Second Partial Derivative Test

In monovarietal functions, determining the nature of a stationary point was via the second derivative $\frac{d^{2}y}{d{x}^2}$. For multivariate functions, a matrix known as the Hessian matrix is used.

Hessian Matrix §

See Hessian Matrix

Definition §

Second Partial Derivative Test ^definition

Let $f$ be a bivariate function that is twice-differentiable. Its Hessian is given by $\det (\mathbf{H})=f_{xx}{f}_{yy}-f_{xy}{f}_{yx}$. Then the second derivative test states that, for any critical point $(a,b)$:

$$\text{(a,b) is a }\{\begin{array}{ll}\text{Local maximum},& \det (\mathbf{H})>0\text{ and }{f}_{xx}(a,b)<0\\ \text{Local minimum},& \det (\mathbf{H})>0\text{ and }{f}_{xx}(a,b)>0\\ \text{Saddle point},& \det (\mathbf{H})<0\\ \text{Inconclusive},& \det (\mathbf{H})=0\end{array}$$

Warning

The second derivative test is inconclusive if $\det (H)=0$, meaning it cannot state the nature of the critical point.

Second Partial Derivative Test (Matrix) ^definition

Let $f$ be a bivariate function that is twice-differentiable. Its Hessian is given by $\det (\mathbf{H})=f_{xx}{f}_{yy}-f_{xy}{f}_{yx}$. Then the matrix version of the second derivative test states that, for any critical point $(a,b)$:

• $\mathbf{H}{|}_{(x,y)=(a,b)}$ is positive-definite matrix : $(a,b)$ is a local minimum
• $\mathbf{H}{|}_{(x,y)=(a,b)}$ is negative-definite matrix : $(a,b)$ is a local maximum
• $\mathbf{H}{|}_{(x,y)=(a,b)}$ is both indefinite and invertible : $(a,b)$ is a saddle point
• $\mathbf{H}{|}_{(x,y)=(a,b)}$ is singular

#todo talk about definiteness of a matrix. And connection to Taylor polynomials