Gradients and Hessians of Quadratic and Linear Functions - Linear Algebra

Card 1 of 48

Didn't Know

Knew It

1 of 2019 left

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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)=-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\&\nabla 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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=- 8ytan(5x) - 13\cdot $5^{(2y)}$$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$$]=a^uduln$(a)\&\nabla f=\begin{bmatrix}- 26\cdot $5^{(2y)}$x - 8y\cdot $(5tan(5x)^{2}$ + 5)\- 8tan(5x) - 26\cdot $5^{(2y)}$$x^{2}$ln(5)\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=-7\cdot $2^{(5y)}$x\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[a^u$$]=a^uduln$$(a)\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&\nabla f=\begin{bmatrix}-7\cdot $2^{(5y)}$\-35\cdot $2^{(5y)}$xln(2)\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y,z)=19z^{2}$$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)\&\nabla $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)}$\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&$\text{Find the vector }$\nabla f $\text{ for }$f(x,y,z)=-16xyln(z)\end{align}$

Tap to reveal answer

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,z)=-16xyln(z)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}$\&\nabla f=\begin{bmatrix}-16yln(z)\-16xln(z)\-$\frac{(16xy)}{z}$\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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)=10sin(2x)sin(6y)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot g'\&d[sin(u)]=cos(u)du\&\nabla f=\begin{bmatrix}20cos(2x)sin(6y)\60cos(6y)sin(2x)\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$}\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\begin{align}&$\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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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)=-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\&\nabla 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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=- 8ytan(5x) - 13\cdot $5^{(2y)}$$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$$]=a^uduln$(a)\&\nabla f=\begin{bmatrix}- 26\cdot $5^{(2y)}$x - 8y\cdot $(5tan(5x)^{2}$ + 5)\- 8tan(5x) - 26\cdot $5^{(2y)}$$x^{2}$ln(5)\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=-7\cdot $2^{(5y)}$x\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[a^u$$]=a^uduln$$(a)\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&\nabla f=\begin{bmatrix}-7\cdot $2^{(5y)}$\-35\cdot $2^{(5y)}$xln(2)\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y,z)=19z^{2}$$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)\&\nabla $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)}$\end{bmatrix}\end{align}$

← Didn't Know|Knew It →

Knew It