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

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

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

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

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

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

The graph of \(\displaystyle f\) has a critical point at \(\displaystyle (0,0)\), so the Hessian matrix test applies.

The Hessian matrix \(\displaystyle H(f)\) is the matrix of partial second derivatives

\(\displaystyle \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}\),

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

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

\(\displaystyle f_{xx}(x,y)= 2\)

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

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

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

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

All four partial second derivatives are constant; the Hessian matrix at \(\displaystyle (0,0)\) is 

\(\displaystyle \begin{bmatrix} 2 & - 3 \\ -3 & -2 \end{bmatrix}\)

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

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

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

\(\displaystyle f(x,y)= e^{x} y^{2}\)

\(\displaystyle f_{x}(x,y)= e^{x} y^{2}\)

\(\displaystyle f_{x}(0,0)= e^{0}\cdot 0^{2} = 0\)

\(\displaystyle f_{y}(x,y)= 2e^{x} y\)

\(\displaystyle f_{y}(0,0) = 2e^{0} \cdot 0 = 0\)

The graph of \(\displaystyle f\) has a critical point at \(\displaystyle (0,0)\), so the Hessian matrix test applies.

The Hessian matrix \(\displaystyle H(f)\) is the matrix of partial second derivatives

\(\displaystyle \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}\),

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

\(\displaystyle f_{xx}(x,y)= e^{x} y^{2}\)

\(\displaystyle f_{xx}(0,0)= e^{0} \cdot 0^{2} = 0\)

\(\displaystyle f_{xy}(x,y)= 2 e^{x} y\)

\(\displaystyle f_{xy}(0,0)= 2 e^{0} \cdot 0 = 0\)

\(\displaystyle f_{yx}(x,y)= 2e^{x} y\)

\(\displaystyle f_{yx}(0,0)= 2 e^{0} \cdot 0 = 0\)

\(\displaystyle f_{yy}(x,y)= 2e^{x}\)

\(\displaystyle f_{yy}(0,0)= 2 e^{0} = 2\)

The Hessian matrix at \(\displaystyle (0,0)\) is 

\(\displaystyle \begin{bmatrix}0 & 0 \\0 & 2 \end{bmatrix}\)

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

\(\displaystyle \begin{vmatrix}0 & 0 \\0 & 2 \end{vmatrix} = 0(2) - 0(0) = 0\)

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

\(\displaystyle f(x,y) = \cos ^{2}x \cos y\)

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

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

\(\displaystyle f_{y}(x,y) = - \cos ^{2}x \sin y\)

\(\displaystyle f_{y}(0,0) = - \cos ^{2}0 \sin 0 = -(1^{2}) (0)= 0\)

The graph of \(\displaystyle f\) has a critical point at \(\displaystyle (0,0)\), so the Hessian matrix test applies.

The Hessian matrix \(\displaystyle H(f)\) is the matrix of partial second derivatives

\(\displaystyle \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}\),

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

\(\displaystyle f_{xx}(x,y) =- 2[ (\sin x) (-\sin x) +( \cos x )(\cos x) ] \cos y\)

\(\displaystyle =- 2 ( \cos ^{2}x- \sin ^{2}x )\cos y\)

\(\displaystyle f(0,0)=- 2 ( \cos ^{2}0- \sin ^{2}0 )\cos 0 = -2(1^{2}-0^{2}) (1) = -2\)

\(\displaystyle f_{xy}(x,y) = 2 \sin x \cos x \sin y\)

\(\displaystyle f_{xy}(0,0) = 2 \sin 0 \cos 0 \sin 0 = 2 (0)(1)(0 ) = 0\)

\(\displaystyle f_{yx}(x,y) = 2 \sin x \cos x \sin y\)

\(\displaystyle f_{yx}(0,0) = 2 \sin 0 \cos 0 \sin 0 = 2 (0)(1)(0 ) = 0\)

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

\(\displaystyle f_{yy}(0,0) = - \cos ^{2}0 \cos 0 = -( 1^{2}) (1 ) = -1\)

The Hessian matrix at \(\displaystyle (0,0)\) is 

\(\displaystyle \begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix}\)

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

\(\displaystyle \begin{vmatrix} -2 & 0 \\ 0 & -1 \end{vmatrix} = (-2)(-1) - (0)(0) = 2\)

The determinant is positive, making \(\displaystyle (0,0)\) a local extremum. Since \(\displaystyle f_{xx}(0,0)\) is negative, \(\displaystyle (0,0)\) is a local maximum. 

First, it must be established that the graph of \(\displaystyle f\) has a critical point at \(\displaystyle (-1,1)\); this holds if \(\displaystyle f_{x}(-1,1)= f_{y}(-1,1) = 0\), so the first partial derivatives of \(\displaystyle f\) must be evaluated at \(\displaystyle (-1,1)\):

\(\displaystyle f(x,y) = (x-y+ 2)^{2}\)

\(\displaystyle f_{x}(x,y) = 2 (x-y+ 2 ) = 2x-2y + 4\)

\(\displaystyle f_{x}(-1, 1) = 2(-1)-2(1) + 4= 0\)

\(\displaystyle f_{y}(x,y) = -2(x-y+ 2) = -2x+2y - 4\)

\(\displaystyle f_{y}(-1, 1) = -2(-1)+2(1) - 4= 0\)

The graph of \(\displaystyle f\) has a critical point at \(\displaystyle (-1,1)\), so the Hessian matrix test applies.

The Hessian matrix \(\displaystyle H(f)\) is the matrix of partial second derivatives

\(\displaystyle \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}\),

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

\(\displaystyle f_{x}(x,y) = 2x-2y + 4\)

\(\displaystyle f_{xx}(x,y) = 2\)

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

\(\displaystyle f_{y}(x,y) = -2x+2y - 4\)

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

\(\displaystyle f_{yy}(x,y) =2\)

All four partial second derivatives are constants. The Hessian matrix at any point, including \(\displaystyle (-1,1)\), is

\(\displaystyle \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}\);

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

\(\displaystyle \begin{vmatrix} 2 & -2 \\ -2 & 2 \end{vmatrix} = 2(2)- (-2)(-2) = 0\)

Since the determinant of the Hessian is 0, the Hessian matrix test is inconclusive.

First, it must be established that \(\displaystyle (1,0)\) is a critical point of the graph of \(\displaystyle f\); 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 \(\displaystyle (1,0)\):

\(\displaystyle f(x, y) = 2x \cos y -x^{2}\)

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

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

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

\(\displaystyle f_{y}(1,0) =- 2(1) \sin 0 = -2(1)(0)= 0\)

Thus, the graph of \(\displaystyle f\) has a critical point at \(\displaystyle (1,0)\).

The Hessian matrix is the matrix of partial second derivatives

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

Find these derivatives and evaluate them at \(\displaystyle (1,0)\):

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

\(\displaystyle f_{xx}(x, y) = -2\)

\(\displaystyle f_{xx}(1,0) = -2\)

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

\(\displaystyle f_{xy}(1,0 ) =- 2 \sin 0 = -2 (0) = 0\)

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

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

\(\displaystyle f_{yx}(1,0 ) =- 2 \sin 0 = -2 (0) = 0\)

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

\(\displaystyle f_{yy}(1,0) =- 2 (1) \cos 0 = -2 (1)(1) = -2\)

At \(\displaystyle (1,0)\), the Hessian matrix is

\(\displaystyle H(f)=\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)

The determinant of this matrix is \(\displaystyle \det H(f) = -2(-2)- 0(0) = 4\)

Since the determinant of the Hessian matrix is positive, the graph of \(\displaystyle f\) has a local extremum at \(\displaystyle (1,0)\); since \(\displaystyle f_{xx}(1,0) = -2\), a negative value, it is a local maximum.

First, it must be established that \(\displaystyle (0,0)\) is a critical point of the graph of \(\displaystyle f\); 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 \(\displaystyle (0,0)\):

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

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

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

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

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

Thus, the graph of \(\displaystyle f\) has a critical point at \(\displaystyle (0,0)\).

The Hessian matrix is the matrix of partial second derivatives

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

Find these derivatives and evaluate them at \(\displaystyle (0,0)\):

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

\(\displaystyle f_{xx}(x,y) = 0\)

\(\displaystyle f_{xx}(0,0) = 0\)

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

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

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

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

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

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

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

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

The Hessian matrix of \(\displaystyle f\) is the matrix of partial second derivatives

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

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

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

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

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

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

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

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

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

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

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

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

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

The Hessian matrix is the matrix of partial second derivatives

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

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

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

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

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

This makes the partial second derivatives easier to find.

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

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

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

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

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

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

The Hessian matrix for \(\displaystyle f\) is

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

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

The Hessian matrix is the matrix of partial second derivatives

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

Since \(\displaystyle f_{yx}= f_{xy} , f_{zx}= f_{xz} , f_{zy}= f_{yz}\), we only need to find six partial second derivatives and compare them to the five choices.

\(\displaystyle f = \cos x \sin y \tan z\)

\(\displaystyle f_{x} =- \sin x \sin y \tan z\)

\(\displaystyle f_{xx} =- \cos x \sin y \tan z\)

\(\displaystyle f_{xy} =- \sin x \cos y \tan z\)

\(\displaystyle f_{xz} =- \sin x \sin y \sec^{2} z\)

\(\displaystyle f_{y} = \cos x \cos y \tan z\)

\(\displaystyle f_{yy} = - \cos x \sin y \tan z\)

\(\displaystyle f_{yz} = \cos x \cos y \sec^{2} z\)

\(\displaystyle f_{z} = \cos x \sin y \sec^{2} z\)

\(\displaystyle f_{zz} =2 \cos x \sin y \tan z \sec^{2} z\)

As stated before,

\(\displaystyle f_{yx}= f_{xy} = - \sin x \cos y \tan z\)

\(\displaystyle f_{zx}= f_{xz} =- \sin x \sin y \sec^{2} z\)

\(\displaystyle f_{zy}= f_{yz} = \cos x \cos y \sec^{2} z\),

so all three expressions will appear twice in the Hessian matrix.

Also note that

\(\displaystyle f_{xx} = f_{yy} =- \cos x \sin y \tan z\),

so this expression appears twice as well.

\(\displaystyle f_{zz} =2 \cos x \sin y \tan z \sec^{2} z\),

however, only appears once. This is the correct choice.

\(\displaystyle \begin{align*}&\text{In asking for H, the question asks for the Hessian.}\\&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=13y^{2}tan(2x)tan(4z)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&Hf=\begin{bmatrix}52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\\13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}\end{align*}\)