Gradients and Hessians of Quadratic and Linear Functions

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Linear Algebra › Gradients and Hessians of Quadratic and Linear Functions

Questions 1 - 10

$\begin{align*}&\text{Calculate the Hessian matrix of the function}\&f(x,y)=2cos(2x)cos(5y) + 6\cdot 2^{(6x)}tan(4y)\end{align*}$

$\begin{bmatrix}216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\end{bmatrix}$

$\begin{bmatrix}20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)\48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)\end{bmatrix}$

Explanation

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=2cos(2x)cos(5y) + 6\cdot 2^{(6x)}tan(4y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[a^u]=a^uduln(a)\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&\&Hf=\begin{bmatrix}216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=-cos(6x)tan(5y)sin(5z)\end{align*}$

$\begin{bmatrix}36cos(6x)tan(5y)sin(5z)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&30cos(5z)sin(6x)tan(5y)\6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&-10cos(6x)tan(5y)sin(5z)\cdot (5tan(5y)^{2} + 5)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)\30cos(5z)sin(6x)tan(5y)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&25cos(6x)tan(5y)sin(5z)\end{bmatrix}$

$\begin{bmatrix}30cos(5z)sin(6x)tan(5y)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&25cos(6x)tan(5y)sin(5z)\6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&-10cos(6x)tan(5y)sin(5z)\cdot (5tan(5y)^{2} + 5)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)\36cos(6x)tan(5y)sin(5z)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&30cos(5z)sin(6x)tan(5y)\end{bmatrix}$

$\begin{bmatrix}30cos(5z)sin(6x)tan(5y)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&36cos(6x)tan(5y)sin(5z)\-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&-10cos(6x)tan(5y)sin(5z)\cdot (5tan(5y)^{2} + 5)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)\25cos(6x)tan(5y)sin(5z)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&30cos(5z)sin(6x)tan(5y)\end{bmatrix}$

$\begin{bmatrix}-10cos(6x)tan(5y)sin(5z)\cdot (5tan(5y)^{2} + 5)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)\6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&36cos(6x)tan(5y)sin(5z)&30cos(5z)sin(6x)tan(5y)\-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&30cos(5z)sin(6x)tan(5y)&25cos(6x)tan(5y)sin(5z)\end{bmatrix}$

Explanation

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=-cos(6x)tan(5y)sin(5z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\&Hf=\begin{bmatrix}36cos(6x)tan(5y)sin(5z)&6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&30cos(5z)sin(6x)tan(5y)\6sin(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)&-10cos(6x)tan(5y)sin(5z)\cdot (5tan(5y)^{2} + 5)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)\30cos(5z)sin(6x)tan(5y)&-5cos(6x)cos(5z)\cdot (5tan(5y)^{2} + 5)&25cos(6x)tan(5y)sin(5z)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find the matrix }Hf \text{ for }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\end{align*}$

$\begin{bmatrix}- 26\cdot 5^{(2y)} - 80ytan(5x)\cdot (5tan(5x)^{2} + 5)&- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40&-52\cdot 5^{(2y)}x^{2}ln(5)^{2}\end{bmatrix}$

$\begin{bmatrix}- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40&-52\cdot 5^{(2y)}x^{2}ln(5)^{2}\- 26\cdot 5^{(2y)} - 80ytan(5x)\cdot (5tan(5x)^{2} + 5)&- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\end{bmatrix}$

$\begin{bmatrix}- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40&- 26\cdot 5^{(2y)} - 80ytan(5x)\cdot (5tan(5x)^{2} + 5)\-52\cdot 5^{(2y)}x^{2}ln(5)^{2}&- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\end{bmatrix}$

$\begin{bmatrix}-52\cdot 5^{(2y)}x^{2}ln(5)^{2}&- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40&- 26\cdot 5^{(2y)} - 80ytan(5x)\cdot (5tan(5x)^{2} + 5)\end{bmatrix}$

Explanation

$\begin{align*}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[a^u]=a^uduln(a)\&Hf=\begin{bmatrix}- 26\cdot 5^{(2y)} - 80ytan(5x)\cdot (5tan(5x)^{2} + 5)&- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40&-52\cdot 5^{(2y)}x^{2}ln(5)^{2}\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Determine the gradient, } abla\text{, of the function}\&f(x,y)=10sin(2x)sin(6y)\end{align*}$

$\begin{bmatrix}20cos(2x)sin(6y)\60cos(6y)sin(2x)\end{bmatrix}$

$\begin{bmatrix}20cos(2x)sin(6y) + 60cos(6y)sin(2x)\end{bmatrix}$

$\begin{bmatrix}-40sin(2x)sin(6y)&120cos(2x)cos(6y)\120cos(2x)cos(6y)&-360sin(2x)sin(6y)\end{bmatrix}$

$\begin{bmatrix}120cos(2x)cos(6y)\end{bmatrix}$

Explanation

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=10sin(2x)sin(6y)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\& abla f=\begin{bmatrix}20cos(2x)sin(6y)\60cos(6y)sin(2x)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Calculate the gradient of the function}\&f(x,y)=-7\cdot 2^{(5y)}x\end{align*}$

$\begin{bmatrix}-7\cdot 2^{(5y)}\-35\cdot 2^{(5y)}xln(2)\end{bmatrix}$

$\begin{bmatrix}- 7\cdot 2^{(5y)} - 35\cdot 2^{(5y)}xln(2)\end{bmatrix}$

$\begin{bmatrix}0&-35\cdot 2^{(5y)}ln(2)\-35\cdot 2^{(5y)}ln(2)&-175\cdot 2^{(5y)}xln(2)^{2}\end{bmatrix}$

$\begin{bmatrix}-35\cdot 2^{(5y)}ln(2)\end{bmatrix}$

Explanation

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=-7\cdot 2^{(5y)}x\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[a^u]=a^uduln(a)\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\& abla f=\begin{bmatrix}-7\cdot 2^{(5y)}\-35\cdot 2^{(5y)}xln(2)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find the vector } abla f \text{ for }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\end{align*}$

$\begin{bmatrix}- 26\cdot 5^{(2y)}x - 8y\cdot (5tan(5x)^{2} + 5)\- 8tan(5x) - 26\cdot 5^{(2y)}x^{2}ln(5)\end{bmatrix}$

$\begin{bmatrix}- 8tan(5x) - 26\cdot 5^{(2y)}x - 8y\cdot (5tan(5x)^{2} + 5) - 26\cdot 5^{(2y)}x^{2}ln(5)\end{bmatrix}$

$\begin{bmatrix}- 40tan(5x)^{2} - 52\cdot 5^{(2y)}xln(5) - 40\end{bmatrix}$

Explanation

$\begin{align*}&\text{In asking for } abla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y)=- 8ytan(5x) - 13\cdot 5^{(2y)}x^{2}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[a^u]=a^uduln(a)\& abla f=\begin{bmatrix}- 26\cdot 5^{(2y)}x - 8y\cdot (5tan(5x)^{2} + 5)\- 8tan(5x) - 26\cdot 5^{(2y)}x^{2}ln(5)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\&f(x,y,z)=9ln(2x)ln(2z)e^{(y)}\end{align*}$

$\begin{bmatrix}-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\end{bmatrix}$

$\begin{bmatrix}\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2z)e^{(y)})}{x}&-\frac{(9ln(2z)e^{(y)})}{x^{2}}\\frac{(9ln(2x)e^{(y)})}{z}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2z)e^{(y)})}{x}\-\frac{(9ln(2x)e^{(y)})}{z^{2}}&\frac{(9ln(2x)e^{(y)})}{z}&\frac{(9e^{(y)})}{(xz)}\end{bmatrix}$

$\begin{bmatrix}9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9ln(2x)e^{(y)})}{z}\\frac{(9ln(2z)e^{(y)})}{x}&-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9e^{(y)})}{(xz)}\\frac{(9ln(2x)e^{(y)})}{z}&\frac{(9e^{(y)})}{(xz)}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}$

Explanation

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=9ln(2x)ln(2z)e^{(y)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}\&d[e^u]=e^udu\&Hf=\begin{bmatrix}-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find the matrix }Hf \text{ for }f(x,y,z)=13y^{2}tan(2x)tan(4z)\end{align*}$

$\begin{bmatrix}52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)\26ytan(2x)\cdot (4tan(4z)^{2} + 4)&26tan(2x)tan(4z)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)\104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}26tan(2x)tan(4z)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\26ytan(2x)\cdot (4tan(4z)^{2} + 4)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}$

Explanation

$\begin{align*}&\text{In asking for H, the question asks for the Hessian.}\&\text{The Hessian of a function is a matrix of second derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\...&...&...\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=13y^{2}tan(2x)tan(4z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&Hf=\begin{bmatrix}52y^{2}tan(2x)tan(4z)\cdot (2tan(2x)^{2} + 2)&26ytan(4z)\cdot (2tan(2x)^{2} + 2)&13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)\26ytan(4z)\cdot (2tan(2x)^{2} + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)\13y^{2}\cdot (2tan(2x)^{2} + 2)\cdot (4tan(4z)^{2} + 4)&26ytan(2x)\cdot (4tan(4z)^{2} + 4)&104y^{2}tan(2x)tan(4z)\cdot (4tan(4z)^{2} + 4)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Determine the gradient, } abla\text{, of the function}\&f(x,y,z)=-cos(6x)tan(5y)sin(5z)\end{align*}$

$\begin{bmatrix}6sin(6x)tan(5y)sin(5z)\-cos(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)\-5cos(6x)cos(5z)tan(5y)\end{bmatrix}$

$\begin{bmatrix}6sin(6x)tan(5y)sin(5z) - 5cos(6x)cos(5z)tan(5y) - cos(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)\end{bmatrix}$

$\begin{bmatrix}30cos(5z)sin(6x)\cdot (5tan(5y)^{2} + 5)\end{bmatrix}$

Explanation

$\begin{align*}&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\ \frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=-cos(6x)tan(5y)sin(5z)\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[cos(u)]=-sin(u)du\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\&d[sin(u)]=cos(u)du\& abla f=\begin{bmatrix}6sin(6x)tan(5y)sin(5z)\-cos(6x)sin(5z)\cdot (5tan(5y)^{2} + 5)\-5cos(6x)cos(5z)tan(5y)\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find the vector } abla f \text{ for }f(x,y,z)=19z^{2}cos(2y)e^{(2x)} - 17\cdot 5^{(2y)}ln(4x)e^{(5z)}\end{align*}$

$\begin{bmatrix}38z^{2}cos(2y)e^{(2x)} -\frac{ (17\cdot 5^{(2y)}e^{(5z)})}{x}\- 38z^{2}sin(2y)e^{(2x)} - 34\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)\38zcos(2y)e^{(2x)} - 85\cdot 5^{(2y)}ln(4x)e^{(5z)}\end{bmatrix}$

$\begin{bmatrix}38zcos(2y)e^{(2x)} - 85\cdot 5^{(2y)}ln(4x)e^{(5z)} -\frac{ (17\cdot 5^{(2y)}e^{(5z)})}{x}+ 38z^{2}cos(2y)e^{(2x)} - 38z^{2}sin(2y)e^{(2x)} - 34\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)\end{bmatrix}$

$\begin{bmatrix}\frac{(17\cdot 5^{(2y)}e^{(5z)})}{x^{2}}+ 76z^{2}cos(2y)e^{(2x)}&- 76z^{2}sin(2y)e^{(2x)} -\frac{ (34\cdot 5^{(2y)}e^{(5z)}ln(5))}{x}&76zcos(2y)e^{(2x)} -\frac{ (85\cdot 5^{(2y)}e^{(5z)})}{x}\- 76z^{2}sin(2y)e^{(2x)} -\frac{ (34\cdot 5^{(2y)}e^{(5z)}ln(5))}{x}&- 76z^{2}cos(2y)e^{(2x)} - 68\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)^{2}&- 76zsin(2y)e^{(2x)} - 170\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)\76zcos(2y)e^{(2x)} -\frac{ (85\cdot 5^{(2y)}e^{(5z)})}{x}&- 76zsin(2y)e^{(2x)} - 170\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)&38cos(2y)e^{(2x)} - 425\cdot 5^{(2y)}ln(4x)e^{(5z)}\end{bmatrix}$

$\begin{bmatrix}- 152zsin(2y)e^{(2x)} -\frac{ (170\cdot 5^{(2y)}e^{(5z)}ln(5))}{x}\end{bmatrix}$

Explanation

$\begin{align*}&\text{In asking for } abla\text{, the question asks for the gradient.}\&\text{The gradient of a function is a vector of first derivatives taken with}\&\text{respect to its constituent variables. The form is as follows:}\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\frac{d^f}{d_{x_2}}\...\\frac{df}{d_{x_n}}\end{bmatrix}\&\text{Considering our function: }f(x,y,z)=19z^{2}cos(2y)e^{(2x)} - 17\cdot 5^{(2y)}ln(4x)e^{(5z)}\&\text{And utilizing derivative rules:}\&(f\circ g )'=(f'\circ g )\cdot g'\&d[e^u]=e^udu\&d[cos(u)]=-sin(u)du\&d[ln(u)]=\frac{du}{u}\&d[a^u]=a^uduln(a)\& abla f=\begin{bmatrix}38z^{2}cos(2y)e^{(2x)} -\frac{ (17\cdot 5^{(2y)}e^{(5z)})}{x}\- 38z^{2}sin(2y)e^{(2x)} - 34\cdot 5^{(2y)}ln(4x)e^{(5z)}ln(5)\38zcos(2y)e^{(2x)} - 85\cdot 5^{(2y)}ln(4x)e^{(5z)}\end{bmatrix}\end{align*}$

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