The Hessian - Linear Algebra

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Question

is a continuous function such that $f_{x} (0,0) = f_{y}(0,0)= 0$ .

The Hessian matrix for , evaluated at , is

$H(f) = \begin{bmatrix} a & 3 \ 3 & 4 \end{bmatrix}$

From the set $\left \{ 1,2,3,4,5\right \}$ , which value(s) can be assigned to so that the graph of has a saddle point at ?

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Answer

The graph of has a saddle point at if and only

$\det H(f) < 0$

when evaluated at this point.

Calculate the determinant of the Hessian at this point in terms of by subtracting the upper-right to lower-left product by from the upper-left to lower-right product; set this less than 0 and solve for .

$\begin{vmatrix} a & 3 \ 3 & 4 \end{vmatrix} < 0$

$a< \frac{9}{4}$

Therefore, the graph of has a saddle point at if $a< \frac{9}{4}$ . The correct choice is therefore $\left \{ 1,2 \right \}$ .

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Question

is a continuous function such that $f_{x} (0,0) = f_{y}(0,0)= 0$ .

The Hessian matrix for , evaluated at , is

$H(f) = \begin{bmatrix} a & 3 \ 3 & 4 \end{bmatrix}$

From the set $\left \{ 1,2,3,4,5\right \}$ , which value(s) can be assigned to so that the graph of has a saddle point at ?

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Answer

The graph of has a saddle point at if and only

$\det H(f) < 0$

when evaluated at this point.

Calculate the determinant of the Hessian at this point in terms of by subtracting the upper-right to lower-left product by from the upper-left to lower-right product; set this less than 0 and solve for .

$\begin{vmatrix} a & 3 \ 3 & 4 \end{vmatrix} < 0$

$a< \frac{9}{4}$

Therefore, the graph of has a saddle point at if $a< \frac{9}{4}$ . The correct choice is therefore $\left \{ 1,2 \right \}$ .

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Question

is a continuous function such that $f_{x} (0,0) = f_{y}(0,0)= 0$ .

The Hessian matrix for , evaluated at , is

$H(f) = \begin{bmatrix} a & 3 \ 3 & 4 \end{bmatrix}$

From the set $\left \{ 1,2,3,4,5\right \}$ , which value(s) can be assigned to so that the graph of has a saddle point at ?

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Answer

The graph of has a saddle point at if and only

$\det H(f) < 0$

when evaluated at this point.

Calculate the determinant of the Hessian at this point in terms of by subtracting the upper-right to lower-left product by from the upper-left to lower-right product; set this less than 0 and solve for .

$\begin{vmatrix} a & 3 \ 3 & 4 \end{vmatrix} < 0$

$a< \frac{9}{4}$

Therefore, the graph of has a saddle point at if $a< \frac{9}{4}$ . The correct choice is therefore $\left \{ 1,2 \right \}$ .

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Question

is a continuous function such that $f_{x} (0,0) = f_{y}(0,0)= 0$ .

The Hessian matrix for , evaluated at , is

$H(f) = \begin{bmatrix} a & 3 \ 3 & 4 \end{bmatrix}$

From the set $\left \{ 1,2,3,4,5\right \}$ , which value(s) can be assigned to so that the graph of has a saddle point at ?

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Answer

The graph of has a saddle point at if and only

$\det H(f) < 0$

when evaluated at this point.

Calculate the determinant of the Hessian at this point in terms of by subtracting the upper-right to lower-left product by from the upper-left to lower-right product; set this less than 0 and solve for .

$\begin{vmatrix} a & 3 \ 3 & 4 \end{vmatrix} < 0$

$a< \frac{9}{4}$

Therefore, the graph of has a saddle point at if $a< \frac{9}{4}$ . The correct choice is therefore $\left \{ 1,2 \right \}$ .

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Question

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

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Answer

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 the partial derivatives as follows:

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

$\frac{\partial^2 f}{\partial x^2} =\frac{1}{y^{2}} \frac{\partial^2 }{\partial x^2} \left ( x^{2} \right )$

$= \frac{1}{y^{2}} \frac{\partial }{\partial x } \left (2x \right )$

$= \frac{1}{y^{2}} \cdot 2$

$= \frac{2 }{y^{2}}$

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

$= \frac{\partial }{\partial y } \left (\frac{1}{y^{2}} \cdot \frac{\partial }{\partial x } \left (x^{2}\right ) \right )$

$= \frac{\partial }{\partial y } \left (\frac{1}{y^{2}} \cdot2x \right )$

$=2x \cdot \frac{\partial }{\partial y } \left (\frac{1}{y^{2}} \right )$

$=2x \cdot \frac{-2}{y^{3}}$

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

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

$= x^{2} \cdot \frac{\partial^2 }{\partial y^2} \left ( \frac{1}{y^{2}} \right )$

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

$= x^{2} \cdot \frac{6}{y^{4}}$

$= \frac{6x^{2} }{y^{4}}$

The Hessian matrix is

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

or

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

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Question

Set up a Hessian Matrix from the following equation,

$f(x,y,z)=12-\frac{5}{2}(x^2-z^2)+3y^2-4z$

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Answer

Recall what a hessian matrix is:

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

Now let's calculate each second order derivative separately, and then put it into the matrix.

$\frac{\partial^2 f}{\partial x \partial x}=-5$

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

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

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

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

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

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

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

$\frac{\partial^2 f}{\partial z \partial z }=5$

Now we put each entry into its place in the Hessian Matrix, and it should look like

$\begin{bmatrix} -5 & 0 & 0 \ 0 & 6 & 0 \ 0 & 0 & 5 \end{bmatrix}$

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Question

Find the Hessian of the following function.

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

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Answer

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

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

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Question

$\begin{align}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y)=3^{(4x)}tan(4y) - 11ye^{(5x)}\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=3^{(4x)}tan(4y) - 11ye^{(5x)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[a^u]=a^uduln(a)\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[e^u]=e^udu\&Hf=\begin{bmatrix}16\cdot 3^{(4x)}tan(4y)ln(3)^{2} - 275ye^{(5x)}&4\cdot 3^{(4x)}ln(3)\cdot (4tan(4y)^{2} + 4) - 55e^{(5x)}\4\cdot 3^{(4x)}ln(3)\cdot (4tan(4y)^{2} + 4) - 55e^{(5x)}&8\cdot 3^{(4x)}tan(4y)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find }H\text{ for the function}f(x,y,z)=3\cdot 5^zcos(3x)e^{(3y)} - 6zcos(4x)e^{(5y)}\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[e^u]=e^udu\&d[a^u]=a^uduln(a)\&\&Hf=\begin{bmatrix}96zcos(4x)e^{(5y)} - 27\cdot 5^zcos(3x)e^{(3y)}&120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)\120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&27\cdot 5^zcos(3x)e^{(3y)} - 150zcos(4x)e^{(5y)}&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}\24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}&3\cdot 5^zcos(3x)e^{(3y)}ln(5)^{2}\end{bmatrix}\end{align}$

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Question

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

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Answer

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

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Question

$\begin{align}&\text{Find the matrix }Hf \text{ for }f(x,y)=- 2\cdot 4^{(6y)} - 5y^{2}ln(2x)\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=- 2\cdot 4^{(6y)} - 5y^{2}ln(2x)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}\&d[cos(u)]=-sin(u)du\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[a^u]=a^uduln(a)\&Hf=\begin{bmatrix}\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\end{bmatrix}\end{align}$

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Question

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

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Answer

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

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Question

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

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=20xy^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&Hf=\begin{bmatrix}0&40y\40y&40x\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the matrix }Hf \text{ for }f(x,y,z)=11\cdot 3^{(4y)}\cdot 4^{(3x)}z\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=11\cdot 3^{(4y)}\cdot 4^{(3x)}z\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[a^u]=a^uduln(a)\&Hf=\begin{bmatrix}99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&0\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Calculate the Hessian matrix of the function}\&f(x,y,z)=14e^{(5x)}e^{(y)}e^{(z)}\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=14e^{(5x)}e^{(y)}e^{(z)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&Hf=\begin{bmatrix}350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=14cos(6z)tan(6x)sin(y)\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=14cos(6z)tan(6x)sin(y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\&d[cos(u)]=-sin(u)du\&Hf=\begin{bmatrix}168cos(6z)tan(6x)sin(y)\cdot (6tan(6x)^{2} + 6)&14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)\14cos(6z)cos(y)\cdot (6tan(6x)^{2} + 6)&-14cos(6z)tan(6x)sin(y)&-84tan(6x)sin(6z)cos(y)\-84sin(6z)sin(y)\cdot (6tan(6x)^{2} + 6)&-84tan(6x)sin(6z)cos(y)&-504cos(6z)tan(6x)sin(y)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Calculate the Hessian matrix of the function}\&f(x,y)=10x^{2}ln(6y)\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=10x^{2}ln(6y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}\&d[e^u]=e^udu\&Hf=\begin{bmatrix}20ln(6y)&\frac{(20x)}{y}\\frac{(20x)}{y}&-\frac{(10x^{2})}{y^{2}}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Calculate the Hessian matrix of the function}\&f(x,y)=10ysin(6x)\end{align}$

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Answer

$\begin{align}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=10ysin(6x)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&Hf=\begin{bmatrix}-360ysin(6x)&60cos(6x)\60cos(6x)&0\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the matrix }Hf \text{ for }f(x,y)=-7cos(3y)ln(4x)\end{align}$

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Answer

$\begin{align}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=-7cos(3y)ln(4x)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}\&d[cos(u)]=-sin(u)du\&Hf=\begin{bmatrix}\frac{(7cos(3y))}{x^{2}}&\frac{(21sin(3y))}{x}\\frac{(21sin(3y))}{x}&63cos(3y)ln(4x)\end{bmatrix}\end{align}$

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Question

Give the Hessian matrix of the function

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

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Answer

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 partial second derivative separately:

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

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

$= \frac{2x }{x^{2}+y}$

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

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

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

$= \frac{(2x^{2}+2y) - 4x^{2}}{(x^{2}+y)^{2}}$

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

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

$=2x \frac{\partial }{\partial y} \left ( x^{2}+y \right )^{-1}$

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

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

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

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

$=\frac{1 }{x^{2}+y}$

$\frac{\partial ^2 f}{\partial y^2} = \frac{\partial }{\partial y} \left ( \frac{1 }{x^{2}+y} \right )$

$= \frac{\partial }{\partial y} \left ( x^{2}+y \right )^{-1}$

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

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

$= \frac{ - 1}{(x^{2}+y)^{2}}$

The Hessian of is

$\begin{bmatrix} \frac{-2x^{2}+2y}{(x^{2}+y)^{2}} & \frac{ - 2x }{(x^{2}+y)^{2}} \ \frac{ - 2x }{(x^{2}+y)^{2}} & \frac{ - 1}{(x^{2}+y)^{2}} \end{bmatrix}$ ,

which can be rewritten as

$\frac{ - 1}{(x^{2}+y)^{2}} \begin{bmatrix} 2x^{2}-2y & 2x \ 2x & 1 \end{bmatrix}$

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