We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Matrix Calculus

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Matrix Calculus

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

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

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

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 critical point at (0,0) , but the Hessian matrix test is inconclusive.

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) .

Report an Error

Example Question #42 : Matrix Calculus

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

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 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 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 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.

Report an Error

Example Question #41 : Matrix Calculus

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

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 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 has a saddle point 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 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.

Report an Error

Example Question #44 : Matrix Calculus

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

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 has a local maximum at (-1,1) .

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

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

The graph of does not have a critical point 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.

Report an Error

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

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

The graph of has a saddle point 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.

Report an Error

Example Question #22 : 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.

Report an Error

Example Question #31 : The Hessian

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:

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

$- xy \cos z$

$2 xz \sec^{3} y$

$- y \sin z + \tan y$

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

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.

Report an Error

Example Question #41 : Matrix Calculus

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

Find the Hessian matrix H(f) .

Possible Answers:

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

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

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

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

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

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

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

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

Report an Error

Example Question #42 : 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:

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

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

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

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

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

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.

Report an Error

Example Question #42 : Linear Algebra

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

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

Report an Error

View Linear Algebra Tutors

Balamurugan
Certified Tutor

University of Delaware, Bachelor of Engineering, Chemical Engineering.

View Linear Algebra Tutors

Reza
Certified Tutor

University of Sherbrooke, Doctor of Philosophy, Mathematics. University of Manitoba, Master of Science, Mathematics.

View Linear Algebra Tutors

Eric
Certified Tutor

Georgia Highlands College, Associate in Science, Computer Science. Kennesaw State University, Bachelor of Science, Computer E...

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SSAT Tutors in Dallas Fort Worth, Statistics Tutors in San Francisco-Bay Area, LSAT Tutors in Phoenix, GRE Tutors in San Diego, Reading Tutors in Los Angeles, Spanish Tutors in Los Angeles, Biology Tutors in Dallas Fort Worth, SSAT Tutors in Denver, ACT Tutors in Dallas Fort Worth, Computer Science Tutors in Dallas Fort Worth

Popular Courses & Classes

MCAT Courses & Classes in Houston, ISEE Courses & Classes in Los Angeles, GRE Courses & Classes in San Diego, ACT Courses & Classes in Seattle, SSAT Courses & Classes in Washington DC, ISEE Courses & Classes in Houston, SSAT Courses & Classes in Los Angeles, LSAT Courses & Classes in Chicago, SAT Courses & Classes in Boston, GRE Courses & Classes in Los Angeles

Popular Test Prep

LSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Chicago, SSAT Test Prep in Atlanta, ISEE Test Prep in Houston, LSAT Test Prep in Atlanta, ISEE Test Prep in Chicago, GMAT Test Prep in Seattle, SSAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Seattle, SSAT Test Prep in Miami

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Matrix Calculus

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #41 : Matrix Calculus

Example Question #42 : Matrix Calculus

Example Question #41 : Matrix Calculus

Example Question #44 : Matrix Calculus

Example Question #41 : Linear Algebra

Example Question #22 : The Hessian

Example Question #31 : The Hessian

Example Question #41 : Matrix Calculus

Example Question #42 : Matrix Calculus

Example Question #42 : Linear Algebra

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: