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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #791 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=\frac{(2ln(3y)e^{(3x)}sin(3z + z^{2}))}{7}\end{align*}$

Possible Answers:

${\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2})\cdot (2z + 3))}{7}}$

${\frac{(4ln(3y)^{2}e^{(6x)}cos(3z + z^{2})^{2}\cdot (2z + 3)^{2})}{49}}$

${\frac{(2ln(3y)e^{(3x)}cos(3z + z^{2})\cdot (2z + 3)\cdot (\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}))}{7}}$

${\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}}$

Correct answer:

${\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2ln(3y)e^{(3x)}sin(3z + z^{2}))}{7}\\&f_{z}=\frac{(2ln(3y)e^{(3x)}cos(3z + z^{2})\cdot (2z + 3))}{7}\\&f_{zz}=\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}\end{align*}$

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Example Question #791 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zzy}\\&\text{Where }f(x,y,z)=5y^{2}tan(x^{2})e^{(z)} - 2^{(3z)}e^{(4x)}e^{(4y)}\end{align*}$

Possible Answers:

${10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}}$

${(5y^{2}tan(x^{2})e^{(z)} - 3\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2))^{2}\cdot (10ytan(x^{2})e^{(z)} - 4\cdot 2^{(3z)}e^{(4x)}e^{(4y)})}$

${(5y^{2}tan(x^{2})e^{(z)} - 3\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2))\cdot (10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2})\cdot (5y^{2}tan(x^{2})e^{(z)} - 9\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2})}$

${10ytan(x^{2})e^{(z)} - 4\cdot 2^{(3z)}e^{(4x)}e^{(4y)} + 10y^{2}tan(x^{2})e^{(z)} - 6\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)}$

Correct answer:

${10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=5y^{2}tan(x^{2})e^{(z)} - 2^{(3z)}e^{(4x)}e^{(4y)}\\&f_{z}=5y^{2}tan(x^{2})e^{(z)} - 3\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)\\&f_{zz}=5y^{2}tan(x^{2})e^{(z)} - 9\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}\\&f_{zzy}=10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}\end{align*}$

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Example Question #3161 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zxy}\\&\text{Where }f(x,y,z)=5y^{4}e^{(x)}cos(z) +\frac{ (2e^{(3x)}sin(y^{3} - 4y))}{(5z^{2})}\end{align*}$

Possible Answers:

${- 20y^{3}e^{(x)}sin(z) -\frac{ (12e^{(3x)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{(5z^{3})}}$

${20y^{3}e^{(x)}cos(z) - 5y^{4}e^{(x)}sin(z) + 5y^{4}e^{(x)}cos(z) +\frac{ (6e^{(3x)}sin(y^{3} - 4y))}{(5z^{2})}-\frac{ (4e^{(3x)}sin(y^{3} - 4y))}{(5z^{3})}+\frac{ (2e^{(3x)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{(5z^{2})}}$

${-(5y^{4}e^{(x)}cos(z) +\frac{ (6e^{(3x)}sin(y^{3} - 4y))}{(5z^{2})})\cdot (5y^{4}e^{(x)}sin(z) +\frac{ (4e^{(3x)}sin(y^{3} - 4y))}{(5z^{3})})\cdot (20y^{3}e^{(x)}cos(z) +\frac{ (2e^{(3x)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{(5z^{2})})}$

${-(5y^{4}e^{(x)}sin(z) +\frac{ (4e^{(3x)}sin(y^{3} - 4y))}{(5z^{3})})\cdot (5y^{4}e^{(x)}sin(z) +\frac{ (12e^{(3x)}sin(y^{3} - 4y))}{(5z^{3})})\cdot (20y^{3}e^{(x)}sin(z) +\frac{ (12e^{(3x)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{(5z^{3})})}$

Correct answer:

${- 20y^{3}e^{(x)}sin(z) -\frac{ (12e^{(3x)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{(5z^{3})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=5y^{4}e^{(x)}cos(z) +\frac{ (2e^{(3x)}sin(y^{3} - 4y))}{(5z^{2})}\\&f_{z}=- 5y^{4}e^{(x)}sin(z) -\frac{ (4e^{(3x)}sin(y^{3} - 4y))}{(5z^{3})}\\&f_{zx}=- 5y^{4}e^{(x)}sin(z) -\frac{ (12e^{(3x)}sin(y^{3} - 4y))}{(5z^{3})}\\&f_{zxy}=- 20y^{3}e^{(x)}sin(z) -\frac{ (12e^{(3x)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{(5z^{3})}\end{align*}$

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Example Question #3162 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xz}\\&\text{Where }f(x,y,z)=\frac{(4^{(3z)}x^{2}e^{(y)})}{5}\end{align*}$

Possible Answers:

${\frac{(2\cdot 4^{(3z)}xe^{(y)})}{5}+\frac{ (3\cdot 4^{(3z)}x^{2}e^{(y)}ln(4))}{5}}$

${\frac{(6\cdot 4^{(3z)}xe^{(y)}ln(4))}{5}}$

${\frac{(12\cdot 4^{(6z)}x^{2}e^{(2y)}ln(4))}{25}}$

Correct answer:

${\frac{(6\cdot 4^{(3z)}xe^{(y)}ln(4))}{5}}$

Explanation:

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Example Question #791 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yyx}\\&\text{Where }f(x,y,z)=\frac{(2\cdot 2^{(4z)}ln(y^{2})e^{(x)})}{9}- 3e^{(y^{2})}ln(x)ln(z)\end{align*}$

Possible Answers:

${\frac{(8\cdot 2^{(4z)}e^{(x)})}{(9y)}-\frac{ (3e^{(y^{2})}ln(z))}{x}+\frac{ (2\cdot 2^{(4z)}ln(y^{2})e^{(x)})}{9}- 12ye^{(y^{2})}ln(x)ln(z)}$

${-(\frac{(3e^{(y^{2})}ln(z))}{x}-\frac{ (2\cdot 2^{(4z)}ln(y^{2})e^{(x)})}{9})\cdot (\frac{(4\cdot 2^{(4z)}e^{(x)})}{(9y)}- 6ye^{(y^{2})}ln(x)ln(z))^{2}}$

${-\frac{ (4\cdot 2^{(4z)}e^{(x)})}{(9y^{2})}-\frac{ (6e^{(y^{2})}ln(z))}{x}-\frac{ (12y^{2}e^{(y^{2})}ln(z))}{x}}$

${(\frac{(4\cdot 2^{(4z)}e^{(x)})}{(9y)}- 6ye^{(y^{2})}ln(x)ln(z))\cdot (\frac{(4\cdot 2^{(4z)}e^{(x)})}{(9y^{2})}+\frac{ (6e^{(y^{2})}ln(z))}{x}+\frac{ (12y^{2}e^{(y^{2})}ln(z))}{x})\cdot (\frac{(4\cdot 2^{(4z)}e^{(x)})}{(9y^{2})}+ 6e^{(y^{2})}ln(x)ln(z) + 12y^{2}e^{(y^{2})}ln(x)ln(z))}$

Correct answer:

${-\frac{ (4\cdot 2^{(4z)}e^{(x)})}{(9y^{2})}-\frac{ (6e^{(y^{2})}ln(z))}{x}-\frac{ (12y^{2}e^{(y^{2})}ln(z))}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 2^{(4z)}ln(y^{2})e^{(x)})}{9}- 3e^{(y^{2})}ln(x)ln(z)\\&f_{y}=\frac{(4\cdot 2^{(4z)}e^{(x)})}{(9y)}- 6ye^{(y^{2})}ln(x)ln(z)\\&f_{yy}=-\frac{ (4\cdot 2^{(4z)}e^{(x)})}{(9y^{2})}- 6e^{(y^{2})}ln(x)ln(z) - 12y^{2}e^{(y^{2})}ln(x)ln(z)\\&f_{yyx}=-\frac{ (4\cdot 2^{(4z)}e^{(x)})}{(9y^{2})}-\frac{ (6e^{(y^{2})}ln(z))}{x}-\frac{ (12y^{2}e^{(y^{2})}ln(z))}{x}\end{align*}$

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Example Question #3164 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yzx}\\&\text{Where }f(x,y,z)=2x^{4}cos(z^{3})sin(y^{3}) -\frac{ (3^yln(3x)cos(z))}{2}\end{align*}$

Possible Answers:

${\frac{(3^yln(3x)sin(z))}{2}+ 8x^{3}cos(z^{3})sin(y^{3}) -\frac{ (3^ycos(z))}{(2x)}+ 6x^{4}y^{2}cos(y^{3})cos(z^{3}) - 6x^{4}z^{2}sin(y^{3})sin(z^{3}) -\frac{ (3^yln(3x)ln(3)cos(z))}{2}}$

${(\frac{(3^yln(3)sin(z))}{(2x)}- 72x^{3}y^{2}z^{2}cos(y^{3})sin(z^{3}))\cdot (\frac{(3^yln(3x)ln(3)sin(z))}{2}- 18x^{4}y^{2}z^{2}cos(y^{3})sin(z^{3}))\cdot (6x^{4}y^{2}cos(y^{3})cos(z^{3}) -\frac{ (3^yln(3x)ln(3)cos(z))}{2})}$

${\frac{(3^yln(3)sin(z))}{(2x)}- 72x^{3}y^{2}z^{2}cos(y^{3})sin(z^{3})}$

${(8x^{3}cos(z^{3})sin(y^{3}) -\frac{ (3^ycos(z))}{(2x)})\cdot (\frac{(3^yln(3x)sin(z))}{2}- 6x^{4}z^{2}sin(y^{3})sin(z^{3}))\cdot (6x^{4}y^{2}cos(y^{3})cos(z^{3}) -\frac{ (3^yln(3x)ln(3)cos(z))}{2})}$

Correct answer:

${\frac{(3^yln(3)sin(z))}{(2x)}- 72x^{3}y^{2}z^{2}cos(y^{3})sin(z^{3})}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=2x^{4}cos(z^{3})sin(y^{3}) -\frac{ (3^yln(3x)cos(z))}{2}\\&f_{y}=6x^{4}y^{2}cos(y^{3})cos(z^{3}) -\frac{ (3^yln(3x)ln(3)cos(z))}{2}\\&f_{yz}=\frac{(3^yln(3x)ln(3)sin(z))}{2}- 18x^{4}y^{2}z^{2}cos(y^{3})sin(z^{3})\\&f_{yzx}=\frac{(3^yln(3)sin(z))}{(2x)}- 72x^{3}y^{2}z^{2}cos(y^{3})sin(z^{3})\end{align*}$

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Example Question #3165 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yzy}\\&\text{Where }f(x,y,z)=\frac{(3cos(z^{2}))}{(x^{2}y)}-\frac{ (y^{2}z^{2})}{(2x)}\end{align*}$

Possible Answers:

${-(\frac{(3cos(z^{2}))}{(x^{2}y^{2})}+\frac{ (yz^{2})}{x})\cdot (\frac{(2yz)}{x}-\frac{ (6zsin(z^{2}))}{(x^{2}y^{2})})\cdot (\frac{(2z)}{x}+\frac{ (12zsin(z^{2}))}{(x^{2}y^{3})})}$

${-\frac{ (6cos(z^{2}))}{(x^{2}y^{2})}-\frac{ (2yz^{2})}{x}-\frac{ (y^{2}z)}{x}-\frac{ (6zsin(z^{2}))}{(x^{2}y)}}$

${-(\frac{(3cos(z^{2}))}{(x^{2}y^{2})}+\frac{ (yz^{2})}{x})^{2}\cdot (\frac{(y^{2}z)}{x}+\frac{ (6zsin(z^{2}))}{(x^{2}y)})}$

${-\frac{ (2z)}{x}-\frac{ (12zsin(z^{2}))}{(x^{2}y^{3})}}$

Correct answer:

${-\frac{ (2z)}{x}-\frac{ (12zsin(z^{2}))}{(x^{2}y^{3})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3cos(z^{2}))}{(x^{2}y)}-\frac{ (y^{2}z^{2})}{(2x)}\\&f_{y}=-\frac{ (3cos(z^{2}))}{(x^{2}y^{2})}-\frac{ (yz^{2})}{x}\\&f_{yz}=\frac{(6zsin(z^{2}))}{(x^{2}y^{2})}-\frac{ (2yz)}{x}\\&f_{yzy}=-\frac{ (2z)}{x}-\frac{ (12zsin(z^{2}))}{(x^{2}y^{3})}\end{align*}$

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Example Question #3166 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zxx}\\&\text{Where }f(x,y,z)=3y^{4}cos(z^{2})e^{(x)} -\frac{ (5\cdot 2^yln(x^{2})e^{(z^{2})})}{2}\end{align*}$

Possible Answers:

${(\frac{(10\cdot 2^yze^{(z^{2})})}{x}+ 6y^{4}zsin(z^{2})e^{(x)})\cdot (\frac{(10\cdot 2^yze^{(z^{2})})}{x^{2}}- 6y^{4}zsin(z^{2})e^{(x)})\cdot (5\cdot 2^yzln(x^{2})e^{(z^{2})} + 6y^{4}zsin(z^{2})e^{(x)})}$

${-(5\cdot 2^yzln(x^{2})e^{(z^{2})} + 6y^{4}zsin(z^{2})e^{(x)})\cdot (\frac{(5\cdot 2^ye^{(z^{2})})}{x}- 3y^{4}cos(z^{2})e^{(x)})^{2}}$

${\frac{(10\cdot 2^yze^{(z^{2})})}{x^{2}}- 6y^{4}zsin(z^{2})e^{(x)}}$

${6y^{4}cos(z^{2})e^{(x)} -\frac{ (10\cdot 2^ye^{(z^{2})})}{x}- 5\cdot 2^yzln(x^{2})e^{(z^{2})} - 6y^{4}zsin(z^{2})e^{(x)}}$

Correct answer:

${\frac{(10\cdot 2^yze^{(z^{2})})}{x^{2}}- 6y^{4}zsin(z^{2})e^{(x)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3y^{4}cos(z^{2})e^{(x)} -\frac{ (5\cdot 2^yln(x^{2})e^{(z^{2})})}{2}\\&f_{z}=- 5\cdot 2^yzln(x^{2})e^{(z^{2})} - 6y^{4}zsin(z^{2})e^{(x)}\\&f_{zx}=-\frac{ (10\cdot 2^yze^{(z^{2})})}{x}- 6y^{4}zsin(z^{2})e^{(x)}\\&f_{zxx}=\frac{(10\cdot 2^yze^{(z^{2})})}{x^{2}}- 6y^{4}zsin(z^{2})e^{(x)}\end{align*}$

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Example Question #792 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xyz}\\&\text{Where }f(x,y,z)=\frac{(2ln(z^{2})cos(y^{3} - 3y)tan(12x + x^{2}))}{7}\end{align*}$

Possible Answers:

${\frac{(16ln(z^{2})^{2}cos(y^{3} - 3y)sin(y^{3} - 3y)^{2}\cdot (2x + 12)^{3}\cdot (tan(12x + x^{2})^{2} + 1)^{3}\cdot (3y^{2} - 3)^{2})}{(343z)}}$

${-\frac{(4sin(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1)\cdot (3y^{2} - 3))}{(7z)}}$

${\frac{(4cos(y^{3} - 3y)tan(12x + x^{2}))}{(7z)}-\frac{ (2ln(z^{2})tan(12x + x^{2})sin(y^{3} - 3y)\cdot (3y^{2} - 3))}{7}+\frac{ (2ln(z^{2})cos(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1))}{7}}$

${-\frac{(16ln(z^{2})^{2}cos(y^{3} - 3y)^{2}tan(12x + x^{2})^{2}sin(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1)\cdot (3y^{2} - 3))}{(343z)}}$

Correct answer:

${-\frac{(4sin(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1)\cdot (3y^{2} - 3))}{(7z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2ln(z^{2})cos(y^{3} - 3y)tan(12x + x^{2}))}{7}\\&f_{x}=\frac{(2ln(z^{2})cos(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1))}{7}\\&f_{xy}=-\frac{(2ln(z^{2})sin(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1)\cdot (3y^{2} - 3))}{7}\\&f_{xyz}=-\frac{(4sin(y^{3} - 3y)\cdot (2x + 12)\cdot (tan(12x + x^{2})^{2} + 1)\cdot (3y^{2} - 3))}{(7z)}\end{align*}$

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Example Question #3168 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zxy}\\&\text{Where }f(x,y,z)=4cos(z^{3})ln(x)sin(y) -\frac{ (ln(4y)e^{(x^{2})}e^{(3z)})}{2}\end{align*}$

Possible Answers:

${-\frac{ (3xe^{(x^{2})}e^{(3z)})}{y}-\frac{ (12z^{2}sin(z^{3})cos(y))}{x}}$

${-(\frac{(12z^{2}sin(z^{3})sin(y))}{x}+ 3xln(4y)e^{(x^{2})}e^{(3z)})\cdot (\frac{(3xe^{(x^{2})}e^{(3z)})}{y}+\frac{ (12z^{2}sin(z^{3})cos(y))}{x})\cdot (\frac{(3ln(4y)e^{(x^{2})}e^{(3z)})}{2}+ 12z^{2}sin(z^{3})ln(x)sin(y))}$

${4cos(z^{3})cos(y)ln(x) -\frac{ (3ln(4y)e^{(x^{2})}e^{(3z)})}{2}+\frac{ (4cos(z^{3})sin(y))}{x}-\frac{ (e^{(x^{2})}e^{(3z)})}{(2y)}- 12z^{2}sin(z^{3})ln(x)sin(y) - xln(4y)e^{(x^{2})}e^{(3z)}}$

${-(\frac{(4cos(z^{3})sin(y))}{x}- xln(4y)e^{(x^{2})}e^{(3z)})\cdot (4cos(z^{3})cos(y)ln(x) -\frac{ (e^{(x^{2})}e^{(3z)})}{(2y)})\cdot (\frac{(3ln(4y)e^{(x^{2})}e^{(3z)})}{2}+ 12z^{2}sin(z^{3})ln(x)sin(y))}$

Correct answer:

${-\frac{ (3xe^{(x^{2})}e^{(3z)})}{y}-\frac{ (12z^{2}sin(z^{3})cos(y))}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=4cos(z^{3})ln(x)sin(y) -\frac{ (ln(4y)e^{(x^{2})}e^{(3z)})}{2}\\&f_{z}=-\frac{ (3ln(4y)e^{(x^{2})}e^{(3z)})}{2}- 12z^{2}sin(z^{3})ln(x)sin(y)\\&f_{zx}=-\frac{ (12z^{2}sin(z^{3})sin(y))}{x}- 3xln(4y)e^{(x^{2})}e^{(3z)}\\&f_{zxy}=-\frac{ (3xe^{(x^{2})}e^{(3z)})}{y}-\frac{ (12z^{2}sin(z^{3})cos(y))}{x}\end{align*}$

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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:

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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)

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