Second derivative test for a function of two variables

This article describes an analogue for functions of multiple variables of the following term/fact/notion for functions of one variable: second derivative test

This article describes a test that can be used to determine whether a point in the domain of a function gives a point of local, endpoint, or absolute (global) maximum or minimum of the function, and/or to narrow down the possibilities for points where such maxima or minima occur.
View a complete list of such tests

Statement

Suppose $f$ is a function of two variables $x, y$ . Suppose $(x_{0}, y_{0})$ is a point in the domain of $f$ such that both the first-order partial derivatives at the point are zero, i.e., $f_{x} (x_{0}, y_{0}) = f_{y} (x_{0}, y_{0}) = 0$ .

Suppose that all the second-order partial derivatives (pure and mixed) for $f$ exist and are continuous at and around $(x_{0}, y_{0})$ . Note that by Clairaut's theorem on equality of mixed partials, this implies that $f_{x y} (x_{0}, y_{0}) = f_{y x} (x_{0}, y_{0})$ .

The second derivative test helps us determine whether $f$ has a local maximum at $(x_{0}, y_{0})$ , a local minimum at $(x_{0}, y_{0})$ , or a saddle point at $(x_{0}, y_{0})$ .

First, consider the Hessian determinant of $f$ at $(x_{0}, y_{0})$ , which we define as:

$D = f_{x x} (x_{0}, y_{0}) f_{y y} (x_{0}, y_{0}) - (f_{x y} (x_{0}, y_{0}))^{2}$

Note that this is the determinant of the Hessian matrix:

$H (f) (x_{0}, y_{0}) = (\begin{matrix} f_{x x} (x_{0}, y_{0}) & f_{x y} (x_{0}, y_{0}) \\ f_{y x} (x_{0}, y_{0}) & f_{y y} (x_{0}, y_{0}) \end{matrix})$

We now have the following:

Case	Local maximum, local minimum, saddle point, or none of these?	Interpretation in terms of second derivative test for a function of multiple variables
$D < 0$	Saddle point	The Hessian matrix is neither positive semidefinite nor negative semidefinite.
$D > 0$ and $f_{x x} (x_{0}, y_{0}) > 0$ (note that these together also force $f_{y y} (x_{0}, y_{0}) > 0$ )	Local minimum (reasoning similar to the single-variable second derivative test)	The Hessian matrix is positive definite.
$D > 0$ and $f_{x x} (x_{0}, y_{0}) < 0$ (note that these together also force $f_{y y} (x_{0}, y_{0}) < 0$ )	Local maximum (reasoning similar to the single-variable second derivative test)	The Hessian matrix is negative definite.
$D = 0$ and one or both of $f_{x x} (x_{0}, y_{0})$ and $f_{y y} (x_{0}, y_{0})$ is positive (note that if one of them is positive, the other one is either positive or zero)	Inconclusive, but we can rule out the possibility of being a local maximum.	The Hessian matrix is positive semidefinite but not positive definite.
$D = 0$ and one or both of $f_{x x} (x_{0}, y_{0})$ and $f_{y y} (x_{0}, y_{0})$ is negative (note that if one of them is negative, the other one is either negative or zero)	Inconclusive, but we can rule out the possibility of being a local minimum	The Hessian matrix is negative semidefinite but not negative definite.
All entries of the Hessian matrix are zero, i.e., $f_{x x} (x_{0}, y_{0}), f_{y y} (x_{0}, y_{0}), f_{x y} (x_{0}, y_{0})$ are all zero	Inconclusive. No possibility can be ruled out.	The Hessian matrix is both positive semidefinite and negative semidefinite. Basically, we can't say anything.

Relation with other tests

Changing the number of variables

Second derivative test: The version for a function of one variable.
Second derivative test for a function of multiple variables: The two-variable case is a special, and relatively tractable, subcase of the multiple-variable case.