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Linear Algebra : The Hessian

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Example Question #1 : The Hessian

Set up a Hessian Matrix from the following equation,

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

Possible Answers:

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

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

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

$\begin{bmatrix} -5 & 0 & 2 \\ 0 & 6 & 0 \\ 2 & 0 & 5 \end{bmatrix}$

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

Correct answer:

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

Explanation:

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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Example Question #2 : The Hessian

Find the Hessian of the following function.

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

Recall the Hessian

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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Example Question #1 : The Hessian

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

Possible Answers:

$\begin{bmatrix}0&0\\0&8\cdot 3^{(4x)}tan(4y)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}8\cdot 3^{(4x)}tan(4y)\cdot (4tan(4y)^{2} + 4)&4\cdot 3^{(4x)}ln(3)\cdot (4tan(4y)^{2} + 4) - 55e^{(5x)}\\4\cdot 3^{(4x)}ln(3)\cdot (4tan(4y)^{2} + 4) - 55e^{(5x)}&16\cdot 3^{(4x)}tan(4y)ln(3)^{2} - 275ye^{(5x)}\end{bmatrix}$

$\begin{bmatrix}8\cdot 3^{(4x)}tan(4y)\cdot (4tan(4y)^{2} + 4)&0\\0&0\end{bmatrix}$

$\begin{bmatrix}16\cdot 3^{(4x)}tan(4y)ln(3)^{2} - 275ye^{(5x)}&4\cdot 3^{(4x)}ln(3)\cdot (4tan(4y)^{2} + 4) - 55e^{(5x)}\\4\cdot 3^{(4x)}ln(3)\cdot (4tan(4y)^{2} + 4) - 55e^{(5x)}&8\cdot 3^{(4x)}tan(4y)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}$

Correct answer:

Explanation:

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

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Example Question #4 : The Hessian

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

Possible Answers:

$\begin{bmatrix}24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)&120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&96zcos(4x)e^{(5y)} - 27\cdot 5^zcos(3x)e^{(3y)}\\9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}&27\cdot 5^zcos(3x)e^{(3y)} - 150zcos(4x)e^{(5y)}&120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}\\3\cdot 5^zcos(3x)e^{(3y)}ln(5)^{2}&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}&24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)\end{bmatrix}$

$\begin{bmatrix}24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}&3\cdot 5^zcos(3x)e^{(3y)}ln(5)^{2}\\120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&27\cdot 5^zcos(3x)e^{(3y)} - 150zcos(4x)e^{(5y)}&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}\\96zcos(4x)e^{(5y)} - 27\cdot 5^zcos(3x)e^{(3y)}&120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)\end{bmatrix}$

$\begin{bmatrix}27\cdot 5^zcos(3x)e^{(3y)} - 150zcos(4x)e^{(5y)}&120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}\\120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&96zcos(4x)e^{(5y)} - 27\cdot 5^zcos(3x)e^{(3y)}&24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)\\9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}&24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)&3\cdot 5^zcos(3x)e^{(3y)}ln(5)^{2}\end{bmatrix}$

$\begin{bmatrix}96zcos(4x)e^{(5y)} - 27\cdot 5^zcos(3x)e^{(3y)}&120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)\\120zsin(4x)e^{(5y)} - 27\cdot 5^zsin(3x)e^{(3y)}&27\cdot 5^zcos(3x)e^{(3y)} - 150zcos(4x)e^{(5y)}&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}\\24sin(4x)e^{(5y)} - 9\cdot 5^zsin(3x)e^{(3y)}ln(5)&9\cdot 5^zcos(3x)e^{(3y)}ln(5) - 30cos(4x)e^{(5y)}&3\cdot 5^zcos(3x)e^{(3y)}ln(5)^{2}\end{bmatrix}$

Correct answer:

Explanation:

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

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Example Question #5 : The Hessian

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

Possible Answers:

$\begin{bmatrix}108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}\\-216sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

$\begin{bmatrix}54sin(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&54sin(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}\\108cos(6z)e^{(3x)}e^{(3y)}&108cos(6z)e^{(3x)}e^{(3y)}&-216sin(6z)e^{(3x)}e^{(3y)}\end{bmatrix}$

Correct answer:

Explanation:

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

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Example Question #1 : The Hessian

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

Possible Answers:

$\begin{bmatrix}\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\end{bmatrix}$

$\begin{bmatrix}-\frac{(10y)}{x}&\frac{(5y^{2})}{x^{2}}\\- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}&-\frac{(10y)}{x}\end{bmatrix}$

$\begin{bmatrix}-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\\\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\end{bmatrix}$

$\begin{bmatrix}- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&\frac{(5y^{2})}{x^{2}}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}\frac{(5y^{2})}{x^{2}}&-\frac{(10y)}{x}\\-\frac{(10y)}{x}&- 10ln(2x) - 72\cdot 4^{(6y)}ln(4)^{2}\end{bmatrix}$

Explanation:

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

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Example Question #7 : The Hessian

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

Possible Answers:

$\begin{bmatrix}36\cdot 4^yze^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&108\cdot 4^yz^{2}e^{(6x)}\\6\cdot 4^yze^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&18\cdot 4^yz^{2}e^{(6x)}ln(4)\\6\cdot 4^ye^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\end{bmatrix}$

$\begin{bmatrix}108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\\36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\end{bmatrix}$

$\begin{bmatrix}36\cdot 4^yze^{(6x)}&6\cdot 4^yze^{(6x)}ln(4)&6\cdot 4^ye^{(6x)}\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&6\cdot 4^yze^{(6x)}ln(4)\\108\cdot 4^yz^{2}e^{(6x)}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}\end{bmatrix}$

$\begin{bmatrix}3\cdot 4^yz^{2}e^{(6x)}ln(4)^{2}&18\cdot 4^yz^{2}e^{(6x)}ln(4)&6\cdot 4^yze^{(6x)}ln(4)\\18\cdot 4^yz^{2}e^{(6x)}ln(4)&108\cdot 4^yz^{2}e^{(6x)}&36\cdot 4^yze^{(6x)}\\6\cdot 4^yze^{(6x)}ln(4)&36\cdot 4^yze^{(6x)}&6\cdot 4^ye^{(6x)}\end{bmatrix}$

Correct answer:

Explanation:

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

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Example Question #1 : The Hessian

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

Possible Answers:

$\begin{bmatrix}40x&40y\\40y&0\end{bmatrix}$

$\begin{bmatrix}0&40y\\40y&40x\end{bmatrix}$

$\begin{bmatrix}40y&40x\\0&40y\end{bmatrix}$

$\begin{bmatrix}40y&0\\40x&40y\end{bmatrix}$

Correct answer:

$\begin{bmatrix}0&40y\\40y&40x\end{bmatrix}$

Explanation:

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Example Question #9 : The Hessian

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

Possible Answers:

$\begin{bmatrix}99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&0\end{bmatrix}$

$\begin{bmatrix}33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}\\44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)\\0&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\end{bmatrix}$

$\begin{bmatrix}176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\\44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&0\end{bmatrix}$

$\begin{bmatrix}33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)&0\\132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&176\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)^{2}&44\cdot 3^{(4y)}\cdot 4^{(3x)}ln(3)\\99\cdot 3^{(4y)}\cdot 4^{(3x)}zln(4)^{2}&132\cdot 3^{(4y)}\cdot 4^{(3x)}zln(3)ln(4)&33\cdot 3^{(4y)}\cdot 4^{(3x)}ln(4)\end{bmatrix}$

Correct answer:

Explanation:

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

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Example Question #10 : The Hessian

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

Possible Answers:

$\begin{bmatrix}14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

$\begin{bmatrix}350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

$\begin{bmatrix}70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\70e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}\\350e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

$\begin{bmatrix}70e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}&350e^{(5x)}e^{(y)}e^{(z)}\\14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\\14e^{(5x)}e^{(y)}e^{(z)}&14e^{(5x)}e^{(y)}e^{(z)}&70e^{(5x)}e^{(y)}e^{(z)}\end{bmatrix}$

Correct answer:

Explanation:

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

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Linear Algebra : The Hessian

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