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Linear Algebra : Matrix Calculus

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Example Question #11 : Linear Algebra

$\begin{align*}&\text{Calculate the gradient of the function}\\&f(x,y)=4ycos(5x) - 13ln(3y)sin(3x)\end{align*}$

Possible Answers:

$\begin{bmatrix}- 20ysin(5x) - 39cos(3x)ln(3y)\\4cos(5x) -\frac{ (13sin(3x))}{y}\end{bmatrix}$

$\begin{bmatrix}117ln(3y)sin(3x) - 100ycos(5x)&- 20sin(5x) -\frac{ (39cos(3x))}{y}\\- 20sin(5x) -\frac{ (39cos(3x))}{y}&\frac{(13sin(3x))}{y^{2}}\end{bmatrix}$

$\begin{bmatrix}4cos(5x) - 20ysin(5x) -\frac{ (13sin(3x))}{y}- 39cos(3x)ln(3y)\end{bmatrix}$

$\begin{bmatrix}117ln(3y)sin(3x) - 100ycos(5x)\\\frac{(13sin(3x))}{y^{2}}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}- 20ysin(5x) - 39cos(3x)ln(3y)\\4cos(5x) -\frac{ (13sin(3x))}{y}\end{bmatrix}$

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=4ycos(5x) - 13ln(3y)sin(3x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&\\&\nabla f=\begin{bmatrix}- 20ysin(5x) - 39cos(3x)ln(3y)\\4cos(5x) -\frac{ (13sin(3x))}{y}\end{bmatrix}\end{align*}$

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Example Question #12 : The Gradient

$\begin{align*}&\text{Determine the gradient, }\nabla\text{, of the function}\\&f(x,y)=13\cdot 2^{(6y)}ln(3x)\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{(13\cdot 2^{(6y)})}{x}\\78\cdot 2^{(6y)}ln(3x)ln(2)\end{bmatrix}$

$\begin{bmatrix}\frac{(13\cdot 2^{(6y)})}{x}+ 78\cdot 2^{(6y)}ln(3x)ln(2)\end{bmatrix}$

$\begin{bmatrix}-\frac{(13\cdot 2^{(6y)})}{x^{2}}&\frac{(78\cdot 2^{(6y)}ln(2))}{x}\\\frac{(78\cdot 2^{(6y)}ln(2))}{x}&468\cdot 2^{(6y)}ln(3x)ln(2)^{2}\end{bmatrix}$

$\begin{bmatrix}-\frac{(13\cdot 2^{(6y)})}{x^{2}}\\468\cdot 2^{(6y)}ln(3x)ln(2)^{2}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}\frac{(13\cdot 2^{(6y)})}{x}\\78\cdot 2^{(6y)}ln(3x)ln(2)\end{bmatrix}$

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=13\cdot 2^{(6y)}ln(3x)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\nabla f=\begin{bmatrix}\frac{(13\cdot 2^{(6y)})}{x}\\78\cdot 2^{(6y)}ln(3x)ln(2)\end{bmatrix}\end{align*}$

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Example Question #13 : The Gradient

$\begin{align*}&\text{Calculate the gradient of the function}\\&f(x,y,z)=-9xsin(6y)e^{(3z)}\end{align*}$

Possible Answers:

$\begin{bmatrix}- 9sin(6y)e^{(3z)} - 54xcos(6y)e^{(3z)} - 27xsin(6y)e^{(3z)}\end{bmatrix}$

$\begin{bmatrix}-9sin(6y)e^{(3z)}\\-54xcos(6y)e^{(3z)}\\-27xsin(6y)e^{(3z)}\end{bmatrix}$

$\begin{bmatrix}0&-54cos(6y)e^{(3z)}&-27sin(6y)e^{(3z)}\\-54cos(6y)e^{(3z)}&324xsin(6y)e^{(3z)}&-162xcos(6y)e^{(3z)}\\-27sin(6y)e^{(3z)}&-162xcos(6y)e^{(3z)}&-81xsin(6y)e^{(3z)}\end{bmatrix}$

$\begin{bmatrix}0\\324xsin(6y)e^{(3z)}\\-81xsin(6y)e^{(3z)}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-9sin(6y)e^{(3z)}\\-54xcos(6y)e^{(3z)}\\-27xsin(6y)e^{(3z)}\end{bmatrix}$

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=-9xsin(6y)e^{(3z)}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&\nabla f=\begin{bmatrix}-9sin(6y)e^{(3z)}\\-54xcos(6y)e^{(3z)}\\-27xsin(6y)e^{(3z)}\end{bmatrix}\end{align*}$

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Example Question #14 : The Gradient

$\begin{align*}&\text{Calculate the gradient of the function}\\&f(x,y,z)=-18x^{2}yln(2z)\end{align*}$

Possible Answers:

$\begin{bmatrix}- 18x^{2}ln(2z) -\frac{ (18x^{2}y)}{z}- 36xyln(2z)\end{bmatrix}$

$\begin{bmatrix}-36yln(2z)\\0\\\frac{(18x^{2}y)}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}-36xyln(2z)\\-18x^{2}ln(2z)\\-\frac{(18x^{2}y)}{z}\end{bmatrix}$

$\begin{bmatrix}-36yln(2z)&-36xln(2z)&-\frac{(36xy)}{z}\\-36xln(2z)&0&-\frac{(18x^{2})}{z}\\-\frac{(36xy)}{z}&-\frac{(18x^{2})}{z}&\frac{(18x^{2}y)}{z^{2}}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-36xyln(2z)\\-18x^{2}ln(2z)\\-\frac{(18x^{2}y)}{z}\end{bmatrix}$

Explanation:

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=-18x^{2}yln(2z)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&\nabla f=\begin{bmatrix}-36xyln(2z)\\-18x^{2}ln(2z)\\-\frac{(18x^{2}y)}{z}\end{bmatrix}\end{align*}$

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Example Question #11 : Matrix Calculus

$f(x,y ) = (x+2y+3 )^{4}$

Evaluate the gradient vector of at (0,0) .

Possible Answers:

(324, 648)

(108, 216)

(27. 27 )

(54,54)

(81, 162)

Correct answer:

(108, 216)

Explanation:

The gradient of f(x, y ) is the vector of partial first derivatives

$\left ( f_{x}, f_{y } \right )$ . Find these derivatives:

$f_{x}(x,y ) = \frac{\partial }{\partial x} (x+2y+3 )^{4}$

$= \frac{\partial }{\partial x} (x+2y+ 3) \cdot4 \cdot (x+2y+3) ^{3}$

$=1 \cdot4 \cdot (x+2y+3) ^{3}$

$=4 (x+2y+3) ^{3}$

$f_{y}(x,y ) = \frac{\partial }{\partial y} (x+2y+3 )^{4}$

$= \frac{\partial }{\partial y} (x+2y+ 3) \cdot 4 \cdot (x+2y+3) ^{3}$

$=2 \cdot 4 \cdot (x+2y+3) ^{3}$

$=8(x+2y+3) ^{3}$

Evaluate $f_{x}(0,0 )$ and $f_{y}(0,0 )$ :

$f_{x}(0,0 ) = 4 (0+2 \cdot 0 +3) ^{3}$

$= 4 \cdot 3^{3 }$

$= 4 \cdot 27$

= 108

$f_{x}(0,0 ) =8 (0+2 \cdot 0 +3) ^{3}$

$= 8 \cdot 3^{3 }$

$= 8 \cdot 27$

= 216

The gradient vector is (108, 216)

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Example Question #16 : The Gradient

Define $\textbf{f} : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ as follows:

$\textbf{f} (x,y) = ( \cos (xy), \cos (x+y ))$

Evaluate the Jacobian matrix of $\textbf{f}$ at (0,0 ) .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The Jacobian matrix of a function $\textbf{f}(x,y) = (f_{1}(x,y), f_{2}(x,y))$ is the matrix of partial first derivatives

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

Find each partial first derivative, and evaluate the expression at (0,0) .

$f_{1}(x,y) = \cos (xy)$

$f_{1x}(x,y) =y \sin (xy)$

$f_{1x}(0,0) =0 \sin (0\cdot 0 ) = 0$

$f_{1y}(x,y) =x \sin (xy)$

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

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

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

$f_{2x} (0,0) = \sin (0+0 ) = 0$

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

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

The Jacobian matrix at (0,0 ) is $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ .

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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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Linear Algebra : Matrix Calculus

Study concepts, example questions & explanations for Linear Algebra

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Example Questions

Example Question #11 : Linear Algebra

Example Question #12 : The Gradient

Example Question #13 : The Gradient

Example Question #14 : The Gradient

Example Question #11 : Matrix Calculus

Example Question #16 : The Gradient

Example Question #1 : The Hessian

Example Question #2 : The Hessian

Example Question #1 : The Hessian

Example Question #4 : The Hessian

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