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Linear Algebra : The Gradient

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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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Linear Algebra : The Gradient

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

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

All Linear Algebra Resources

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