The Gradient

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Linear Algebra › The Gradient

Questions 1 - 10

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

Evaluate the gradient vector of at .

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

What is the the gradient vector of the following function?

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

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

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

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

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

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

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

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