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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #791 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

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Example Question #792 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #793 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #794 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #795 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

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Example Question #796 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #797 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #798 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #799 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

Explanation:

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

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Example Question #800 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Partial Derivatives

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Example Question #791 : Partial Derivatives

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Example Question #799 : Partial Derivatives

Example Question #800 : Partial Derivatives

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