist

Recall Self-test your readiness

What does $f (x, y)$ return, and what picture goes with it? ::: One number = the height of a surface above the map-point $(x, y)$ . What does the coordinate $z$ stand for? ::: The vertical altitude above the $x y$ -plane; the surface is $z = f (x, y)$ . What does $f_{x}$ measure geometrically? ::: The slope of the surface if you walk only in the $+ x$ direction, freezing $y$ . What is the gradient $\nabla f$ ? ::: The pair $(f_{x}, f_{y})$ ; an arrow on the map pointing steepest-uphill. When is $(a, b)$ a critical point? ::: When $f_{x} (a, b) = 0$ and $f_{y} (a, b) = 0$ — flat in both directions, hence all. What does $f_{xx}$ measure, versus $f_{x}$ ? ::: $f_{xx}$ is curving (how the slope changes), not slope itself. What does the mixed partial $f_{x y}$ represent? ::: The tilt/coupling between the two directions. Why is the Hessian symmetric? ::: Clairaut's theorem: $f_{x y} = f_{y x}$ when $f$ is $C^{2}$ (second partials continuous). Compute $det (a c b d)$ . ::: $a d - b c$ . Write $D$ for the Hessian. ::: $D = f_{xx} f_{y y} - (f_{x y})^{2}$ . What is the quadratic form $Q (u, v)$ ? ::: $f_{xx} u^{2} + 2 f_{x y} uv + f_{y y} v^{2}$ — the local shape near a critical point.

Connections

Second derivative test — Hessian determinant — the parent this page feeds into.
Critical points and gradient — solving $\nabla f = 0$ .
Taylor series multivariable — where the quadratic form comes from.
Hessian matrix — the matrix and its eigenvalues.
Quadratic forms and definiteness — when $Q$ keeps one sign.
Clairaut's theorem — why $f_{x y} = f_{y x}$ .
Lagrange multipliers — the constrained cousin.