Linear Algebra - The Hessian

Give the Hessian matrix for the function $f(x,y) = xy^{2}e^{x}$ .

$e^{x}\begin{bmatrix} x y^{2}+ 2 y^{2} & 2 x y +2 y \ 2xy+2y &2x \end{bmatrix}$

$e^{x}\begin{bmatrix} x y^{2}- 2 y^{2} & 2 x y - 2y \ 2xy-2y &2x \end{bmatrix}$

$e^{x}y^{2}\begin{bmatrix} x + 2 & 2 x + 1 \ 2x+1 &2x \end{bmatrix}$

$e^{x}y^{2}\begin{bmatrix} x + 2 & 2 x -2 \ 2x-2 &2x \end{bmatrix}$

$e^{x}y \begin{bmatrix} x y - 2 y & 2 x - 2 \ 2x -2 &2x \end{bmatrix}$

Explanation

The Hessian matrix of a function is the matrix of partial second derivatives

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y } \ \ \frac{\partial^2 f}{\partial y\partial x } & \frac{\partial^2 f}{ \partial y^2 } \end{bmatrix}$

Find each of these derivatives as follows:

$\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}(xy^{2}e^{x})$

$=y^{2} \frac{\partial }{\partial x}(xe^{x})$

$=y^{2}\left ( x \frac{\partial }{\partial x} e^{x}+ e^{x} \frac{\partial }{\partial x}x \right )$

$=y^{2}\left ( x \cdot e^{x}+ e^{x} \cdot 1 \right )$

$=y^{2}\left ( x e^{x}+ e^{x} \right )$

$\frac{ \partial^2 f}{\partial x^2}= \frac{\partial }{\partial x}(y^{2}\left ( x e^{x}+ e^{x} \right ))$

$=y^{2} \frac{\partial }{\partial x} ( x e^{x}+ e^{x} )$

$=y^{2} \left (\frac{\partial }{\partial x} ( x e^{x})+ \frac{\partial }{\partial x} (e^{x} ) \right )$

$=y^{2} ( x e^{x}+ e^{x} + e^{x} )$

$=y^{2} ( x e^{x}+ 2 e^{x} )$

$=( x + 2 )e^{x} y^{2}$

$\frac{\partial^2 f}{\partial x\partial y }= \frac{\partial^2 f}{\partial y\partial x } = \frac{\partial }{\partial y}(y^{2}\left ( x e^{x}+ e^{x} \right ))$

$=\left ( x e^{x}+ e^{x} \right ) \frac{\partial }{\partial y}(y^{2})$

$= \left ( x e^{x}+ e^{x} \right ) \cdot 2y$

$=2 \left ( x + 1 \right )e^{x}y$

$\frac{\partial f}{\partial y} = \frac{\partial }{\partial y}(xy^{2}e^{x})$

$=xe^{x} \frac{\partial }{\partial y}y^{2}$

$=xe^{x} \cdot 2y$

$=2xe^{x} y$

$\frac{\partial f^{2}}{\partial^{2} y} = \frac{\partial }{\partial y}(2xe^{x} y)$

$=2xe^{x} \frac{\partial }{\partial y}y$

$=2xe^{x} \cdot 1$

$=2xe^{x }$

The Hessian matrix is

$\begin{bmatrix} ( x + 2 )e^{x} y^{2} & 2 \left ( x + 1 \right )e^{x}y \ 2 \left ( x + 1 \right )e^{x}y &2xe^{x } \end{bmatrix}$ ,

which can be rewritten, after a little algebra, as

$e^{x}\begin{bmatrix} x y^{2}+ 2 y^{2} & 2 x y +2 y \ 2xy+2y &2x \end{bmatrix}$ .

Find the Hessian of the following function.

$f(x,y,z)=\frac{1}{2}xy^2+\ln(yz^2)+x^3$

$\begin{bmatrix} 6x & y & 0\ \ y & x-\frac{1}{y^2} & 0 \ \ 0 & 0 & -\frac{2}{z^2} \end{bmatrix}$

$\begin{bmatrix} 6x & y & 0\ \ 0 & x-\frac{1}{y^2} & 0 \ \ 0 & 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 6x & 0 & 0\ \ y & 0& 0 \ \ 0 & 0 & -\frac{2}{z^2} \end{bmatrix}$

Explanation

Recall the Hessian

$\begin{bmatrix} \frac{\partial^2 f}{\partial x \partial x} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \ \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y \partial y} & \frac{\partial^2 f}{\partial y \partial z} \ \ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z \partial z} \end{bmatrix}$

So lets find the partial derivatives, and then put them into matrix form.

$\frac{\partial^2 f}{\partial x \partial x} =6x$

$\frac{\partial^2 f}{\partial x \partial y} =y$

$\frac{\partial^2 f}{\partial x \partial z} =0$

$\frac{\partial^2 f}{\partial y \partial x} =y$

$\frac{\partial^2 f}{\partial y \partial y} =x-\frac{1}{y^2}$

$\frac{\partial^2 f}{\partial y \partial z} =0$

$\frac{\partial^2 f}{\partial z \partial x} =0$

$\frac{\partial^2 f}{\partial z \partial y} =0$

$\frac{\partial^2 f}{\partial z \partial z} =-\frac{2}{z^2}$

Now lets put them into the matrix

$\begin{bmatrix} 6x & y & 0\ \ y & x-\frac{1}{y^2} & 0 \ \ 0 & 0 & -\frac{2}{z^2} \end{bmatrix}$

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y)=20xy^{2}\end{align*}$

$\begin{bmatrix}0&40y\40y&40x\end{bmatrix}$

$\begin{bmatrix}40y&40x\0&40y\end{bmatrix}$

$\begin{bmatrix}40y&0\40x&40y\end{bmatrix}$

$\begin{bmatrix}40x&40y\40y&0\end{bmatrix}$

Explanation

Give the Hessian matrix of the function $f(x, y) = \frac{1}{x^{2}y^{3}}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&6xy\ \ 6xy&12x^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6x^{2}&6xy\ \ 6xy&12y^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6x^{2}&-6xy\ \ -6xy&12y^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&-6xy\ \- 6xy&12x^{2}\end{bmatrix}$

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6x^{2}& 6xy\ \ -6xy&12y^{2}\end{bmatrix}$

Explanation

The Hessian matrix of a function is the matrix of partial second derivatives

$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y } \ \ \frac{\partial^2 f}{\partial y\partial x } & \frac{\partial^2 f}{ \partial y^2 } \end{bmatrix}$

First, rewrite

$f(x, y) = \frac{1}{x^{2}y^{3}}$

$f(x, y) = x^{-2}y^{-3}$

Find each partial second derivative separately:

$\frac{\partial f }{\partial x} =\frac{\partial }{\partial x} ( x^{-2}y^{-3})$

$=-2 x^{-3}y^{-3}$

$\frac{\partial^2 f}{\partial x^2} =\frac{\partial }{\partial x} \left ( -2 x^{-3}y^{-3} \right )$

$= -2 (-3)x^{-4}y^{-3} \right )$

$=6x^{-4}y^{-3}$

$=\frac{6}{x^{4}y^{3}}$

$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} =\frac{\partial }{\partial y } \left ( -2 x^{-3}y^{-3} \right )$

$=-2 (-3)x^{-3}y^{-4}$

$=\frac{6}{x^{3}y^{4}}$

$\frac{\partial f }{\partial y} =\frac{\partial }{\partial y} ( x^{-2}y^{-3})$

$= -3 x^{-2}y^{-4}$

$\frac{\partial ^2 f}{\partial y^2} = \frac{\partial }{\partial y} \left ( -3 x^{-2}y^{-4} \right )$

$=-3 (-4) x^{-2}y^{-5}$

$=12x^{-2}y^{-5}$

$=\frac{12}{x^{2}y^{5}}$

The Hessian of is

$\begin{bmatrix}\frac{6}{x^{4}y^{3}} &\frac{6}{x^{3}y^{4}}\ \ \frac{6}{x^{3}y^{4}}&\frac{12}{x^{2}y^{5}}\end{bmatrix}$ ,

which can be rewritten as

$\frac{1}{x^{4}y^{5}}\begin{bmatrix}6y^{2}&6xy\ \ 6xy&12x^{2}\end{bmatrix}$ .

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=6sin(6z)e^{(3x)}e^{(3y)}\end{align*}$

$\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}\108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}\-216sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

Explanation

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=6sin(6z)e^{(3x)}e^{(3y)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[sin(u)]=cos(u)du\&Hf=\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}\end{align*}$

$f(x,y)= x^{2} - 3xy - y^{2}$

Use the Hessian matrix , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at ?

The graph of has a saddle point at .

The graph of has a critical point at , but the Hessian matrix test is inconclusive.

The graph of does not have a critical point at .

The graph of has a local minimum at .

The graph of has a local maximum at .

Explanation

First, it must be established that the graph of has a critical point at ; this holds if $f_{x}(0,0) = f_{y}(0,0) = 0$ , so the first partial derivatives of must be evaluated at :

$f(x,y)= x^{2} - 3xy - y^{2}$

$f_{x}(x,y)= 2x - 3y$

$f_{x}(0,0)= 2(0) - 3(0) = 0$

$f_{y}(x,y)= - 3x- 2y$

$f_{y}(0,0)= - 3(0) -2(0) = 0$

The graph of has a critical point at , so the Hessian matrix test applies.

The Hessian matrix is the matrix of partial second derivatives

$\begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}$ ,

the determinant of which can be used to determine whether a critical point of is a local maximum, a local minimum, or a saddle point. Find the partial second derivatives of :

$f_{x}(x,y)= 2x - 3y$

$f_{xx}(x,y)= 2$

$f_{xy}(x,y)= -3$

$f_{y}(x,y)= - 3x- 2y$

$f_{yx}(x,y)= - 3$

$f_{y}(x,y)= - 2$

All four partial second derivatives are constant; the Hessian matrix at is

$\begin{bmatrix} 2 & - 3 \ -3 & -2 \end{bmatrix}$

Its determinant is the upper-left to lower-right product minus the upper-right to lower-left product:

$\begin{vmatrix} 2 & - 3 \ -3 & -2 \end{vmatrix} = 2(-2) - (-3)(-3) = -13$

The determinant of the Hessian is negative, so the graph of has a saddle point at .

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=3\cdot 4^yz^{2}e^{(6x)}\end{align*}$

$\begin{bmatrix}108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\end{bmatrix}$

$\begin{bmatrix}36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\end{bmatrix}$

$\begin{bmatrix}36\cdot 4^yze^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&108\cdot 4^yz^{2}e^{(6x)}\6\cdot 4^yze^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&18\cdot 4^yz^{2}e^{(6x)}ln(4)\6\cdot 4^ye^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\end{bmatrix}$

$\begin{bmatrix}3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&6\cdot 4^yze^{(6x)}ln(4)\18\cdot 4^yz^{2}e^{(6x)}ln(4)&108\cdot 4^yz^{2}e^{(6x)}&36\cdot 4^yze^{(6x)}\6\cdot 4^yze^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}&6\cdot 4^ye^{(6x)}\end{bmatrix}$

Explanation

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=3\cdot 4^yz^{2}e^{(6x)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[a^u]=a^uduln(a)\&Hf=\begin{bmatrix}108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\end{bmatrix}\end{align*}$

Let $f(x, y) =2^{x}3^{y}$ .

Find the Hessian matrix .

$2^{x}3^{y} \begin{bmatrix} (\ln 2)^{2} &(\ln 2)(\ln 3) \ (\ln 2)(\ln 3) & (\ln 3)^{2} \end{bmatrix}$

$2^{x}3^{y} \begin{bmatrix} (\ln 2)^{2} &1 \1& (\ln 3)^{2} \end{bmatrix}$

$2^{x}3^{y} \begin{bmatrix} 1 &(\ln 2)(\ln 3) \ (\ln 2)(\ln 3) & 1 \end{bmatrix}$

$2^{x}3^{y} \begin{bmatrix} 1& 1 \ 1 & 1 \end{bmatrix}$

$2^{x}3^{y} \begin{bmatrix} 1 & 0 \ 0& 1 \end{bmatrix}$

Explanation

The Hessian matrix is the matrix of partial second derivatives

$H(f)=\begin{bmatrix} f_{xx}(x,y) & f_{xy}(x,y) \ f_{yx}(x,y) & f_{yy}(x,y) \end{bmatrix}$ .

$f(x, y) =2^{x}3^{y}$ can be rewritten as

$f(x, y) =(e^{ \ln 2} )^{x}\cdot (e^{ \ln 3})^{y}$

$f(x, y) =e^{x \ln 2} \cdot e^{y \ln 3}$ , then

$f(x, y) =e^{x \ln 2+ y \ln 3}$

This makes the partial second derivatives easier to find.

$f_{x}(x, y) =(\ln 2) e^{x \ln 2+ y \ln 3}$

$f_{xx}(x, y) =(\ln 2) (\ln 2)e^{x \ln 2+ y \ln 3} = 2^{x}3^{y}(\ln 2)^{2}$

$f_{xy}(x, y) =(\ln 2) (\ln 3)e^{x \ln 2+ y \ln 3}= 2^{x}3^{y}(\ln 2)(\ln 3)$

$f_{y}(x, y) =(\ln 3) e^{x \ln 2+ y \ln 3}$

$f_{yx}(x, y) =(\ln 3) (\ln 2)e^{x \ln 2+ y \ln 3}= 2^{x}3^{y}(\ln 2)(\ln 3)$

$f_{yy}(x, y) =(\ln 3) (\ln 3)e^{x \ln 2+ y \ln 3}= 2^{x}3^{y}(\ln 3)^{2}$

The Hessian matrix for is

$H(f)=\begin{bmatrix} 2^{x}3^{y}(\ln 2)^{2} & 2^{x}3^{y}(\ln 2)(\ln 3) \ 2^{x}3^{y}(\ln 2)(\ln 3) & 2^{x}3^{y}(\ln 3)^{2} \end{bmatrix}$

$= 2^{x}3^{y} \begin{bmatrix} (\ln 2)^{2} &(\ln 2)(\ln 3) \ (\ln 2)(\ln 3) & (\ln 3)^{2} \end{bmatrix}$

Let $f(x,y, z) = xy \cos z + xz \tan y$

Which of the following does not appear in the Hessian matrix of ?

$2 xz \sec^{3} y$

$- xy \cos z$

$- x \sin z + x \sec^{2} y$

$- y \sin z + \tan y$

$\cos z + z \sec^{2} y$

Explanation

The Hessian matrix of is the matrix of partial second derivatives

$H(f) = \begin{bmatrix} f_{xx} & f_{xy} & f_{xz} \f_{yx} & f_{yy} & f_{yz} \ f_{zx} & f_{zy} & f_{zz} \end{bmatrix}$

To identify which choice does not give an entry of the matrix, we need to find all nine partial derivatives; however, since $f_{yx}= f_{xy}$ , $f_{zx}= f_{xz}$ , and $f_{zy}= f_{yz}$ , we need only find six such derivatives. They are as follows:

$f(x,y, z) = xy \cos z + xz \tan y$

$f_{x}(x,y, z) = y \cos z + z \tan y$

$f_{xx}(x,y, z) =0$

$f_{xy}(x,y, z) = \cos z + z \sec^{2} y$

$f_{xz}(x,y, z) =- y \sin z + \tan y$

$f_{y}(x,y, z) = x \cos z + xz \sec^{2} y$

$f_{yy}(x,y, z) = 2 xz \tan y \sec^{2} y$

$f_{yz}(x,y, z) = - x \sin z + x \sec^{2} y$

$f_z(x,y, z) =- xy \sin z + x \tan y$

$f_{zz}(x,y, z) =- xy \cos z$

Of the five given choices, only $2 xz \sec^{3} y$ is not one of the partial second derivatives. This is the correct choice.

Consider the function $f(x, y) = 2x \cos y - 2x$ .

Determine whether the graph of the function has a critical point at ; if so, use the Hessian matrix to identify as a local maximum, a local minimum, or a saddle point.

The graph of has a critical point at , but the Hessian matrix test is inconclusive.

The graph of has a saddle point at .

The graph of has a local maximum at .

The graph of has a local minimum at .

The graph of does not have a critical point at .

Explanation

First, it must be established that is a critical point of the graph of ; this holds if and only if both first partial derivatives are equal to 0 at this point. Find the partial derivatives and evaluate them at :

$f(x, y) = 2x \cos y - 2x$

$f_{x}(x,y) = 2 \cos y - 2$

$f_{x}(0,0) = 2 \cos 0 - 2 = 2(1)- 2 = 0$

$f_{y}(x, y) = -2x \sin y$

$f_{y}(0,0) = -2 (0) \sin 0=0$

Thus, the graph of has a critical point at .

The Hessian matrix is the matrix of partial second derivatives

$H(f)=\begin{bmatrix} f_{xx}(x,y) & f_{xy}(x,y) \ f_{yx}(x,y) & f_{yy}(x,y) \end{bmatrix}$ ;

Find these derivatives and evaluate them at :

$f_{x}(x,y) = 2 \cos y - 2$

$f_{xx}(x,y) = 0$

$f_{xx}(0,0) = 0$

$f_{xy}(x,y) =- 2 \sin y$

$f_{xy}(0,0) =- 2 \sin 0 = -2(0)= 0$

$f_{y}(x, y) = -2x \sin y$

$f_{yx}(x, y) = -2 \sin y$

$f_{yx}(0,0) =- 2 \sin 0 = -2(0)= 0$

$f_{yy}(x, y) = -2x \cos y$

$f_{yy}(0,0) = -2 (0) \cos 0 = 0$

The Hessian matrix, evaluated at , ends up being the matrix $\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$ . The determinant of the matrix is 0, which means that the Hessian matrix test is inconclusive.

The Hessian

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