The Gradient - Linear Algebra

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Question

What is the the gradient vector of the following function?

Answer

Recall that

$\nabla_x f(x)=\begin{bmatrix} \frac{\partial f}{\partial x_1}\ \ \frac{\partial f}{\partial x_2} \ \ \frac{\partial f}{\partial x_3} \ \vdots \ \frac{\partial f}{\partial x_n} \end{bmatrix}$

All we need to do is calculate 3 partial derivatives, and put them into this form.

$\frac{\partial f}{\partial x}=18yz+100xz$

$\frac{\partial f}{\partial y}=18xz+30y$

$\frac{\partial f}{\partial z}=18xy+50x^2$

Put these into vector form to get

$\nabla_x f(x)=\begin{bmatrix} 18yz+100xz\ 18xz+30y \ 18xy+50x^2 \end{bmatrix}$

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Question

Find the gradient vector of the following function.

$f(x,y)=x^2y^2+10\ln(xy)$

Answer

To find the gradient vector, we need to find the partial derivatives in respect to x and y.

$\frac{\partial f}{\partial x}=2xy^2+\frac{10}{x}$

$\frac{\partial f}{\partial y}=2yx^2+\frac{10}{y}$

Then our final answer looks like

$\begin{bmatrix} 2xy^2+\frac{10}{x}\ \ 2yx^2+\frac{10}{y} \end{bmatrix}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y)=5xcos(y)\end{align}$

Answer

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

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&d[a^u]=a^uduln(a)\&d[ln(u)]=\frac{du}{u}\&d[e^u]=e^udu\&\nabla f=\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y,z)=12ln(3y)ln(2z)tan(2x)\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=12ln(3y)ln(2z)tan(2x)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[ln(u)]=\frac{du}{u}\&\nabla f=\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\frac{(12ln(2z)tan(2x))}{y}\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Find the vector }\nabla f \text{ for }f(x,y)=9x^{2}tan(2y)\end{align}$

Answer

$\begin{align}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=9x^{2}tan(2y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\nabla f=\begin{bmatrix}18xtan(2y)\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y,z)=- xcos(3z)e^{(5y)} - 20\cdot 2^{(3z)}y^{2}ln(4x)\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=- xcos(3z)e^{(5y)} - 20\cdot 2^{(3z)}y^{2}ln(4x)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[cos(u)]=-sin(u)du\&d[ln(u)]=\frac{du}{u}\&d[a^u]=a^uduln(a)\&\nabla f=\begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y)=-18e^{(5x)}tan(y)\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=-18e^{(5x)}tan(y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\nabla f=\begin{bmatrix}-90e^{(5x)}tan(y)\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Calculate the gradient of the function}\&f(x,y,z)=14tan(5y)e^{(4z)}tan(x) - 13\cdot 5^{(4x)}ln(2z)tan(2y)\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=14tan(5y)e^{(4z)}tan(x) - 13\cdot 5^{(4x)}ln(2z)tan(2y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[e^u]=e^udu\&d[a^u]=a^uduln(a)\&d[ln(u)]=\frac{du}{u}\&\nabla f=\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2)\56tan(5y)e^{(4z)}tan(x) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y)=11ysin(5x) + 11x^{2}tan(y)\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=11ysin(5x) + 11x^{2}tan(y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\&\nabla f=\begin{bmatrix}55ycos(5x) + 22xtan(y)\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Evaluate the gradient vector of at .

Answer

The gradient of 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 $f_{y}(0,0 )$ :

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

$= 4 \cdot 3^{3 }$

$= 4 \cdot 27$

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

$= 8 \cdot 3^{3 }$

$= 8 \cdot 27$

The gradient vector is

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Question

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 .

Answer

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 .

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

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

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

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

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

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

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

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

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

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

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

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Question

$\begin{align}&\text{Find the vector }\nabla f \text{ for }f(x,y)=9x^{2}tan(2y)\end{align}$

Answer

$\begin{align}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=9x^{2}tan(2y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\nabla f=\begin{bmatrix}18xtan(2y)\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=15\cdot 4^{(2x)}ln(3y) - 15xy^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&d[a^u]=a^uduln(a)\&d[ln(u)]=\frac{du}{u}\&d[e^u]=e^udu\&\nabla f=\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}\end{align}$

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Question

$\begin{align}&\text{Determine the gradient, }\nabla\text{, of the function}\&f(x,y,z)=12ln(3y)ln(2z)tan(2x)\end{align}$

Answer

$\begin{align}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=12ln(3y)ln(2z)tan(2x)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[ln(u)]=\frac{du}{u}\&\nabla f=\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\frac{(12ln(2z)tan(2x))}{y}\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}\end{align}$

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Question

What is the the gradient vector of the following function?

Answer

Recall that

$\nabla_x f(x)=\begin{bmatrix} \frac{\partial f}{\partial x_1}\ \ \frac{\partial f}{\partial x_2} \ \ \frac{\partial f}{\partial x_3} \ \vdots \ \frac{\partial f}{\partial x_n} \end{bmatrix}$

All we need to do is calculate 3 partial derivatives, and put them into this form.

$\frac{\partial f}{\partial x}=18yz+100xz$

$\frac{\partial f}{\partial y}=18xz+30y$

$\frac{\partial f}{\partial z}=18xy+50x^2$

Put these into vector form to get

$\nabla_x f(x)=\begin{bmatrix} 18yz+100xz\ 18xz+30y \ 18xy+50x^2 \end{bmatrix}$

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