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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #3111 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{(2\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{2})}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{(2\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{2})}\end{align*}$

Possible Answers:

${\frac{(4\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{3})}-\frac{ (6\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{2})}}$ $\displaystyle {\frac{(4\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{3})}-\frac{ (6\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{2})}}$

${-\frac{(24\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{5})}}$ $\displaystyle {-\frac{(24\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{5})}}$

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

${\frac{(48\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{6})}}$ $\displaystyle {\frac{(48\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{6})}}$

Correct answer:

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

Explanation:

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

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Example Question #3112 : Calculus 3

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

Possible Answers:

${-\frac{(32\cdot 2^{(8x)}zcos(z^{2})ln(4y)sin(z^{2}))}{y}}$ $\displaystyle {-\frac{(32\cdot 2^{(8x)}zcos(z^{2})ln(4y)sin(z^{2}))}{y}}$

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

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

${\frac{(8\cdot 2^{(4x)}zsin(z^{2}))}{y}}$ $\displaystyle {\frac{(8\cdot 2^{(4x)}zsin(z^{2}))}{y}}$

Correct answer:

${\frac{(8\cdot 2^{(4x)}zsin(z^{2}))}{y}}$ $\displaystyle {\frac{(8\cdot 2^{(4x)}zsin(z^{2}))}{y}}$

Explanation:

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

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Example Question #3113 : Calculus 3

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

Possible Answers:

${-\frac{(2tan(y^{3}))}{(5xz)}}$ $\displaystyle {-\frac{(2tan(y^{3}))}{(5xz)}}$

${tan(\frac{y^{3})^{2}}{(25x^{2}z^{2})}}$ $\displaystyle {tan(\frac{y^{3})^{2}}{(25x^{2}z^{2})}}$

${\frac{tan(y^{3})}{(5x^{2}z)}}$ $\displaystyle {\frac{tan(y^{3})}{(5x^{2}z)}}$

${-tan(\frac{y^{3})^{2}}{(25x^{3}z^{2})}}$ $\displaystyle {-tan(\frac{y^{3})^{2}}{(25x^{3}z^{2})}}$

Correct answer:

${\frac{tan(y^{3})}{(5x^{2}z)}}$ $\displaystyle {\frac{tan(y^{3})}{(5x^{2}z)}}$

Explanation:

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

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Example Question #3114 : Calculus 3

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

Possible Answers:

${\frac{(2x^{3}ze^{(z^{2})})}{y^{2}}+\frac{ (4zcos(z^{2})tan(x^{4})sin(4y + y^{2})\cdot (2y + 4))}{5}}$ $\displaystyle {\frac{(2x^{3}ze^{(z^{2})})}{y^{2}}+\frac{ (4zcos(z^{2})tan(x^{4})sin(4y + y^{2})\cdot (2y + 4))}{5}}$

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

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

${-(\frac{(2x^{3}ze^{(z^{2})})}{y^{2}}+\frac{ (4zcos(z^{2})tan(x^{4})sin(4y + y^{2})\cdot (2y + 4))}{5})\cdot (\frac{(2x^{3}ze^{(z^{2})})}{y}+\frac{ (4zcos(z^{2})tan(x^{4})cos(4y + y^{2}))}{5})}$ $\displaystyle {-(\frac{(2x^{3}ze^{(z^{2})})}{y^{2}}+\frac{ (4zcos(z^{2})tan(x^{4})sin(4y + y^{2})\cdot (2y + 4))}{5})\cdot (\frac{(2x^{3}ze^{(z^{2})})}{y}+\frac{ (4zcos(z^{2})tan(x^{4})cos(4y + y^{2}))}{5})}$

Correct answer:

Explanation:

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

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Example Question #3115 : Calculus 3

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

Possible Answers:

${\frac{(9\cdot 4^{(2y)}e^{(6z)})}{(16x^{2})}}$ $\displaystyle {\frac{(9\cdot 4^{(2y)}e^{(6z)})}{(16x^{2})}}$

${\frac{(27\cdot 4^{(2y)}e^{(6z)})}{(16x^{2})}}$ $\displaystyle {\frac{(27\cdot 4^{(2y)}e^{(6z)})}{(16x^{2})}}$

${\frac{(3\cdot 4^ye^{(3z)})}{(2x)}}$ $\displaystyle {\frac{(3\cdot 4^ye^{(3z)})}{(2x)}}$

${\frac{(9\cdot 4^ye^{(3z)})}{(4x)}}$ $\displaystyle {\frac{(9\cdot 4^ye^{(3z)})}{(4x)}}$

Correct answer:

${\frac{(9\cdot 4^ye^{(3z)})}{(4x)}}$ $\displaystyle {\frac{(9\cdot 4^ye^{(3z)})}{(4x)}}$

Explanation:

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

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Example Question #3116 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{(2y^{2}sin(x^{4} - 4x)tan(12z + z^{4}))}{9}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{(2y^{2}sin(x^{4} - 4x)tan(12z + z^{4}))}{9}\end{align*}$

Possible Answers:

${-\frac{ (4ysin(x^{4} - 4x)tan(12z + z^{4}))}{9}-\frac{ (2y^{2}cos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$ $\displaystyle {-\frac{ (4ysin(x^{4} - 4x)tan(12z + z^{4}))}{9}-\frac{ (2y^{2}cos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$

${-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$ $\displaystyle {-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$

${\frac{(16y^{2}cos(x^{4} - 4x)sin(x^{4} - 4x)tan(12z + z^{4})^{2}\cdot (4x^{3} - 4))}{81}}$ $\displaystyle {\frac{(16y^{2}cos(x^{4} - 4x)sin(x^{4} - 4x)tan(12z + z^{4})^{2}\cdot (4x^{3} - 4))}{81}}$

${\frac{(8y^{3}cos(x^{4} - 4x)sin(x^{4} - 4x)tan(12z + z^{4})^{2}\cdot (4x^{3} - 4))}{81}}$ $\displaystyle {\frac{(8y^{3}cos(x^{4} - 4x)sin(x^{4} - 4x)tan(12z + z^{4})^{2}\cdot (4x^{3} - 4))}{81}}$

Correct answer:

${-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$ $\displaystyle {-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$

Explanation:

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

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Example Question #3117 : Calculus 3

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

Possible Answers:

${\frac{(2^{(z^{2})}xln(y)\cdot (3y^{3}z^{2}e^{(z^{3})} +\frac{ (2^{(z^{2})}x^{2}zln(2)ln(y))}{2}))}{2}}$ $\displaystyle {\frac{(2^{(z^{2})}xln(y)\cdot (3y^{3}z^{2}e^{(z^{3})} +\frac{ (2^{(z^{2})}x^{2}zln(2)ln(y))}{2}))}{2}}$

${2^{(z^{2})}xzln(2)ln(y)}$ $\displaystyle {2^{(z^{2})}xzln(2)ln(y)}$

${3y^{3}z^{2}e^{(z^{3})} +\frac{ (2^{(z^{2})}xln(y))}{2}+\frac{ (2^{(z^{2})}x^{2}zln(2)ln(y))}{2}}$ $\displaystyle {3y^{3}z^{2}e^{(z^{3})} +\frac{ (2^{(z^{2})}xln(y))}{2}+\frac{ (2^{(z^{2})}x^{2}zln(2)ln(y))}{2}}$

${\frac{(2^{(2z^{2})}x^{2}zln(2)ln(y)^{2})}{2}}$ $\displaystyle {\frac{(2^{(2z^{2})}x^{2}zln(2)ln(y)^{2})}{2}}$

Correct answer:

${2^{(z^{2})}xzln(2)ln(y)}$ $\displaystyle {2^{(z^{2})}xzln(2)ln(y)}$

Explanation:

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

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Example Question #3118 : Calculus 3

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

Possible Answers:

${(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))^{2}}$ $\displaystyle {(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))^{2}}$

${-(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))\cdot (2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x))}$ $\displaystyle {-(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))\cdot (2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x))}$

${32y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 4\cdot 4^{(3x)}cos(4z + z^{4})cos(y)}$ $\displaystyle {32y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 4\cdot 4^{(3x)}cos(4z + z^{4})cos(y)}$

${2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)}$ $\displaystyle {2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})sin(y)\\&f_{y}=16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})cos(y)\\&f_{yy}=2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})sin(y)\\&f_{y}=16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})cos(y)\\&f_{yy}=2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)\end{align*}$

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Example Question #3119 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xz}\\&\text{Where }f(x,y,z)=3xln(z^{2})e^{(y^{2})} +\frac{ (e^{(4y)}tan(z^{2} - 11z)cos(x))}{5}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{xz}\\&\text{Where }f(x,y,z)=3xln(z^{2})e^{(y^{2})} +\frac{ (e^{(4y)}tan(z^{2} - 11z)cos(x))}{5}\end{align*}$

Possible Answers:

${(3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5})\cdot (\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5})}$ $\displaystyle {(3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5})\cdot (\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5})}$

${3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5}+\frac{ (6xe^{(y^{2})})}{z}+\frac{ (e^{(4y)}cos(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$ $\displaystyle {3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5}+\frac{ (6xe^{(y^{2})})}{z}+\frac{ (e^{(4y)}cos(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$

${\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$ $\displaystyle {\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$

${(\frac{(6xe^{(y^{2})})}{z}+\frac{ (e^{(4y)}cos(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5})\cdot (3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5})}$ $\displaystyle {(\frac{(6xe^{(y^{2})})}{z}+\frac{ (e^{(4y)}cos(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5})\cdot (3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5})}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=3xln(z^{2})e^{(y^{2})} +\frac{ (e^{(4y)}tan(z^{2} - 11z)cos(x))}{5}\\&f_{x}=3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5}\\&f_{xz}=\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=3xln(z^{2})e^{(y^{2})} +\frac{ (e^{(4y)}tan(z^{2} - 11z)cos(x))}{5}\\&f_{x}=3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5}\\&f_{xz}=\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}\end{align*}$

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Example Question #3120 : Calculus 3

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

Possible Answers:

${(\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3}))^{2}}$ $\displaystyle {(\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3}))^{2}}$

${\frac{(4\cdot 4^{(3x)}e^{(z)})}{3}+ 10x^{4}cos(y^{2})cos(z^{3}) - 20x^{4}y^{2}cos(z^{3})sin(y^{2})}$ $\displaystyle {\frac{(4\cdot 4^{(3x)}e^{(z)})}{3}+ 10x^{4}cos(y^{2})cos(z^{3}) - 20x^{4}y^{2}cos(z^{3})sin(y^{2})}$

${\frac{(8\cdot 4^{(3x)}ye^{(z)})}{3}+ 20x^{4}ycos(y^{2})cos(z^{3})}$ $\displaystyle {\frac{(8\cdot 4^{(3x)}ye^{(z)})}{3}+ 20x^{4}ycos(y^{2})cos(z^{3})}$

${(\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3}))\cdot (\frac{(4\cdot 4^{(3x)}e^{(z)})}{3}+ 10x^{4}cos(y^{2})cos(z^{3}) - 20x^{4}y^{2}cos(z^{3})sin(y^{2}))}$ $\displaystyle {(\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3}))\cdot (\frac{(4\cdot 4^{(3x)}e^{(z)})}{3}+ 10x^{4}cos(y^{2})cos(z^{3}) - 20x^{4}y^{2}cos(z^{3})sin(y^{2}))}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 4^{(3x)}y^{2}e^{(z)})}{3}+ 5x^{4}cos(z^{3})sin(y^{2})\\&f_{y}=\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3})\\&f_{yy}=\frac{(4\cdot 4^{(3x)}e^{(z)})}{3}+ 10x^{4}cos(y^{2})cos(z^{3}) - 20x^{4}y^{2}cos(z^{3})sin(y^{2})\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 4^{(3x)}y^{2}e^{(z)})}{3}+ 5x^{4}cos(z^{3})sin(y^{2})\\&f_{y}=\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3})\\&f_{yy}=\frac{(4\cdot 4^{(3x)}e^{(z)})}{3}+ 10x^{4}cos(y^{2})cos(z^{3}) - 20x^{4}y^{2}cos(z^{3})sin(y^{2})\end{align*}$

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Calculus 3 : Calculus 3

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