The Hessian - Linear Algebra

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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:

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

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

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

$= $\frac{\partial }{\partial y }$ \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 $$\frac{-2}{y^{3}$$}$

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

$= $x^{2}$ \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

.

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Question

Set up a Hessian Matrix from the following equation,

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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.

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.

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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.

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)}$__MATH_EXPR_9__24sin(4x)e^{(5y)}$ - 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)}$$e^{(3y)}$__MATH_EXPR_1054sin(6z)e^{(3x)}$$e^{(3y)}$MATH_EXPR_12108cos(6z)e^{(3x)}$$e^{(3y)}$$\54sin(6z)e^{(3x)}$$e^{(3y)}$MATH_EXPR_1654sin(6z)e^{(3x)}$$e^{(3y)}$MATH_EXPR_18108cos(6z)e^{(3x)}$$e^{(3y)}$$\108cos(6z)e^{(3x)}$$e^{(3y)}$MATH_EXPR_22108cos(6z)e^{(3x)}$$e^{(3y)}$MATH_EXPR_24__-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)}$$e^{(z)}$__MATH_EXPR_870e^{(5x)}$$e^{(y)}$$e^{(z)}$MATH_EXPR_970e^{(5x)}$$e^{(y)}$$e^{(z)}$$\70e^{(5x)}$$e^{(y)}$$e^{(z)}$MATH_EXPR_1114e^{(5x)}$$e^{(y)}$$e^{(z)}$MATH_EXPR_1214e^{(5x)}$$e^{(y)}$$e^{(z)}$$\70e^{(5x)}$$e^{(y)}$$e^{(z)}$MATH_EXPR_1414e^{(5x)}$$e^{(y)}$$e^{(z)}$MATH_EXPR_15__14e^{(5x)}$$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}$

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

$=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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Question

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

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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}$

First, rewrite

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

as

$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}$$y^{-3}$$

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

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

The Hessian of is

,

which can be rewritten as

.

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Question

Give the Hessian matrix for the function $f(x, y) = \sin (xy)+ \cos (xy)$ .

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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 of these derivatives as follows:

$$\frac{\partial f}{\partial x}$ = $\frac{\partial }{\partial x}$ ( \sin (xy)+\cos (xy ))$

$= $\frac{\partial }{\partial x}$(xy) \cdot ( \cos (xy)-\sin (xy ))$

$=y( \cos (xy)-\sin (xy ))$

$= y \cdot $\frac{\partial }{\partial x}$ ( \cos (xy)-\sin (xy ))$

$= y \cdot $\frac{\partial }{\partial x}$(xy ) \cdot ( -\sin(xy)-\cos (xy ))$

$= y \cdot y \cdot ( -\sin(xy)-\cos (xy ))$

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

$$\frac{\partial $f^{2}$$}{\partial y\partial x } = $\frac{\partial $f^{2}$$}{\partial x\partial y } =\frac{\partial }{\partial y}$ (y( \cos (xy)-\sin (xy )))$

$=y \cdot $\frac{\partial }{\partial y}$ ( ( \cos (xy)-\sin (xy ))) + $\frac{\partial }{\partial y}$ (y) ( \cos (xy)-\sin (xy ))$

$=y \cdot $\frac{\partial }{\partial y}$(xy)\cdot ( -\sin(xy)-\cos (xy )) + 1 ( \cos (xy)-\sin (xy ))$

$=y \cdot x\cdot ( -\sin(xy)-\cos (xy )) + ( \cos (xy)-\sin (xy ))$

$= \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy ))$

$$\frac{\partial f}{\partial y}$ = $\frac{\partial }{\partial y}$ ( \sin (xy)+\cos (xy ))$

$= $\frac{\partial }{\partial y}$(xy) \cdot ( \cos (xy)-\sin (xy ))$

$=x( \cos (xy)-\sin (xy ))$

$= x \cdot $\frac{\partial }{\partial y}$ ( \cos (xy)-\sin (xy ))$

$= x \cdot $\frac{\partial }{\partial y}$(xy ) \cdot ( -\sin(xy)-\cos (xy ))$

$= x \cdot x \cdot ( -\sin(xy)-\cos (xy ))$

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

The Hessian matrix is

$\begin{bmatrix}- y ^{2}( \sin(xy)+\cos (xy )) & \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) \ \cos (xy ) - \sin(xy ) -xy (\sin(xy)+\cos(xy )) & -x ^{2}( \sin(xy)+\cos (xy )) \end{bmatrix}$

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Question

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

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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 of these derivatives as follows:

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

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

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

$=\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}$)$

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

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 __MATH_EXPR_20__2xe^{x }$ \end{bmatrix}$ ,

which can be rewritten, after a little algebra, as

.

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Question

Give the Hessian matrix of the function $f(x,y) = \cos x \sin y$ .

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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}$ .

To get the entries, find these derivatives as follows:

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

$=\sin y ( - \cos x )$

$= -\cos x \sin y$

$= $\frac{\partial }{\partial x}$(\cos x $\frac{\partial }{\partial y}$ ( \sin y))$

$= $\frac{\partial }{\partial x}$(\cos x \cos y)$

$= \cos y $\frac{\partial }{\partial x}$(\cos x )$

$= \cos y (- \sin x )$

$= - \sin x \cos y$

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

$=\cos x $\frac{\partial }{\partial y }$ ( \cos y)$

$=\cos x ( - \sin y)$

$= -\cos x \sin y$

The Hessian matrix is $\begin{bmatrix} -\cos x \sin y & - \sin x \cos y \ - \sin x \cos y & -\cos x \sin y \end{bmatrix}$ .

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