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线性代数:二次函数和线性函数的梯度和Hessians

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4诊断测试 108练习测试今日问题抽认卡从概念中学习

例子问题

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例子问题1:二次函数和线性函数的梯度和Hessians

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

可能的答案:

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

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

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

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

正确答案:

解释：

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

报告错误

问题51:矩阵微积分

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

可能的答案:

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

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

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

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

正确答案:

解释：

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

报告错误

例子问题1:二次函数和线性函数的梯度和Hessians

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

可能的答案:

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

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

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

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

正确答案:

解释：

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

报告错误

问题53:矩阵微积分

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

可能的答案:

正确答案:

解释：

报告错误

例子问题1:二次函数和线性函数的梯度和Hessians

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

可能的答案:

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

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

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

正确答案:

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

解释：

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

报告错误

问题55:矩阵微积分

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

可能的答案:

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

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

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

正确答案:

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

解释：

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

报告错误

问题56:矩阵微积分

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

可能的答案:

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

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

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

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

正确答案:

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

解释：

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

报告错误

问题57:矩阵微积分

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

可能的答案:

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

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

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

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

正确答案:

解释：

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

报告错误

问题58:矩阵微积分

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

可能的答案:

$\begin{bmatrix}- 16xln(z) - 16yln(z) -\frac{ (16xy)}{z}\end{bmatrix}$

$\begin{bmatrix}0&-16ln(z)&-\frac{(16y)}{z}\\-16ln(z)&0&-\frac{(16x)}{z}\\-\frac{(16y)}{z}&-\frac{(16x)}{z}&\frac{(16xy)}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}-\frac{16}{z}\end{bmatrix}$

$\begin{bmatrix}-16yln(z)\\-16xln(z)\\-\frac{(16xy)}{z}\end{bmatrix}$

正确答案:

$\begin{bmatrix}-16yln(z)\\-16xln(z)\\-\frac{(16xy)}{z}\end{bmatrix}$

解释：

$\begin{align*}&\text{In asking for }\nabla\text{, the question asks for the gradient.}\\&\text{The gradient of a function is a vector of first derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^f}{d_{x_1}}\\\frac{d^f}{d_{x_2}}\\...\\\frac{df}{d_{x_n}}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=-16xyln(z)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&\nabla f=\begin{bmatrix}-16yln(z)\\-16xln(z)\\-\frac{(16xy)}{z}\end{bmatrix}\end{align*}$

报告错误

问题59:矩阵微积分

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

可能的答案:

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

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

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

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

正确答案:

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

解释：

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

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便雅悯
认证导师

亚利桑那州立大学，物理学学士。夏威夷大学马诺阿分校，物理学硕士。

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追逐
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西部州长大学，理学学士，数学。

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约旦
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路易斯安那州杜兰大学，理学学士，数学。路易斯安那州杜兰大学，物理学士。

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4诊断测试 108练习测试今日问题抽认卡从概念中学习

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洛杉矶的数学导师，圣地亚哥的微积分导师，圣地亚哥的GRE导师，圣地亚哥的生物导师，西雅图的ACT导师，旧金山湾区的SSAT导师，费城的生物导师，纽约市的GMAT家教，旧金山湾区的SAT辅导老师，旧金山湾区的GRE导师

常用备考

圣地亚哥SSAT考试准备中心，在迈阿密准备GMAT考试，在费城准备GMAT考试，在休斯顿准备SSAT考试，在亚特兰大准备SSAT考试，在旧金山湾区的MCAT考试准备，在迈阿密准备GRE考试，在华盛顿特区准备GRE考试，在亚特兰大准备GMAT考试，在达拉斯沃斯堡的ACT考试准备

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