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Example Question #25 : The Hessian

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

Use the Hessian matrix H(f) , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at (0,0) ?

Possible Answers:

The graph of has a local maximum at (0,0) .

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

The graph of does not have a critical point at (0,0) .

The graph of has a saddle point at (0,0) .

The graph of has a local minimum at (0,0) .

Correct answer:

The graph of has a saddle point at (0,0) .

Explanation:

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

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

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

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

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

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

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

The Hessian matrix H(f) 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 (0,0) 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 (0,0) .

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Example Question #26 : The Hessian

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

Use the Hessian matrix H(f) , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at (0,0) ?

Possible Answers:

The graph of has a local minimum at (0,0) .

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

The graph of has a saddle point at (0,0) .

The graph of has a local maximum at (0,0) .

The graph of does not have a critical point at (0,0) .

Correct answer:

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

Explanation:

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

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

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

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

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

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

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

The Hessian matrix H(f) 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 :

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

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

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

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

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

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

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

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

The Hessian matrix at (0,0) is

$\begin{bmatrix}0 & 0 \\0 & 2 \end{bmatrix}$

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

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

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

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Example Question #27 : The Hessian

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

Use the Hessian matrix H(f) , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at (0,0) ?

Possible Answers:

The graph of has a saddle point at (0,0) .

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

The graph of has a local minimum at (0,0) .

The graph of does not have a critical point at (0,0) .

The graph of has a local maximum at (0,0) .

Correct answer:

The graph of has a local maximum at (0,0) .

Explanation:

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

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

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

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

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

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

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

The Hessian matrix H(f) 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 :

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

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

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

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

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

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

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

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

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

The Hessian matrix at (0,0) is

$\begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix}$

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

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

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

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Example Question #28 : The Hessian

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

Use the Hessian matrix H(f) , if applicable, to answer this question:

Does the graph of have a local maximum, a local minimum, or a saddle point at (-1,1) ?

Possible Answers:

The graph of does not have a critical point at (-1,1) .

The graph of has a saddle point at (-1,1) .

The graph of has a local maximum at (-1,1) .

The graph of has a local minimum at (-1,1) .

The graph of has a critical point at (-1,1) , but the Hessian matrix test is inconclusive.

Correct answer:

The graph of has a critical point at (-1,1) , but the Hessian matrix test is inconclusive.

Explanation:

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

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

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

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

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

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

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

The Hessian matrix H(f) 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 :

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

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

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

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

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

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

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

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

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

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

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

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Example Question #41 : Linear Algebra

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

Determine whether the graph of the function has a critical point at (1,0) ; if so, use the Hessian matrix H(f) to identify (1,0) as a local maximum, a local minimum, or a saddle point.

Possible Answers:

The graph of has a saddle point at (0,0) .

The graph of does not have a critical point at (0,0) .

The graph of has a local maximum at (0,0) .

The graph of has a local minimum at (0,0) .

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

Correct answer:

The graph of has a local maximum at (0,0) .

Explanation:

First, it must be established that (1,0) 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 (1,0) :

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

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

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

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

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

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

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 (1,0) :

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

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

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

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

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

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

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

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

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

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

At (1,0) , the Hessian matrix is

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

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

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

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Example Question #30 : The Hessian

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

Determine whether the graph of the function has a critical point at (0,0) ; if so, use the Hessian matrix H(f) to identify (0,0) as a local maximum, a local minimum, or a saddle point.

Possible Answers:

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

The graph of has a saddle point at (0,0) .

The graph of has a local maximum at (0,0) .

The graph of has a local minimum at (0,0) .

The graph of does not have a critical point at (0,0) .

Correct answer:

The graph of has a critical point at (0,0) , but the Hessian matrix test is inconclusive.

Explanation:

First, it must be established that (0,0) 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 (0,0) :

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

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

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

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

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

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

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 (0,0) :

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

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

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

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

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

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

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

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

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

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

The Hessian matrix, evaluated at (0,0) , 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.

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Example Question #41 : Linear Algebra

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

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

Possible Answers:

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

$- xy \cos z$

$- y \sin z + \tan y$

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

$2 xz \sec^{3} y$

Correct answer:

$2 xz \sec^{3} y$

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.

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Example Question #32 : The Hessian

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

Find the Hessian matrix H(f) .

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #41 : Matrix Calculus

Consider the function $f(x, y, z ) = \cos x \sin y \tan z$ .

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

Possible Answers:

$- \cos x \sin y \tan z$

$2 \cos x \sin y \tan z \sec^{2} z$

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

$- \sin x \cos y \tan z$

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

Correct answer:

$2 \cos x \sin y \tan z \sec^{2} z$

Explanation:

The Hessian matrix is the matrix of partial second derivatives

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

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

$f = \cos x \sin y \tan z$

$f_{x} =- \sin x \sin y \tan z$

$f_{xx} =- \cos x \sin y \tan z$

$f_{xy} =- \sin x \cos y \tan z$

$f_{xz} =- \sin x \sin y \sec^{2} z$

$f_{y} = \cos x \cos y \tan z$

$f_{yy} = - \cos x \sin y \tan z$

$f_{yz} = \cos x \cos y \sec^{2} z$

$f_{z} = \cos x \sin y \sec^{2} z$

$f_{zz} =2 \cos x \sin y \tan z \sec^{2} z$

As stated before,

$f_{yx}= f_{xy} = - \sin x \cos y \tan z$

$f_{zx}= f_{xz} =- \sin x \sin y \sec^{2} z$

$f_{zy}= f_{yz} = \cos x \cos y \sec^{2} z$ ,

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

Also note that

$f_{xx} = f_{yy} =- \cos x \sin y \tan z$ ,

so this expression appears twice as well.

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

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

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Example Question #1 : Gradients And Hessians Of Quadratic And Linear Functions

$\begin{align*}&\text{Find the matrix }Hf \text{ for }f(x,y,z)=13y^{2}tan(2x)tan(4z)\end{align*}$

Possible Answers:

$\begin{bmatrix}13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\\52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}26tan(2x)tan(4z)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\\26ytan(2x)\cdot (4tan(4z)^{2} + 4)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\\13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)\\26ytan(2x)\cdot (4tan(4z)^{2} + 4)&26tan(2x)tan(4z)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)\\104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

Correct answer:

Explanation:

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

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