Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : The Gradient

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : The Gradient

What is the the gradient vector of the following function?

f(x,y,z)=18xyz+50x^2z+15y^2

Possible Answers:

$\nabla_x f(x)=\begin{bmatrix} 1 \\1 \\18yz+100xz\end{bmatrix}$

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

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

$\nabla_x f(x)=\begin{bmatrix} 0 \\18xy+50x^2 \\0\end{bmatrix}$

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

Correct answer:

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

Explanation:

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

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

Report an Error

Example Question #2 : The Gradient

Find the gradient vector of the following function.

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

Possible Answers:

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

$\begin{bmatrix} 1+\frac{10}{x}\\ \\ 1+\frac{10}{y} \end{bmatrix}$

$\begin{bmatrix} 2xy^2\\ \\ 2yx^2 \end{bmatrix}$

$\begin{bmatrix} \frac{10}{x}\\ \\ \frac{10}{y} \end{bmatrix}$

Correct answer:

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

Explanation:

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

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

Report an Error

Example Question #1 : The Gradient

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

Possible Answers:

$\begin{bmatrix}5cos(y)\\-5xsin(y)\end{bmatrix}$

$\begin{bmatrix}-5sin(y)\end{bmatrix}$

$\begin{bmatrix}0&-5sin(y)\\-5sin(y)&-5xcos(y)\end{bmatrix}$

$\begin{bmatrix}5cos(y) - 5xsin(y)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}5cos(y)\\-5xsin(y)\end{bmatrix}$

Explanation:

Report an Error

Example Question #2 : The Gradient

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

Possible Answers:

$\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}$

$\begin{bmatrix}\frac{(15\cdot 4^{(2x)})}{y}- 30xy - 15y^{2} + 30\cdot 4^{(2x)}ln(3y)ln(4)\end{bmatrix}$

$\begin{bmatrix}60\cdot 4^{(2x)}ln(3y)ln(4)^{2}&\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y\\\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y&- 30x -\frac{ (15\cdot 4^{(2x)})}{y^{2}}\end{bmatrix}$

$\begin{bmatrix}\frac{(30\cdot 4^{(2x)}ln(4))}{y}- 30y\end{bmatrix}$

Correct answer:

$\begin{bmatrix}30\cdot 4^{(2x)}ln(3y)ln(4) - 15y^{2}\\\frac{(15\cdot 4^{(2x)})}{y}- 30xy\end{bmatrix}$

Explanation:

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

Report an Error

Example Question #3 : Matrix Calculus

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

Possible Answers:

$\begin{bmatrix}48ln(3y)ln(2z)tan(2x)\cdot (2tan(2x)^{2} + 2)\\-\frac{(12ln(2z)tan(2x))}{y^{2}}\\-\frac{(12ln(3y)tan(2x))}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2) +\frac{ (12ln(2z)tan(2x))}{y}+\frac{ (12ln(3y)tan(2x))}{z}\end{bmatrix}$

$\begin{bmatrix}48ln(3y)ln(2z)tan(2x)\cdot (2tan(2x)^{2} + 2)&\frac{(12ln(2z)\cdot (2tan(2x)^{2} + 2))}{y}&\frac{(12ln(3y)\cdot (2tan(2x)^{2} + 2))}{z}\\\frac{(12ln(2z)\cdot (2tan(2x)^{2} + 2))}{y}&-\frac{(12ln(2z)tan(2x))}{y^{2}}&\frac{(12tan(2x))}{(yz)}\\\frac{(12ln(3y)\cdot (2tan(2x)^{2} + 2))}{z}&\frac{(12tan(2x))}{(yz)}&-\frac{(12ln(3y)tan(2x))}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\\frac{(12ln(2z)tan(2x))}{y}\\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}12ln(3y)ln(2z)\cdot (2tan(2x)^{2} + 2)\\\frac{(12ln(2z)tan(2x))}{y}\\\frac{(12ln(3y)tan(2x))}{z}\end{bmatrix}$

Explanation:

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

Report an Error

Example Question #1 : Linear Algebra

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

Possible Answers:

$\begin{bmatrix}18tan(2y)\\36x^{2}tan(2y)\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

$\begin{bmatrix}18tan(2y)&18x\cdot (2tan(2y)^{2} + 2)\\18x\cdot (2tan(2y)^{2} + 2)&36x^{2}tan(2y)\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

$\begin{bmatrix}18xtan(2y) + 9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

$\begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}18xtan(2y)\\9x^{2}\cdot (2tan(2y)^{2} + 2)\end{bmatrix}$

Explanation:

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

Report an Error

Example Question #3 : Linear Algebra

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

Possible Answers:

$\begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$

$\begin{bmatrix}3xsin(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x) - cos(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$

$\begin{bmatrix}\frac{(20\cdot 2^{(3z)}y^{2})}{x^{2}}\\- 40\cdot 2^{(3z)}ln(4x) - 25xcos(3z)e^{(5y)}\\9xcos(3z)e^{(5y)} - 180\cdot 2^{(3z)}y^{2}ln(4x)ln(2)^{2}\end{bmatrix}$

$\begin{bmatrix}\frac{(20\cdot 2^{(3z)}y^{2})}{x^{2}}&- 5cos(3z)e^{(5y)} -\frac{ (40\cdot 2^{(3z)}y)}{x}&3sin(3z)e^{(5y)} -\frac{ (60\cdot 2^{(3z)}y^{2}ln(2))}{x}\\- 5cos(3z)e^{(5y)} -\frac{ (40\cdot 2^{(3z)}y)}{x}&- 40\cdot 2^{(3z)}ln(4x) - 25xcos(3z)e^{(5y)}&15xsin(3z)e^{(5y)} - 120\cdot 2^{(3z)}yln(4x)ln(2)\\3sin(3z)e^{(5y)} -\frac{ (60\cdot 2^{(3z)}y^{2}ln(2))}{x}&15xsin(3z)e^{(5y)} - 120\cdot 2^{(3z)}yln(4x)ln(2)&9xcos(3z)e^{(5y)} - 180\cdot 2^{(3z)}y^{2}ln(4x)ln(2)^{2}\end{bmatrix}$

Correct answer:

$\begin{bmatrix}- cos(3z)e^{(5y)} -\frac{ (20\cdot 2^{(3z)}y^{2})}{x}\\- 5xcos(3z)e^{(5y)} - 40\cdot 2^{(3z)}yln(4x)\\3xsin(3z)e^{(5y)} - 60\cdot 2^{(3z)}y^{2}ln(4x)ln(2)\end{bmatrix}$

Explanation:

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

Report an Error

Example Question #2 : Matrix Calculus

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

Possible Answers:

$\begin{bmatrix}-450e^{(5x)}tan(y)&-90e^{(5x)}\cdot (tan(y)^{2} + 1)\\-90e^{(5x)}\cdot (tan(y)^{2} + 1)&-36e^{(5x)}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}-450e^{(5x)}tan(y)\\-36e^{(5x)}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}- 18e^{(5x)}\cdot (tan(y)^{2} + 1) - 90e^{(5x)}tan(y)\end{bmatrix}$

$\begin{bmatrix}-90e^{(5x)}tan(y)\\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-90e^{(5x)}tan(y)\\-18e^{(5x)}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Explanation:

Report an Error

Example Question #6 : Linear Algebra

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

Possible Answers:

$\begin{bmatrix}28tan(5y)e^{(4z)}tan(x)\cdot (tan(x)^{2} + 1) - 208\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)^{2}&14e^{(4z)}\cdot (5tan(5y)^{2} + 5)\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)ln(5)\cdot (2tan(2y)^{2} + 2)&56tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) -\frac{ (52\cdot 5^{(4x)}tan(2y)ln(5))}{z}\\14e^{(4z)}\cdot (5tan(5y)^{2} + 5)\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)ln(5)\cdot (2tan(2y)^{2} + 2)&140tan(5y)e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)\cdot (2tan(2y)^{2} + 2)&56e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) -\frac{ (13\cdot 5^{(4x)}\cdot (2tan(2y)^{2} + 2))}{z}\\56tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) -\frac{ (52\cdot 5^{(4x)}tan(2y)ln(5))}{z}&56e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) -\frac{ (13\cdot 5^{(4x)}\cdot (2tan(2y)^{2} + 2))}{z}&\frac{(13\cdot 5^{(4x)}tan(2y))}{z^{2}}+ 224tan(5y)e^{(4z)}tan(x)\end{bmatrix}$

$\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\\14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2)\\56tan(5y)e^{(4z)}tan(x) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}\end{bmatrix}$

$\begin{bmatrix}14tan(5y)e^{(4z)}\cdot (tan(x)^{2} + 1) - 13\cdot 5^{(4x)}ln(2z)\cdot (2tan(2y)^{2} + 2) -\frac{ (13\cdot 5^{(4x)}tan(2y))}{z}+ 56tan(5y)e^{(4z)}tan(x) + 14e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)\end{bmatrix}$

$\begin{bmatrix}28tan(5y)e^{(4z)}tan(x)\cdot (tan(x)^{2} + 1) - 208\cdot 5^{(4x)}ln(2z)tan(2y)ln(5)^{2}\\140tan(5y)e^{(4z)}tan(x)\cdot (5tan(5y)^{2} + 5) - 52\cdot 5^{(4x)}ln(2z)tan(2y)\cdot (2tan(2y)^{2} + 2)\\\frac{(13\cdot 5^{(4x)}tan(2y))}{z^{2}}+ 224tan(5y)e^{(4z)}tan(x)\end{bmatrix}$

Correct answer:

Explanation:

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

Report an Error

Example Question #2 : The Gradient

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

Possible Answers:

$\begin{bmatrix}22tan(y) - 275ysin(5x)\\22x^{2}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}11sin(5x) + 55ycos(5x) + 22xtan(y) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}22tan(y) - 275ysin(5x)&55cos(5x) + 22x\cdot (tan(y)^{2} + 1)\\55cos(5x) + 22x\cdot (tan(y)^{2} + 1)&22x^{2}tan(y)\cdot (tan(y)^{2} + 1)\end{bmatrix}$

$\begin{bmatrix}55ycos(5x) + 22xtan(y)\\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Correct answer:

$\begin{bmatrix}55ycos(5x) + 22xtan(y)\\11sin(5x) + 11x^{2}\cdot (tan(y)^{2} + 1)\end{bmatrix}$

Explanation:

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

Report an Error

View Linear Algebra Tutors

Ying
Certified Tutor

Iowa State University, Master of Science, Mathematics.

View Linear Algebra Tutors

Scott
Certified Tutor

Florida State University, Bachelor of Science, Applied Mathematics. Florida State University, Bachelor of Science, Astrophysi...

View Linear Algebra Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Chicago, GMAT Tutors in Phoenix, ISEE Tutors in Chicago, ACT Tutors in Houston, Math Tutors in San Diego, French Tutors in Dallas Fort Worth, SAT Tutors in Boston, LSAT Tutors in Boston, Computer Science Tutors in San Francisco-Bay Area, SSAT Tutors in Washington DC

Popular Courses & Classes

ACT Courses & Classes in Atlanta, GMAT Courses & Classes in San Diego, MCAT Courses & Classes in Philadelphia, SAT Courses & Classes in Houston, GRE Courses & Classes in San Diego, SAT Courses & Classes in Washington DC, SAT Courses & Classes in Philadelphia, SSAT Courses & Classes in Houston, LSAT Courses & Classes in San Diego, GMAT Courses & Classes in Boston

Popular Test Prep

MCAT Test Prep in Los Angeles, MCAT Test Prep in Miami, GMAT Test Prep in Miami, MCAT Test Prep in Boston, LSAT Test Prep in Houston, GMAT Test Prep in Seattle, ACT Test Prep in Atlanta, MCAT Test Prep in Houston, GRE Test Prep in San Diego, SAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : The Gradient

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : The Gradient

Example Question #2 : The Gradient

Example Question #1 : The Gradient

Example Question #2 : The Gradient

Example Question #3 : Matrix Calculus

Example Question #1 : Linear Algebra

Example Question #3 : Linear Algebra

Example Question #2 : Matrix Calculus

Example Question #6 : Linear Algebra

Example Question #2 : The Gradient

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: