How to Use the Second Derivative Test for Multivariable Functions
The second derivative test for multivariable functions classifies each critical point of a function $f(x,y)$ — a point where both first partial derivatives vanish — as a local minimum, a local maximum, or a saddle point, by examining the determinant of the Hessian matrix $H$ and the value of $f_{xx}$ at that point. If $D = f_{xx}f_{yy} - (f_{xy})^2 > 0$ and $f_{xx} > 0$, the point is a local minimum; if $D > 0$ and $f_{xx} < 0$, it is a local maximum; if $D < 0$, it is a saddle; and if $D = 0$, the test says nothing. Below, we build the Hessian, run the decision rules, and work through one full example.
Why the Test Works
In one variable, the second derivative tells you about concavity: a positive $f’'$ bends the graph upward, a negative one bends it downward (Wikipedia). At a stationary point, bending up means a minimum, bending down means a maximum. In two or more variables, “concavity” is a direction-dependent idea, so a single number can’t capture it — you need the whole matrix of second partial derivatives.
That matrix is the Hessian, and its determinant plays the role that $f’'$ plays in one dimension. The Hessian describes the local curvature of the surface in every direction at once. Its determinant measures whether that curvature bends the same way across directions (det positive: consistent bowl or dome) or flips sign between directions (det negative: a saddle). This is the core logic behind the multivariable second derivative test.
More precisely, a Taylor expansion near a critical point reduces to a quadratic form built from the Hessian. Whether that form is always positive, always negative, or mixed — decided by the determinant and the leading entry $f_{xx}$ — determines the shape of the surface locally. A positive-definite form gives a bowl, a negative-definite form gives a dome, and an indefinite form gives a saddle.
Hessian and Critical Points
The first step is always the same: find the critical points. A point $(a,b)$ is critical when the gradient is zero, meaning $f_x(a,b) = 0$ and $f_y(a,b) = 0$. Solve that system; the solutions are the only candidates for local extrema and saddles.
Next, compute the four second partial derivatives $f_{xx}$, $f_{xy}$, $f_{yx}$, and $f_{yy}$. For any function with continuous second partials, the mixed partials agree ($f_{xy} = f_{yx}$, Clairaut’s theorem), so the Hessian is symmetric:
$$H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{pmatrix}$$
Its determinant is $D = f_{xx}f_{yy} - (f_{xy})^2$. Evaluating $D$ at each critical point is the heart of the procedure.
Applying the Test
Evaluate $D$ and $f_{xx}$ at each critical point, then read off the classification:
| Condition at $(a,b)$ | Classification |
|---|---|
| $D > 0$ and $f_{xx} > 0$ | Local minimum |
| $D > 0$ and $f_{xx} < 0$ | Local maximum |
| $D < 0$ | Saddle point |
| $D = 0$ | Inconclusive |
Two quick sanity checks help. When $D > 0$, $f_{xx}$ and $f_{yy}$ must share the same sign, so testing $f_{xx}$ alone is enough to tell a bowl from a dome. When $D < 0$, the two pure second partials disagree in sign, which is exactly the signature of a saddle.
The determinant equals the product of the Hessian’s eigenvalues. Both eigenvalues positive means upward curvature in every direction (minimum); both negative means downward curvature everywhere (maximum); opposite signs means the surface rises one way and falls another (saddle). When $D = 0$, at least one eigenvalue is zero, the quadratic form is degenerate, and higher-order terms must decide — so the basic test is silent (Paul’s Online Math Notes).
Worked Example
Classify the critical points of $f(x,y) = x^2 + y^2 - 2x - 6y + 3$.
First partials: $f_x = 2x - 2$ and $f_y = 2y - 6$. Setting both to zero gives the single critical point $(1,3)$. Second partials: $f_{xx} = 2$, $f_{yy} = 2$, $f_{xy} = 0$. So $D = (2)(2) - 0 = 4 > 0$ and $f_{xx} = 2 > 0$. The point $(1,3)$ is a local minimum.
A contrast helps fix the saddle case. For $f(x,y) = x^2 - y^2$, the partials $f_x = 2x$ and $f_y = -2y$ make $(0,0)$ the only critical point. Here $f_{xx} = 2$, $f_{yy} = -2$, $f_{xy} = 0$, so $D = (2)(-2) - 0 = -4 < 0$: a saddle, where the surface rises along the $x$-axis and falls along the $y$-axis.
Practical Tips
Differentiate carefully. The mixed partial is the step most often botched, and a wrong $f_{xy}$ flips the sign of $D$. Compute the second derivatives as functions first, then substitute the critical point — that way the mixed-partial term stays clean. And remember that “local” means only in a neighborhood: a function can have several minima and maxima at once, so evaluate each critical point separately rather than assuming one shape fits them all. When $D = 0$, you must reason from the function itself, perhaps via a graph or a direct inequality; MIT OpenCourseWare’s multivariable calculus materials (MIT OpenCourseWare) and Khan Academy’s multivariable series (Khan Academy) walk through such cases if you want extra practice.
Key Takeaways
- The multivariable second derivative test uses the Hessian determinant $D = f_{xx}f_{yy} - (f_{xy})^2$ plus the sign of $f_{xx}$ to classify critical points.
- $D > 0$ with $f_{xx} > 0$ means a local minimum; $D > 0$ with $f_{xx} < 0$ means a local maximum.
- $D < 0$ signals a saddle point, where the surface curves oppositely in different directions.
- $D = 0$ is inconclusive — fall back on higher-order reasoning, a graph, or direct evaluation.
- The determinant works because it encodes the signs of the Hessian’s eigenvalues, generalizing the single-variable concavity test to every direction at once.
