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Calculus 3 : Calculus 3

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Example Questions

Example Question #2941 : Calculus 3

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

Possible Answers:

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

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

${-64cos{(4y)}ln{(z^2)}^2sin{(4x)}^2sin{(4y)}}$ $\displaystyle {-64cos{(4y)}ln{(z^2)}^2sin{(4x)}^2sin{(4y)}}$

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

Correct answer:

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

Explanation:

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

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Example Question #2942 : Calculus 3

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

Possible Answers:

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

${\frac{-{(48z^4e^{(8y)})}}{x^7}}$ $\displaystyle {\frac{-{(48z^4e^{(8y)})}}{x^7}}$

${\frac{{(16z^4e^{(8y)})}}{x^6}}$ $\displaystyle {\frac{{(16z^4e^{(8y)})}}{x^6}}$

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

Correct answer:

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

Explanation:

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

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Example Question #2943 : Calculus 3

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

Possible Answers:

${-\frac{ {(3\cdot4^{(3y)}cos{(3z)}e^{(x^2)})}}{2} - 4^{(3y)}xsin{(3z)}e^{(x^2)}}$ $\displaystyle {-\frac{ {(3\cdot4^{(3y)}cos{(3z)}e^{(x^2)})}}{2} - 4^{(3y)}xsin{(3z)}e^{(x^2)}}$

${\frac{{(3\cdot4^{(6y)}xcos{(3z)}sin{(3z)}e^{(2x^2)})}}{2}}$ $\displaystyle {\frac{{(3\cdot4^{(6y)}xcos{(3z)}sin{(3z)}e^{(2x^2)})}}{2}}$

${-3\cdot4^{(3y)}xcos{(3z)}e^{(x^2)}}$ $\displaystyle {-3\cdot4^{(3y)}xcos{(3z)}e^{(x^2)}}$

${\frac{{(9\cdot4^{(6y)}xcos{(3z)}^2e^{(2x^2)})}}{2}}$ $\displaystyle {\frac{{(9\cdot4^{(6y)}xcos{(3z)}^2e^{(2x^2)})}}{2}}$

Correct answer:

${-3\cdot4^{(3y)}xcos{(3z)}e^{(x^2)}}$ $\displaystyle {-3\cdot4^{(3y)}xcos{(3z)}e^{(x^2)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(4^{(3y)}sin{(3z)}e^{(x^2)})}}{2}\\&f_{z}=\frac{-{(3\cdot4^{(3y)}cos{(3z)}e^{(x^2)})}}{2}\\&f_{zx}=-3\cdot4^{(3y)}xcos{(3z)}e^{(x^2)}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(4^{(3y)}sin{(3z)}e^{(x^2)})}}{2}\\&f_{z}=\frac{-{(3\cdot4^{(3y)}cos{(3z)}e^{(x^2)})}}{2}\\&f_{zx}=-3\cdot4^{(3y)}xcos{(3z)}e^{(x^2)}\end{align*}$

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Example Question #2944 : Calculus 3

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

Possible Answers:

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

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

${\frac{-{(4cos{(4y)}cos{(3z)}sin{(4y)}^2sin{(3z)}^2)}}{{(9x^4)}}}$ $\displaystyle {\frac{-{(4cos{(4y)}cos{(3z)}sin{(4y)}^2sin{(3z)}^2)}}{{(9x^4)}}}$

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

Correct answer:

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

Explanation:

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

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Example Question #2945 : Calculus 3

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

Possible Answers:

${18z^3cos{(3x)}^2cos{(4y)}^2}$ $\displaystyle {18z^3cos{(3x)}^2cos{(4y)}^2}$

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

$\displaystyle {6zcos{(3x)}cos{(4y)}}$

${2zcos{(4y)}sin{(3x)} + 3z^2cos{(3x)}cos{(4y)}}$ $\displaystyle {2zcos{(4y)}sin{(3x)} + 3z^2cos{(3x)}cos{(4y)}}$

Correct answer:

$\displaystyle {6zcos{(3x)}cos{(4y)}}$

Explanation:

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

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Example Question #2946 : Calculus 3

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

Possible Answers:

${\frac{-{(z^2cos{(3y)}^3ln{(3x)})}}{{(8x^2)}}}$ $\displaystyle {\frac{-{(z^2cos{(3y)}^3ln{(3x)})}}{{(8x^2)}}}$

${\frac{-cos{(3y)}}{{(2x)}}}$ $\displaystyle {\frac{-cos{(3y)}}{{(2x)}}}$

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

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

Correct answer:

${\frac{-cos{(3y)}}{{(2x)}}}$ $\displaystyle {\frac{-cos{(3y)}}{{(2x)}}}$

Explanation:

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

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Example Question #2947 : Calculus 3

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

Possible Answers:

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

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

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

${\frac{-{(ze^{(3y)})}}{x}}$ $\displaystyle {\frac{-{(ze^{(3y)})}}{x}}$

Correct answer:

${\frac{-{(ze^{(3y)})}}{x}}$ $\displaystyle {\frac{-{(ze^{(3y)})}}{x}}$

Explanation:

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

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Example Question #2948 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{-{(2\cdot4^xln{(y)}sin{(z)})}}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{-{(2\cdot4^xln{(y)}sin{(z)})}}{7}\end{align*}$

Possible Answers:

${-\frac{ {(2\cdot4^xsin{(z)})}}{{(7y)}} -\frac{ {(2\cdot4^xln{(4)}ln{(y)}sin{(z)})}}{7}}$ $\displaystyle {-\frac{ {(2\cdot4^xsin{(z)})}}{{(7y)}} -\frac{ {(2\cdot4^xln{(4)}ln{(y)}sin{(z)})}}{7}}$

${\frac{{(4\cdot4^{(2x)}ln{(4)}sin{(z)}^2)}}{{(49y^2)}}}$ $\displaystyle {\frac{{(4\cdot4^{(2x)}ln{(4)}sin{(z)}^2)}}{{(49y^2)}}}$

${\frac{-{(2\cdot4^xln{(4)}sin{(z)})}}{{(7y)}}}$ $\displaystyle {\frac{-{(2\cdot4^xln{(4)}sin{(z)})}}{{(7y)}}}$

${\frac{{(4\cdot4^{(2x)}ln{(4)}ln{(y)}sin{(z)}^2)}}{{(49y)}}}$ $\displaystyle {\frac{{(4\cdot4^{(2x)}ln{(4)}ln{(y)}sin{(z)}^2)}}{{(49y)}}}$

Correct answer:

${\frac{-{(2\cdot4^xln{(4)}sin{(z)})}}{{(7y)}}}$ $\displaystyle {\frac{-{(2\cdot4^xln{(4)}sin{(z)})}}{{(7y)}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2\cdot4^xln{(y)}sin{(z)})}}{7}\\&f_{y}=\frac{-{(2\cdot4^xsin{(z)})}}{{(7y)}}\\&f_{yx}=\frac{-{(2\cdot4^xln{(4)}sin{(z)})}}{{(7y)}}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2\cdot4^xln{(y)}sin{(z)})}}{7}\\&f_{y}=\frac{-{(2\cdot4^xsin{(z)})}}{{(7y)}}\\&f_{yx}=\frac{-{(2\cdot4^xln{(4)}sin{(z)})}}{{(7y)}}\end{align*}$

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Example Question #2949 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xxzy}\\&\text{Where }f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{xxzy}\\&\text{Where }f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\end{align*}$

Possible Answers:

${\frac{-{(192\cdot4^{(12y)}x^6ln{(4)}cos{(z)}^3sin{(z)})}}{2401}}$ $\displaystyle {\frac{-{(192\cdot4^{(12y)}x^6ln{(4)}cos{(z)}^3sin{(z)})}}{2401}}$

${\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}}$ $\displaystyle {\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}}$

${\frac{{(768\cdot4^{(12y)}xln{(4)}cos{(z)}^2sin{(z)}^2)}}{2401}}$ $\displaystyle {\frac{{(768\cdot4^{(12y)}xln{(4)}cos{(z)}^2sin{(z)}^2)}}{2401}}$

${\frac{{(8\cdot4^{(3y)}xcos{(z)})}}{7} -\frac{ {(2\cdot4^{(3y)}x^2sin{(z)})}}{7} +\frac{ {(6\cdot4^{(3y)}x^2ln{(4)}cos{(z)})}}{7}}$ $\displaystyle {\frac{{(8\cdot4^{(3y)}xcos{(z)})}}{7} -\frac{ {(2\cdot4^{(3y)}x^2sin{(z)})}}{7} +\frac{ {(6\cdot4^{(3y)}x^2ln{(4)}cos{(z)})}}{7}}$

Correct answer:

${\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}}$ $\displaystyle {\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\\&f_{x}=\frac{{(4\cdot4^{(3y)}xcos{(z)})}}{7}\\&f_{xx}=\frac{{(4\cdot4^{(3y)}cos{(z)})}}{7}\\&f_{xxz}=\frac{-{(4\cdot4^{(3y)}sin{(z)})}}{7}\\&f_{xxzy}=\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\\&f_{x}=\frac{{(4\cdot4^{(3y)}xcos{(z)})}}{7}\\&f_{xx}=\frac{{(4\cdot4^{(3y)}cos{(z)})}}{7}\\&f_{xxz}=\frac{-{(4\cdot4^{(3y)}sin{(z)})}}{7}\\&f_{xxzy}=\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}\end{align*}$

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Example Question #2950 : Calculus 3

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

Possible Answers:

${108\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^3}$ $\displaystyle {108\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^3}$

${12\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}}$ $\displaystyle {12\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}}$

${18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$ $\displaystyle {18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$

${36\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^2}$ $\displaystyle {36\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^2}$

Correct answer:

${18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$ $\displaystyle {18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$

Explanation:

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

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