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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3091 : Calculus 3

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

Possible Answers:

${-\frac{(4z^{2}e^{(6x)}e^{(2z^{2})}cos(y)sin(y))}{9}}$

${-\frac{(2ze^{(6x)}e^{(2z^{2})}cos(y)sin(y))}{9}}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #3092 : Calculus 3

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

Possible Answers:

${-4y^{2}z^{3}cos(x)\cdot (tan(z^{4})^{2} + 1)}$

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

${16y^{4}z^{6}cos(x)sin(x)\cdot (tan(z^{4})^{2} + 1)^{2}}$

${4y^{4}z^{3}tan(z^{4})cos(x)sin(x)\cdot (tan(z^{4})^{2} + 1)}$

Correct answer:

${-4y^{2}z^{3}cos(x)\cdot (tan(z^{4})^{2} + 1)}$

Explanation:

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

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Example Question #3093 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3094 : Calculus 3

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

Possible Answers:

${144\cdot 4^{(2z)}x^{6}e^{(6y)}\cdot (tan(x^{4})^{2} + 1)^{2}}$

${-24\cdot 4^zx^{3}e^{(3y)}\cdot (tan(x^{4})^{2} + 1)}$

${12\cdot 4^zx^{3}e^{(3y)}\cdot (tan(x^{4})^{2} + 1)\cdot (36\cdot 4^zx^{2}e^{(3y)}\cdot (tan(x^{4})^{2} + 1) + 96\cdot 4^zx^{6}tan(x^{4})e^{(3y)}\cdot (tan(x^{4})^{2} + 1))}$

${- 36\cdot 4^zx^{2}e^{(3y)}\cdot (tan(x^{4})^{2} + 1) - 96\cdot 4^zx^{6}tan(x^{4})e^{(3y)}\cdot (tan(x^{4})^{2} + 1)}$

Correct answer:

${- 36\cdot 4^zx^{2}e^{(3y)}\cdot (tan(x^{4})^{2} + 1) - 96\cdot 4^zx^{6}tan(x^{4})e^{(3y)}\cdot (tan(x^{4})^{2} + 1)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-3\cdot 4^ztan(x^{4})e^{(3y)}\\&f_{x}=-12\cdot 4^zx^{3}e^{(3y)}\cdot (tan(x^{4})^{2} + 1)\\&f_{xx}=- 36\cdot 4^zx^{2}e^{(3y)}\cdot (tan(x^{4})^{2} + 1) - 96\cdot 4^zx^{6}tan(x^{4})e^{(3y)}\cdot (tan(x^{4})^{2} + 1)\end{align*}$

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Example Question #3095 : Calculus 3

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

Possible Answers:

${\frac{24}{(y^{4}z^{2})}-\frac{ (4yz^{2}e^{(y^{2})}sin(x))}{3}}$

${(\frac{12}{(y^{4}z^{2})}-\frac{ (2yz^{2}e^{(y^{2})}sin(x))}{3})^{2}}$

${-(\frac{12}{(y^{4}z^{2})}-\frac{ (2yz^{2}e^{(y^{2})}sin(x))}{3})\cdot (\frac{48}{(y^{5}z^{2})}+\frac{ (2z^{2}e^{(y^{2})}sin(x))}{3}+\frac{ (4y^{2}z^{2}e^{(y^{2})}sin(x))}{3})}$

${-\frac{ 48}{(y^{5}z^{2})}-\frac{ (2z^{2}e^{(y^{2})}sin(x))}{3}-\frac{ (4y^{2}z^{2}e^{(y^{2})}sin(x))}{3}}$

Correct answer:

${-\frac{ 48}{(y^{5}z^{2})}-\frac{ (2z^{2}e^{(y^{2})}sin(x))}{3}-\frac{ (4y^{2}z^{2}e^{(y^{2})}sin(x))}{3}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ 4}{(y^{3}z^{2})}-\frac{ (z^{2}e^{(y^{2})}sin(x))}{3}\\&f_{y}=\frac{12}{(y^{4}z^{2})}-\frac{ (2yz^{2}e^{(y^{2})}sin(x))}{3}\\&f_{yy}=-\frac{ 48}{(y^{5}z^{2})}-\frac{ (2z^{2}e^{(y^{2})}sin(x))}{3}-\frac{ (4y^{2}z^{2}e^{(y^{2})}sin(x))}{3}\end{align*}$

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Example Question #3096 : Calculus 3

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

Possible Answers:

${\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}+\frac{ (8sin(y))}{(x^{2}z^{2})}}$

${(\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (8sin(y))}{(xz^{2})})\cdot (\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}+\frac{ (8sin(y))}{(x^{2}z^{2})})}$

${\frac{(2\cdot 2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (16sin(y))}{(xz^{2})}}$

${(\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (8sin(y))}{(xz^{2})})^{2}}$

Correct answer:

${\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}+\frac{ (8sin(y))}{(x^{2}z^{2})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (4ln(x^{2})sin(y))}{z^{2}}\\&f_{x}=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}-\frac{ (8sin(y))}{(xz^{2})}\\&f_{xx}=\frac{(2^{(4z)}ln(3y)e^{(x)})}{5}+\frac{ (8sin(y))}{(x^{2}z^{2})}\end{align*}$

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Example Question #3097 : Calculus 3

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

Possible Answers:

${\frac{(24z^{3}e^{(6x)}e^{(6y)})}{49}}$

${-\frac{ (4ze^{(3x)}e^{(3y)})}{7}-\frac{ (6z^{2}e^{(3x)}e^{(3y)})}{7}}$

${-\frac{(12ze^{(3x)}e^{(3y)})}{7}}$

${\frac{(48z^{2}e^{(6x)}e^{(6y)})}{49}}$

Correct answer:

${-\frac{(12ze^{(3x)}e^{(3y)})}{7}}$

Explanation:

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

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Example Question #3098 : Calculus 3

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

Possible Answers:

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

${\frac{(4\cdot 4^{(8x)}ytan(z^{3})^{2})}{9}}$

${\frac{(4\cdot 4^{(4x)}ytan(z^{3}))}{3}}$

${\frac{(4\cdot 4^{(8x)}y^{2}tan(z^{3})^{2})}{9}}$

Correct answer:

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

Explanation:

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

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Example Question #3099 : Calculus 3

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

Possible Answers:

${-\frac{(10cos(x)cos(y))}{z^{3}}}$

${-\frac{(50cos(x)^{2}cos(y)^{2})}{z^{5}}}$

${-\frac{(50cos(x)cos(y)^{2}sin(x))}{z^{5}}}$

${\frac{(5cos(x)cos(y))}{z^{2}}-\frac{ (10cos(y)sin(x))}{z^{3}}}$

Correct answer:

${-\frac{(10cos(x)cos(y))}{z^{3}}}$

Explanation:

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

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Example Question #3100 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=\frac{(2sin(x^{2})e^{(4z)})}{(7y)}+ 3x^{4}z^{3}e^{(y)}\end{align*}$

Possible Answers:

${18x^{4}ze^{(y)} +\frac{ (32sin(x^{2})e^{(4z)})}{(7y)}}$

${(\frac{(8sin(x^{2})e^{(4z)})}{(7y)}+ 9x^{4}z^{2}e^{(y)})\cdot (18x^{4}ze^{(y)} +\frac{ (32sin(x^{2})e^{(4z)})}{(7y)})}$

${(\frac{(8sin(x^{2})e^{(4z)})}{(7y)}+ 9x^{4}z^{2}e^{(y)})^{2}}$

${\frac{(16sin(x^{2})e^{(4z)})}{(7y)}+ 18x^{4}z^{2}e^{(y)}}$

Correct answer:

${18x^{4}ze^{(y)} +\frac{ (32sin(x^{2})e^{(4z)})}{(7y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2sin(x^{2})e^{(4z)})}{(7y)}+ 3x^{4}z^{3}e^{(y)}\\&f_{z}=\frac{(8sin(x^{2})e^{(4z)})}{(7y)}+ 9x^{4}z^{2}e^{(y)}\\&f_{zz}=18x^{4}ze^{(y)} +\frac{ (32sin(x^{2})e^{(4z)})}{(7y)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3091 : Calculus 3

Example Question #3092 : Calculus 3

Example Question #3093 : Calculus 3

Example Question #3094 : Calculus 3

Example Question #3095 : Calculus 3

Example Question #3096 : Calculus 3

Example Question #3097 : Calculus 3

Example Question #3098 : Calculus 3

Example Question #3099 : Calculus 3

Example Question #3100 : Calculus 3

All Calculus 3 Resources

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