Second derivative test — Hessian determinant

WHY does this work? (Derivation from scratch)

We want to know the shape of $f$ near a critical point $(a,b)$ . Taylor expand to second order. Since $f_x=f_y=0$ there, the linear terms vanish and the quadratic part controls everything. Let $u=x-a$ , $v=y-b$ :

$f(a+u,b+v) \approx f(a,b) + \tfrac12\big(f_{xx}u^2 + 2f_{xy}uv + f_{yy}v^2\big)$

The shape depends entirely on the quadratic form $Q(u,v)=f_{xx}u^2 + 2f_{xy}uv + f_{yy}v^2$ . We ask: is $Q$ always positive (bowl), always negative (dome), or both signs (saddle)?

Step — complete the square. Assume $f_{xx}\neq 0$ :

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

Why this step? Pulling $f_{xx}$ out lets us complete the square in $u$ , turning a messy cross-term into something we can sign-analyse.

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

Why this step? The squared bracket is $\ge 0$ always. So the sign behaviour now hinges on the two coefficients.

Multiply the second coefficient out: $f_{yy} - \frac{f_{xy}^2}{f_{xx}} = \frac{f_{xx}f_{yy} - f_{xy}^2}{f_{xx}} = \frac{D}{f_{xx}}$

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

Now read off the cases:

If $D>0$ and $f_{xx}>0$ : both coefficients $f_{xx}$ and $D/f_{xx}$ are positive → $Q>0$ always → minimum.
If $D>0$ and $f_{xx}<0$ : both coefficients negative → $Q<0$ always → maximum. (Note $D>0$ forces $f_{xx}$ and $f_{yy}$ to have the same sign, so checking $f_{xx}$ alone suffices.)
If $D<0$ : $f_{xx}$ and $D/f_{xx}$ have opposite signs → $Q$ can be made positive (pick $v=0$ ) or negative (kill the bracket) → saddle.
If $D=0$ : one coefficient vanishes, quadratic info is flat along a direction → inconclusive.

Figure — Second derivative test — Hessian determinant

Worked examples

Worked example Example 3:

$f(x,y)=x^3-3xy+y^3$ $f_x=3x^2-3y=0\Rightarrow y=x^2$ ; $f_y=-3x+3y^2=0\Rightarrow x=y^2$ . Substitute: $x=(x^2)^2=x^4\Rightarrow x^4-x=0\Rightarrow x(x^3-1)=0\Rightarrow x=0,1$ . Critical points: $(0,0)$ and $(1,1)$ . Why two? The system is nonlinear — multiple solutions. $f_{xx}=6x,\ f_{yy}=6y,\ f_{xy}=-3$ .

At $(0,0)$ : $D=(0)(0)-(-3)^2=-9<0$ → saddle.
At $(1,1)$ : $D=(6)(6)-9=27>0$ , $f_{xx}=6>0$ → local minimum.

Common mistake Steel-manning the common errors

Mistake 1: Forgetting the $(f_{xy})^2$ term. Writing $D=f_{xx}f_{yy}$ . Why it feels right: product of "the two curvatures" seems natural. Fix: the cross-term $f_{xy}$ encodes how the curvature axes are tilted — ignoring it is like ignoring the tilt of the Pringle. The minus sign comes from completing the square.

Mistake 2: Using $f_{yy}$ when $D>0$ gives a different answer. Why it feels right: "just pick either second derivative." Fix: when $D>0$ , $f_{xx}$ and $f_{yy}$ are forced to have the same sign, so they agree — but if your algebra gives them different signs, you made an arithmetic error. Use $f_{xx}$ as the standard.

Mistake 3: Treating $D=0$ as "no extremum." Why it feels right: feels like a "fail = none." Fix: $D=0$ means inconclusive — there might still be a min, max, or saddle (e.g. $f=x^4+y^4$ has a min but $D=0$ at origin). Investigate directly.

Mistake 4: Applying the test at a non-critical point. The Taylor expansion only drops the linear terms because the gradient is zero. Always solve $\nabla f=0$ first.

Recall Feynman: explain it to a 12-year-old

Imagine you're standing on a hill where the ground is perfectly flat under your feet (no slope in any direction). Are you at the bottom of a valley, the top of a hill, or on a mountain pass? Look at how the ground curves. If it curves up in every direction, you're in a valley (minimum). If it curves down everywhere, you're on a peak (maximum). If it curves up one way but down another — like sitting on a horse's saddle — you're at a pass. The "Hessian determinant" is just a number we compute from the curving that tells us which of these three it is, like a magic detector.

Active recall

What is the Hessian determinant

D

for

f(x,y)

D = f_{xx}f_{yy} - (f_{xy})^2

What must be true before applying the second derivative test?

The point must be critical:

f_x=f_y=0

D>0

and

f_{xx}>0

implies what?

Local minimum.

D>0

and

f_{xx}<0

implies what?

Local maximum.

D<0

implies what?

Saddle point.

D=0

implies what?

Test is inconclusive — investigate directly.

Why does the test only use second-order Taylor terms?

At a critical point the gradient (linear terms) is zero, so the quadratic form controls local shape.

What is

D

in terms of the Hessian's eigenvalues?

D=\lambda_1\lambda_2

(product of curvatures).

Why must we subtract

(f_{xy})^2

and not ignore it?

It comes from completing the square;

f_{xy}

encodes the tilt/coupling of the two curvature directions.

For

f=x^2-y^2

at origin, what is

D

and the conclusion?

D=-4<0

, saddle point.

Connections

Critical points and gradient — finding where $\nabla f = 0$ .
Taylor series multivariable — the foundation of this derivation.
Hessian matrix — eigenvalues, definiteness, $n$ -dimensional generalisation.
Quadratic forms and definiteness — positive/negative definite ↔ min/max.
Clairaut's theorem — why $f_{xy}=f_{yx}$ .
Lagrange multipliers — constrained optimisation cousin.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab humein kisi function $f(x,y)$ ka maximum ya minimum nikalna hota hai, sabse pehle hum critical point dhoondhte hain — yani jahan $f_x=0$ aur $f_y=0$ ho (surface ki tangent flat ho jaye). Lekin yeh critical point teen cheez ho sakta hai: ek katora (minimum), ek pahaad ki choti (maximum), ya ek ghode ki saddle (na max na min). In teeno mein farak kaise karein? Yahin Hessian determinant kaam aata hai.

Hessian determinant hai $D = f_{xx}f_{yy} - (f_{xy})^2$ . Rule simple hai: agar $D>0$ aur $f_{xx}>0$ to minimum; $D>0$ aur $f_{xx}<0$ to maximum; $D<0$ to saddle; aur $D=0$ to test fail — khud check karna padega. Yeh sab kahan se aaya? Taylor expansion se. Critical point pe linear term zero ho jaate hain, to bachta hai sirf quadratic term, aur usko "complete the square" karke dekhte hain ki woh hamesha positive hai (bowl), hamesha negative hai (dome), ya dono signs leta hai (saddle).

Sabse common galti yeh hoti hai ki students $(f_{xy})^2$ wala term bhool jaate hain ya $D=0$ ko "no extremum" samajh lete hain. Yaad rakho — $D=0$ ka matlab "pata nahi", "nahi hai" nahi. Aur intuition ke liye: $D$ actually Hessian ke do eigenvalues ka product hai, jo do directions mein curvature batate hain. Same sign curvature = bowl/dome, opposite sign = saddle. Bas itna samajh lo, baaki sab plug-and-chug hai.

Second derivative test — Hessian determinant

WHAT is the Hessian?

HOW the test works

WHY does this work? (Derivation from scratch)

Worked examples

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections