Visual walkthrough — Second derivative test — Hessian determinant

Everything below assumes only two things: you know what a function $f(x,y)$ is (a height above the floor at each point $(x,y)$), and you know what a partial derivative is (the slope of that height if you walk in only one direction, freezing the other). Everything else we build here.

Step 1 — The flat spot we are standing on

WHAT. We pick a point $(a,b)$ on the floor where the surface is flat in every direction. "Flat" means: if you walk east–west the ground has zero slope, and if you walk north–south the ground has zero slope. In symbols,

$f_x(a,b)=0\text{and}{f}_y(a,b)=0.$

Here $f_x$ is the east–west slope and $f_y$ the north–south slope. A point where both are zero is called a critical point (see Critical points and gradient).

WHY start here. A flat spot is the only place where a max, min, or saddle can hide. On a slope you can always walk downhill, so you are at none of them. Flatness is our permission slip to look at curvature next.

PICTURE. Three flat spots — a bowl bottom, a dome top, a saddle pass — all share the same horizontal tangent plane (the amber sheet). The slopes are zero; only the bending differs.

Step 2 — Zoom in: the linear part dies, curvature is all that's left

WHAT. We want the surface's shape near $(a,b)$, so we write $f$ as a sum of a flat part plus a slope part plus a bending part. This is the Taylor expansion to second order. Let $u=x-a$ and $v=y-b$ measure how far we stepped from the flat spot:

$f(a+u,b+v)\approx \underset{\text{height at centre}}{\underbrace{f(a,b)}}+\underset{\text{slope part}}{\underbrace{f_{x}u+f_{y}v}}+\underset{\text{bending part}}{\underbrace{\frac{1}{2}(f_{xx}{u}^2+2f_{xy}uv+f_{yy}{v}^2)}}.$

• $u,v$ — tiny steps east and north.
• $f_{xx}$ — how the east–west slope itself changes as you go east (curvature along $u$).
• $f_{yy}$ — curvature along $v$.
• $f_{xy}$ — the coupling: how the east–west slope changes as you walk north. It measures a diagonal twist.

WHY. Because we are at a critical point, $f_x=f_y=0$, so the entire slope part vanishes. The constant $f(a,b)$ just shifts the whole surface up or down and never changes its shape. So the only thing that decides bowl vs dome vs saddle is the bending part.

PICTURE. Same surface, blown up under a magnifying glass. Far out it looks curved; zoomed in the flat sheet and the parabolic bending are all that survive — the slope arrows shrink to nothing.

Step 3 — Name the survivor: the quadratic form $Q$

WHAT. Strip away the constant and the dead linear terms. What remains (times $\frac{1}{2}$) is a single expression in $u$ and $v$:

$Q(u,v)=f_{xx}{u}^2+2f_{xy}uv+f_{yy}{v}^2.$

This is a quadratic form (see Quadratic forms and definiteness): every term is degree two in $u,v$. Reading it left to right:

• $f_{xx}{u}^2$ — pure east–west bending, always same sign as $f_{xx}$.
• $2f_{xy}uv$ — the twist; it is positive in some corners of the $(u,v)$ plane and negative in others because $uv$ flips sign across the axes.
• $f_{yy}{v}^2$ — pure north–south bending.

WHY. The height near the flat spot is $f(a,b)+\frac{1}{2}Q$. So the surface curves up where $Q>0$ and down where $Q<0$. Our whole problem shrinks to one question:

Is $Q$ always positive, always negative, or does it take both signs?

Always positive → valley (minimum). Always negative → peak (maximum). Both signs → saddle.

PICTURE. The $(u,v)$ floor coloured by the sign of $Q$: cyan where $Q>0$, amber where $Q<0$. A bowl is all cyan; a saddle splits the plane into cyan and amber wedges.

Step 4 — The obstacle: that middle term $2f_{xy}uv$

WHAT. If the twist term were gone, life would be easy: $Q=f_{xx}{u}^2+f_{yy}{v}^2$, two clean squares whose signs we just read off. The trouble is $2f_{xy}uv$ mixes $u$ and $v$, so we cannot tell the sign at a glance.

WHY we need a tool. We reach for completing the square — the classic move that rewrites $a{u}^2+bu+c$ as $a(u+\frac{b}{2a}{)}^2+\text{leftover}$. Why this tool and not, say, factoring? Because a perfect square $(\dots {)}^2$ is guaranteed $\ge 0$; once we bottle the messy cross-term inside one square, whatever is left over becomes a clean, isolated coefficient we can sign-test. Factoring would need us to know the roots first; completing the square never asks that.

We must assume $f_{xx}\ne 0$ to pull it out as a common factor. (Step 8 handles $f_{xx}=0$ separately.)

PICTURE. The tilted trough of $Q$: the ridge/valley line runs at an angle because of $f_{xy}$. Completing the square is really rotating our coordinates to line up with that tilt.

Step 5 — Do the algebra: the square appears

WHAT. Factor $f_{xx}$ out of the two terms containing $u$:

$Q=f_{xx}\left(u^2+\frac{2f_{xy}}{f_{xx}}uv\right)+f_{yy}{v}^2.$

Now complete the square inside the bracket. The pattern $u^2+2kuv=(u+kv{)}^2-k^2{v}^2$ with $k=\frac{f_{xy}}{f_{xx}}$ gives:

$Q=f_{xx}{\left(u+\frac{f_{xy}}{f_{xx}}v\right)}^2+\left(f_{yy}-\frac{f_{xy}^2}{f_{xx}}\right)v^2.$

Term by term:

• ${\left(u+\frac{f_{xy}}{f_{xx}}v\right)}^2$ — a perfect square, never negative. Its coefficient is $f_{xx}$.
• $\left(f_{yy}-\frac{f_{xy}^2}{f_{xx}}\right)v^2$ — a second square $v^2$ (never negative) times a leftover coefficient. That subtracted $\frac{f_{xy}^2}{f_{xx}}$ is exactly the price the twist charged us.

WHY. We have now written $Q$ as (positive coefficient?) times (square) + (leftover coefficient) times (square). Since both squares are $\ge 0$, the sign of $Q$ is decided purely by the two coefficients out front.

PICTURE. The two coefficient "dials". One reads $f_{xx}$; the other reads the leftover. Their signs light up green (positive) or red (negative).

Step 6 — The number $D$ is born

WHAT. Tidy the leftover coefficient over a common denominator:

$f_{yy}-\frac{f_{xy}^2}{f_{xx}}=\frac{f_{xx}{f}_{yy}-f_{xy}^2}{f_{xx}}=\frac{D}{f_{xx}},\boxed{D:=f_{xx}{f}_{yy}-f_{xy}^2}.$

That numerator $f_{xx}{f}_{yy}-f_{xy}^2$ is precisely the determinant of the Hessian matrix $H=\left(\begin{array}{cc}{f}_{xx}& f_{xy}\\ f_{xy}& f_{yy}\end{array}\right)$. (The two off-diagonal entries are equal by Clairaut's theorem.) So $Q$ becomes:

$Q=f_{xx}{\left(u+\frac{f_{xy}}{f_{xx}}v\right)}^2+\frac{D}{f_{xx}}{v}^2.$

Now every symbol is earned. The two coefficients are $f_{xx}$ and $\frac{D}{f_{xx}}$.

WHY the subtraction, not a product. Notice $D$ is $f_{xx}{f}_{yy}$ minus $f_{xy}^2$. The $-f_{xy}^2$ is the memory of the twist we absorbed. Ignore it (Mistake 1 in the parent) and you throw away the tilt entirely.

PICTURE. The two-by-two grid of second derivatives with the "diagonal product minus off-diagonal square" arrows drawn on top — the determinant made visible.

Step 7 — Read off the three main cases

WHAT. $Q=f_{xx}({\text{square}}_1)+\frac{D}{f_{xx}}({\text{square}}_2)$. Both squares $\ge 0$. So:

WHY it's airtight. For $D<0$ we didn't wave hands — we constructed one direction where $Q>0$ and another where $Q<0$. That is the honest definition of a saddle.

PICTURE. Three side-by-side height plots keyed to the coefficient signs: all-up bowl, all-down dome, and the split saddle with its two coloured escape directions arrowed.

Step 8 — The degenerate edges: $D=0$ and $f_{xx}=0$

WHAT. Two loose ends the tidy algebra skipped.

Case $f_{xx}=0$ (couldn't factor). Then $Q=2f_{xy}uv+f_{yy}{v}^2=v(2f_{xy}u+f_{yy}v)$. If $f_{xy}\ne 0$ this is a product of two independent linear factors, so it changes sign as you cross either line → saddle (consistent with $D=f_{xx}{f}_{yy}-f_{xy}^2=-f_{xy}^2<0$). If also $f_{xy}=0$, then $Q=f_{yy}{v}^2$ — flat along the entire $u$-axis → degenerate, and indeed $D=0$.

Case $D=0$. One coefficient in Step 6 vanishes, so $Q$ is a single square (or zero) — it stays $\ge 0$ (or $\le 0$) but touches zero along a whole line, not just a point. Along that valley-floor line the quadratic term gives no information; the true shape is decided by third- or fourth-order terms we threw away. Hence inconclusive. Example: $f=x^4+y^4$ has a genuine minimum yet $D=0$ at the origin.

WHY these deserve their own step. The contract: the reader must never hit a scenario we didn't show. $f_{xx}=0$ breaks the factoring; $D=0$ breaks the strict inequality. Both are real inputs, so both get handled.

PICTURE. Left: $x^4+y^4$, a flat-bottomed bowl whose base is a point but curves like a plate — $D=0$, still a min. Right: a "monkey saddle" gutter where $Q$ vanishes along a line. Both defeat the second-order test.

The one-picture summary

WHAT. The whole logic on a single map: flat spot → drop linear terms → quadratic form $Q$ → complete the square → coefficients $f_{xx}$ and $D/f_{xx}$ → sign combinations → verdict.

Active recall

Connections

• Second derivative test — Hessian determinant — the parent rule this page derives.
• Critical points and gradient — Step 1, where $\nabla f=0$.
• Taylor series multivariable — Step 2, the expansion.
• Hessian matrix — Step 6, $D=\det H$.
• Quadratic forms and definiteness — Steps 3 & 7, sign analysis of $Q$.
• Clairaut's theorem — why $f_{xy}=f_{yx}$, making $H$ symmetric.
• Lagrange multipliers — the constrained cousin of this optimisation.

Concept Map