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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #2921 : Calculus 3

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

Possible Answers:

${\frac{{(12x^2cos{(3z)}ln{(3y)}^2sin{(3z)}e^{(2x^2)})}}{25}}$ $\displaystyle {\frac{{(12x^2cos{(3z)}ln{(3y)}^2sin{(3z)}e^{(2x^2)})}}{25}}$

${\frac{-{(6xcos{(3z)}ln{(3y)}e^{(x^2)})}}{5}}$ $\displaystyle {\frac{-{(6xcos{(3z)}ln{(3y)}e^{(x^2)})}}{5}}$

${-\frac{ {(3cos{(3z)}ln{(3y)}e^{(x^2)})}}{5} -\frac{ {(2xln{(3y)}sin{(3z)}e^{(x^2)})}}{5}}$ $\displaystyle {-\frac{ {(3cos{(3z)}ln{(3y)}e^{(x^2)})}}{5} -\frac{ {(2xln{(3y)}sin{(3z)}e^{(x^2)})}}{5}}$

${\frac{{(6xcos{(3z)}ln{(3y)}^2sin{(3z)}e^{(2x^2)})}}{25}}$ $\displaystyle {\frac{{(6xcos{(3z)}ln{(3y)}^2sin{(3z)}e^{(2x^2)})}}{25}}$

Correct answer:

${\frac{-{(6xcos{(3z)}ln{(3y)}e^{(x^2)})}}{5}}$ $\displaystyle {\frac{-{(6xcos{(3z)}ln{(3y)}e^{(x^2)})}}{5}}$

Explanation:

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

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Example Question #2922 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zxz}\\&\text{Where }f(x,y,z)=\frac{-{(2\cdot3^{(3z)}cos{(y^2)}e^{(3x)})}}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{zxz}\\&\text{Where }f(x,y,z)=\frac{-{(2\cdot3^{(3z)}cos{(y^2)}e^{(3x)})}}{7}\end{align*}$

Possible Answers:

${\frac{-{(54\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)}^2)}}{7}}$ $\displaystyle {\frac{-{(54\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)}^2)}}{7}}$

${-\frac{ {(6\cdot3^{(3z)}cos{(y^2)}e^{(3x)})}}{7} -\frac{ {(12\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)})}}{7}}$ $\displaystyle {-\frac{ {(6\cdot3^{(3z)}cos{(y^2)}e^{(3x)})}}{7} -\frac{ {(12\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)})}}{7}}$

${\frac{-{(5832\cdot3^{(9z)}cos{(y^2)}^3e^{(9x)}ln{(3)}^4)}}{343}}$ $\displaystyle {\frac{-{(5832\cdot3^{(9z)}cos{(y^2)}^3e^{(9x)}ln{(3)}^4)}}{343}}$

${\frac{-{(216\cdot3^{(9z)}cos{(y^2)}^3e^{(9x)}ln{(3)}^2)}}{343}}$ $\displaystyle {\frac{-{(216\cdot3^{(9z)}cos{(y^2)}^3e^{(9x)}ln{(3)}^2)}}{343}}$

Correct answer:

${\frac{-{(54\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)}^2)}}{7}}$ $\displaystyle {\frac{-{(54\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)}^2)}}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2\cdot3^{(3z)}cos{(y^2)}e^{(3x)})}}{7}\\&f_{z}=\frac{-{(6\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)})}}{7}\\&f_{zx}=\frac{-{(18\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)})}}{7}\\&f_{zxz}=\frac{-{(54\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)}^2)}}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2\cdot3^{(3z)}cos{(y^2)}e^{(3x)})}}{7}\\&f_{z}=\frac{-{(6\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)})}}{7}\\&f_{zx}=\frac{-{(18\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)})}}{7}\\&f_{zxz}=\frac{-{(54\cdot3^{(3z)}cos{(y^2)}e^{(3x)}ln{(3)}^2)}}{7}\end{align*}$

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Example Question #2923 : Calculus 3

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

Possible Answers:

${-\frac{ {(ze^{(x)})}}{{(2y^2)}} -\frac{ {(z^2e^{(x)})}}{{(4y^2)}}}$ $\displaystyle {-\frac{ {(ze^{(x)})}}{{(2y^2)}} -\frac{ {(z^2e^{(x)})}}{{(4y^2)}}}$

${\frac{{(z^3e^{(2x)})}}{{(8y^4)}}}$ $\displaystyle {\frac{{(z^3e^{(2x)})}}{{(8y^4)}}}$

${\frac{-{(ze^{(x)})}}{{(2y^2)}}}$ $\displaystyle {\frac{-{(ze^{(x)})}}{{(2y^2)}}}$

${\frac{{(z^2e^{(2x)})}}{{(4y^4)}}}$ $\displaystyle {\frac{{(z^2e^{(2x)})}}{{(4y^4)}}}$

Correct answer:

${\frac{-{(ze^{(x)})}}{{(2y^2)}}}$ $\displaystyle {\frac{-{(ze^{(x)})}}{{(2y^2)}}}$

Explanation:

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

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Example Question #2924 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zxz}\\&\text{Where }f(x,y,z)=2\cdot3^{(3x)}y^2cos{(z)}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{zxz}\\&\text{Where }f(x,y,z)=2\cdot3^{(3x)}y^2cos{(z)}\end{align*}$

Possible Answers:

${-72\cdot3^{(9x)}y^6ln{(3)}^2cos{(z)}sin{(z)}^2}$ $\displaystyle {-72\cdot3^{(9x)}y^6ln{(3)}^2cos{(z)}sin{(z)}^2}$

${-6\cdot3^{(3x)}y^2ln{(3)}cos{(z)}}$ $\displaystyle {-6\cdot3^{(3x)}y^2ln{(3)}cos{(z)}}$

${6\cdot3^{(3x)}y^2ln{(3)}cos{(z)} - 4\cdot3^{(3x)}y^2sin{(z)}}$ $\displaystyle {6\cdot3^{(3x)}y^2ln{(3)}cos{(z)} - 4\cdot3^{(3x)}y^2sin{(z)}}$

${24\cdot3^{(9x)}y^6ln{(3)}cos{(z)}sin{(z)}^2}$ $\displaystyle {24\cdot3^{(9x)}y^6ln{(3)}cos{(z)}sin{(z)}^2}$

Correct answer:

${-6\cdot3^{(3x)}y^2ln{(3)}cos{(z)}}$ $\displaystyle {-6\cdot3^{(3x)}y^2ln{(3)}cos{(z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=2\cdot3^{(3x)}y^2cos{(z)}\\&f_{z}=-2\cdot3^{(3x)}y^2sin{(z)}\\&f_{zx}=-6\cdot3^{(3x)}y^2ln{(3)}sin{(z)}\\&f_{zxz}=-6\cdot3^{(3x)}y^2ln{(3)}cos{(z)}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=2\cdot3^{(3x)}y^2cos{(z)}\\&f_{z}=-2\cdot3^{(3x)}y^2sin{(z)}\\&f_{zx}=-6\cdot3^{(3x)}y^2ln{(3)}sin{(z)}\\&f_{zxz}=-6\cdot3^{(3x)}y^2ln{(3)}cos{(z)}\end{align*}$

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Example Question #2925 : Calculus 3

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

Possible Answers:

${\frac{{(192y^6cos{(3z)}sin{(3z)}^3cos{(x)}sin{(x)}^3)}}{625}}$ $\displaystyle {\frac{{(192y^6cos{(3z)}sin{(3z)}^3cos{(x)}sin{(x)}^3)}}{625}}$

${\frac{-{(12cos{(3z)}cos{(x)})}}{5}}$ $\displaystyle {\frac{-{(12cos{(3z)}cos{(x)})}}{5}}$

${-\frac{ {(8ysin{(3z)}sin{(x)})}}{5} -\frac{ {(6y^2cos{(3z)}sin{(x)})}}{5} -\frac{ {(2y^2sin{(3z)}cos{(x)})}}{5}}$ $\displaystyle {-\frac{ {(8ysin{(3z)}sin{(x)})}}{5} -\frac{ {(6y^2cos{(3z)}sin{(x)})}}{5} -\frac{ {(2y^2sin{(3z)}cos{(x)})}}{5}}$

${\frac{{(5184y^5cos{(3z)}^4cos{(x)}^3sin{(x)})}}{625}}$ $\displaystyle {\frac{{(5184y^5cos{(3z)}^4cos{(x)}^3sin{(x)})}}{625}}$

Correct answer:

${\frac{-{(12cos{(3z)}cos{(x)})}}{5}}$ $\displaystyle {\frac{-{(12cos{(3z)}cos{(x)})}}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2y^2sin{(3z)}sin{(x)})}}{5}\\&f_{z}=\frac{-{(6y^2cos{(3z)}sin{(x)})}}{5}\\&f_{zx}=\frac{-{(6y^2cos{(3z)}cos{(x)})}}{5}\\&f_{zxy}=\frac{-{(12ycos{(3z)}cos{(x)})}}{5}\\&f_{zxyy}=\frac{-{(12cos{(3z)}cos{(x)})}}{5}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2y^2sin{(3z)}sin{(x)})}}{5}\\&f_{z}=\frac{-{(6y^2cos{(3z)}sin{(x)})}}{5}\\&f_{zx}=\frac{-{(6y^2cos{(3z)}cos{(x)})}}{5}\\&f_{zxy}=\frac{-{(12ycos{(3z)}cos{(x)})}}{5}\\&f_{zxyy}=\frac{-{(12cos{(3z)}cos{(x)})}}{5}\end{align*}$

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Example Question #2926 : Calculus 3

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

Possible Answers:

${-2592z^2cos{(z^2)}^2sin{(z^2)}e^{(9y)}ln{(x)}^3}$ $\displaystyle {-2592z^2cos{(z^2)}^2sin{(z^2)}e^{(9y)}ln{(x)}^3}$

${-36zcos{(z^2)}e^{(3y)}ln{(x)}}$ $\displaystyle {-36zcos{(z^2)}e^{(3y)}ln{(x)}}$

${- 12sin{(z^2)}e^{(3y)}ln{(x)} - 4zcos{(z^2)}e^{(3y)}ln{(x)}}$ $\displaystyle {- 12sin{(z^2)}e^{(3y)}ln{(x)} - 4zcos{(z^2)}e^{(3y)}ln{(x)}}$

${-144zcos{(z^2)}sin{(z^2)}^2e^{(9y)}ln{(x)}^3}$ $\displaystyle {-144zcos{(z^2)}sin{(z^2)}^2e^{(9y)}ln{(x)}^3}$

Correct answer:

${-36zcos{(z^2)}e^{(3y)}ln{(x)}}$ $\displaystyle {-36zcos{(z^2)}e^{(3y)}ln{(x)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-2sin{(z^2)}e^{(3y)}ln{(x)}\\&f_{y}=-6sin{(z^2)}e^{(3y)}ln{(x)}\\&f_{yz}=-12zcos{(z^2)}e^{(3y)}ln{(x)}\\&f_{yzy}=-36zcos{(z^2)}e^{(3y)}ln{(x)}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-2sin{(z^2)}e^{(3y)}ln{(x)}\\&f_{y}=-6sin{(z^2)}e^{(3y)}ln{(x)}\\&f_{yz}=-12zcos{(z^2)}e^{(3y)}ln{(x)}\\&f_{yzy}=-36zcos{(z^2)}e^{(3y)}ln{(x)}\end{align*}$

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Example Question #2927 : Calculus 3

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

Possible Answers:

${\frac{{(4z^6ln{(4x)}^2sin{(y)}^4)}}{x^2}}$ $\displaystyle {\frac{{(4z^6ln{(4x)}^2sin{(y)}^4)}}{x^2}}$

${- 4zln{(4x)}sin{(y)} -\frac{ {(2z^2sin{(y)})}}{x}}$ $\displaystyle {- 4zln{(4x)}sin{(y)} -\frac{ {(2z^2sin{(y)})}}{x}}$

${\frac{-{(16zln{(4x)}^2sin{(y)}^4)}}{x^3}}$ $\displaystyle {\frac{-{(16zln{(4x)}^2sin{(y)}^4)}}{x^3}}$

${\frac{{(2sin{(y)})}}{x^2}}$ $\displaystyle {\frac{{(2sin{(y)})}}{x^2}}$

Correct answer:

${\frac{{(2sin{(y)})}}{x^2}}$ $\displaystyle {\frac{{(2sin{(y)})}}{x^2}}$

Explanation:

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

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Example Question #2928 : Calculus 3

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

Possible Answers:

$\displaystyle {-6ycos{(3x)}ln{(z)}}$

${\frac{{(2ycos{(3x)}ln{(z)})}}{3} - 2y^2sin{(3x)}ln{(z)}}$ $\displaystyle {\frac{{(2ycos{(3x)}ln{(z)})}}{3} - 2y^2sin{(3x)}ln{(z)}}$

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

${\frac{{(2y^5cos{(3x)}sin{(3x)}^2ln{(z)}^3)}}{3}}$ $\displaystyle {\frac{{(2y^5cos{(3x)}sin{(3x)}^2ln{(z)}^3)}}{3}}$

Correct answer:

$\displaystyle {-6ycos{(3x)}ln{(z)}}$

Explanation:

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

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Example Question #2929 : Calculus 3

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

Possible Answers:

${\frac{-{(32cos{(4y)}^2sin{(4y)}sin{(z^2)}^3)}}{{(125x^5)}}}$ $\displaystyle {\frac{-{(32cos{(4y)}^2sin{(4y)}sin{(z^2)}^3)}}{{(125x^5)}}}$

${-\frac{ {(4cos{(4y)}sin{(z^2)})}}{{(5x^2)}} -\frac{ {(8sin{(4y)}sin{(z^2)})}}{{(5x)}}}$ $\displaystyle {-\frac{ {(4cos{(4y)}sin{(z^2)})}}{{(5x^2)}} -\frac{ {(8sin{(4y)}sin{(z^2)})}}{{(5x)}}}$

${\frac{{(128cos{(4y)}^2sin{(4y)}sin{(z^2)}^3)}}{{(125x^8)}}}$ $\displaystyle {\frac{{(128cos{(4y)}^2sin{(4y)}sin{(z^2)}^3)}}{{(125x^8)}}}$

${\frac{-{(16sin{(4y)}sin{(z^2)})}}{{(5x^3)}}}$ $\displaystyle {\frac{-{(16sin{(4y)}sin{(z^2)})}}{{(5x^3)}}}$

Correct answer:

${\frac{-{(16sin{(4y)}sin{(z^2)})}}{{(5x^3)}}}$ $\displaystyle {\frac{-{(16sin{(4y)}sin{(z^2)})}}{{(5x^3)}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2cos{(4y)}sin{(z^2)})}}{{(5x)}}\\&f_{x}=\frac{-{(2cos{(4y)}sin{(z^2)})}}{{(5x^2)}}\\&f_{xx}=\frac{{(4cos{(4y)}sin{(z^2)})}}{{(5x^3)}}\\&f_{xxy}=\frac{-{(16sin{(4y)}sin{(z^2)})}}{{(5x^3)}}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2cos{(4y)}sin{(z^2)})}}{{(5x)}}\\&f_{x}=\frac{-{(2cos{(4y)}sin{(z^2)})}}{{(5x^2)}}\\&f_{xx}=\frac{{(4cos{(4y)}sin{(z^2)})}}{{(5x^3)}}\\&f_{xxy}=\frac{-{(16sin{(4y)}sin{(z^2)})}}{{(5x^3)}}\end{align*}$

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Example Question #2930 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=\frac{{(2\cdot4^{(4z)}y^2sin{(x^2)})}}{5}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=\frac{{(2\cdot4^{(4z)}y^2sin{(x^2)})}}{5}\end{align*}$

Possible Answers:

${\frac{{(64\cdot4^{(8z)}y^4sin{(x^2)}^2ln{(4)}^2)}}{25}}$ $\displaystyle {\frac{{(64\cdot4^{(8z)}y^4sin{(x^2)}^2ln{(4)}^2)}}{25}}$

${\frac{{(32\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)}^2)}}{5}}$ $\displaystyle {\frac{{(32\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)}^2)}}{5}}$

${\frac{{(256\cdot4^{(8z)}y^4sin{(x^2)}^2ln{(4)}^3)}}{25}}$ $\displaystyle {\frac{{(256\cdot4^{(8z)}y^4sin{(x^2)}^2ln{(4)}^3)}}{25}}$

${\frac{{(16\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)})}}{5}}$ $\displaystyle {\frac{{(16\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)})}}{5}}$

Correct answer:

${\frac{{(32\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)}^2)}}{5}}$ $\displaystyle {\frac{{(32\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)}^2)}}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(4z)}y^2sin{(x^2)})}}{5}\\&f_{z}=\frac{{(8\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)})}}{5}\\&f_{zz}=\frac{{(32\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)}^2)}}{5}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(4z)}y^2sin{(x^2)})}}{5}\\&f_{z}=\frac{{(8\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)})}}{5}\\&f_{zz}=\frac{{(32\cdot4^{(4z)}y^2sin{(x^2)}ln{(4)}^2)}}{5}\end{align*}$

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