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 $f_{xy}(x_{0},y_{0})=f_{yx}(x_{0},y_{0})$.

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:

$D=f_{xx}(x_{0},y_{0})f_{yy}(x_{0},y_{0})-(f_{xy}(x_{0},y_{0}))^{2}$

Note that this is the determinant of the Hessian matrix:

$H(f)(x_{0},y_{0})={\begin{pmatrix}f_{xx}(x_{0},y_{0})&f_{xy}(x_{0},y_{0})\\f_{yx}(x_{0},y_{0})&f_{yy}(x_{0},y_{0})\\\end{pmatrix}}$

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 $f_{xx}(x_{0},y_{0})>0$ (note that these together also force $f_{yy}(x_{0},y_{0})>0$)	Local minimum (reasoning similar to the single-variable second derivative test)	The Hessian matrix is positive definite.
$D>0$ and $f_{xx}(x_{0},y_{0})<0$ (note that these together also force $f_{yy}(x_{0},y_{0})<0$)	Local maximum (reasoning similar to the single-variable second derivative test)	The Hessian matrix is negative definite.
$D=0$ and one or both of $f_{xx}(x_{0},y_{0})$ and $f_{yy}(x_{0},y_{0})$ 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 $f_{xx}(x_{0},y_{0})$ and $f_{yy}(x_{0},y_{0})$ 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., $f_{xx}(x_{0},y_{0}),f_{yy}(x_{0},y_{0}),f_{xy}(x_{0},y_{0})$ 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.

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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.