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Calculus 3 : Calculus 3

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

Example Questions

Example Question #601 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=4x^{4}cos(z^{3})ln(y^{2}) - 2\cdot2^{(4y)}\cdot4^xcos(4z)\end{align*}$

Possible Answers:

${32\cdot2^{(4y)}\cdot4^xcos(4z) - 24x^{4}zln(y^{2})sin(z^{3}) - 36x^{4}z^{4}cos(z^{3})ln(y^{2})}$

${-(8\cdot2^{(4y)}\cdot4^xsin(4z) - 12x^{4}z^{2}ln(y^{2})sin(z^{3}))\cdot(24x^{4}zln(y^{2})sin(z^{3}) - 32\cdot2^{(4y)}\cdot4^xcos(4z) + 36x^{4}z^{4}cos(z^{3})ln(y^{2}))}$

${16\cdot2^{(4y)}\cdot4^xsin(4z) - 24x^{4}z^{2}ln(y^{2})sin(z^{3})}$

${(8\cdot2^{(4y)}\cdot4^xsin(4z) - 12x^{4}z^{2}ln(y^{2})sin(z^{3}))^{2}}$

Correct answer:

${32\cdot2^{(4y)}\cdot4^xcos(4z) - 24x^{4}zln(y^{2})sin(z^{3}) - 36x^{4}z^{4}cos(z^{3})ln(y^{2})}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=4x^{4}cos(z^{3})ln(y^{2}) - 2\cdot2^{(4y)}\cdot4^xcos(4z)\\&f_{z}=8\cdot2^{(4y)}\cdot4^xsin(4z) - 12x^{4}z^{2}ln(y^{2})sin(z^{3})\\&f_{zz}=32\cdot2^{(4y)}\cdot4^xcos(4z) - 24x^{4}zln(y^{2})sin(z^{3}) - 36x^{4}z^{4}cos(z^{3})ln(y^{2})\end{align*}$

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Example Question #604 : Partial Derivatives

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

Possible Answers:

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

${(\frac{(4z^{2}ln(y))}{x}- 2xln(z^{2})e^{(x^{2})}e^{(4y)})^{2}}$

${\frac{(8z^{2}ln(y))}{x}- 4xln(z^{2})e^{(x^{2})}e^{(4y)}}$

${-(\frac{(4z^{2}ln(y))}{x}- 2xln(z^{2})e^{(x^{2})}e^{(4y)})\cdot(2ln(z^{2})e^{(x^{2})}e^{(4y)} +\frac{ (4z^{2}ln(y))}{x^{2}}+ 4x^{2}ln(z^{2})e^{(x^{2})}e^{(4y)})}$

Correct answer:

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

Explanation:

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

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Example Question #605 : Partial Derivatives

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

Possible Answers:

${-(ln(x^{2})sin(y^{3})e^{(z)} - 2x^{2}zsin(4y)e^{(z^{2})})\cdot(4x^{2}cos(4y)e^{(z^{2})} - 3y^{2}cos(y^{3})ln(x^{2})e^{(z)})}$

${(8x^{2}zcos(4y)e^{(z^{2})} - 3y^{2}cos(y^{3})ln(x^{2})e^{(z)})\cdot(4x^{2}cos(4y)e^{(z^{2})} - 3y^{2}cos(y^{3})ln(x^{2})e^{(z)})}$

${3y^{2}cos(y^{3})ln(x^{2})e^{(z)} - 8x^{2}zcos(4y)e^{(z^{2})}}$

${ln(x^{2})sin(y^{3})e^{(z)} - 4x^{2}cos(4y)e^{(z^{2})} - 2x^{2}zsin(4y)e^{(z^{2})} + 3y^{2}cos(y^{3})ln(x^{2})e^{(z)}}$

Correct answer:

${3y^{2}cos(y^{3})ln(x^{2})e^{(z)} - 8x^{2}zcos(4y)e^{(z^{2})}}$

Explanation:

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

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Example Question #606 : Partial Derivatives

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

Possible Answers:

${\frac{(3sin(x))}{y}-\frac{ (9e^{(3x)})}{(yz^{3})}}$

${3ln(y)sin(x) +\frac{ (9e^{(3x)})}{(2yz^{2})}+\frac{ (3zsin(x))}{y}-\frac{ (9ln(4y)e^{(3x)})}{z^{3}}}$

${(3ln(y)sin(x) -\frac{ (9ln(4y)e^{(3x)})}{z^{3}})\cdot(\frac{(3sin(x))}{y}-\frac{ (9e^{(3x)})}{(yz^{3})})}$

${(3ln(y)sin(x) -\frac{ (9ln(4y)e^{(3x)})}{z^{3}})\cdot(\frac{(9e^{(3x)})}{(2yz^{2})}+\frac{ (3zsin(x))}{y})}$

Correct answer:

${\frac{(3sin(x))}{y}-\frac{ (9e^{(3x)})}{(yz^{3})}}$

Explanation:

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

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Example Question #607 : Partial Derivatives

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

Possible Answers:

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

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

${- 12y^{2}sin(y^{3})e^{(z)} -\frac{ (16sin(4y)e^{(3x)})}{(3z)}}$

${4cos(y^{3})e^{(z)} -\frac{ (8ln(z^{2})sin(4y)e^{(3x)})}{3}- 12y^{2}sin(y^{3})e^{(z)} +\frac{ (4cos(4y)e^{(3x)})}{(3z)}}$

Correct answer:

${- 12y^{2}sin(y^{3})e^{(z)} -\frac{ (16sin(4y)e^{(3x)})}{(3z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4cos(y^{3})e^{(z)} +\frac{ (2cos(4y)ln(z^{2})e^{(3x)})}{3}\\&f_{z}=4cos(y^{3})e^{(z)} +\frac{ (4cos(4y)e^{(3x)})}{(3z)}\\&f_{zy}=- 12y^{2}sin(y^{3})e^{(z)} -\frac{ (16sin(4y)e^{(3x)})}{(3z)}\end{align*}$

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Example Question #608 : Partial Derivatives

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

Possible Answers:

${10cos(y^{2})cos(z) -\frac{ (2ln(3x))}{(y^{2}z^{2})}- 20y^{2}sin(y^{2})cos(z)}$

${(\frac{(2ln(3x))}{(yz^{2})}+ 10ycos(y^{2})cos(z))^{2}}$

${-(\frac{(2ln(3x))}{(yz^{2})}+ 10ycos(y^{2})cos(z))\cdot(\frac{(2ln(3x))}{(y^{2}z^{2})}- 10cos(y^{2})cos(z) + 20y^{2}sin(y^{2})cos(z))}$

${\frac{(4ln(3x))}{(yz^{2})}+ 20ycos(y^{2})cos(z)}$

Correct answer:

${10cos(y^{2})cos(z) -\frac{ (2ln(3x))}{(y^{2}z^{2})}- 20y^{2}sin(y^{2})cos(z)}$

Explanation:

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

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Example Question #609 : Partial Derivatives

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

Possible Answers:

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

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

${-(\frac{(2zsin(y))}{x^{3}}+\frac{ (4yze^{(y^{2})})}{x})\cdot(\frac{(2zcos(y))}{x^{3}}-\frac{ (2ze^{(y^{2})})}{x})}$

${-\frac{ (2zsin(y))}{x^{3}}-\frac{ (4yze^{(y^{2})})}{x}}$

Correct answer:

${-\frac{ (2zsin(y))}{x^{3}}-\frac{ (4yze^{(y^{2})})}{x}}$

Explanation:

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

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Example Question #610 : Partial Derivatives

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

Possible Answers:

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

${\frac{(5ze^{(y)})}{x}}$

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

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

Correct answer:

${\frac{(5ze^{(y)})}{x}}$

Explanation:

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

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Example Question #611 : Partial Derivatives

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

Possible Answers:

${2x^{2}y^{2} - 6zcos(y^{3})sin(z^{3})e^{(x)} - 9z^{4}cos(y^{3})cos(z^{3})e^{(x)}}$

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

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

${-(2x^{2}y^{2}z - 3z^{2}cos(y^{3})sin(z^{3})e^{(x)})\cdot(6zcos(y^{3})sin(z^{3})e^{(x)} - 2x^{2}y^{2} + 9z^{4}cos(y^{3})cos(z^{3})e^{(x)})}$

Correct answer:

${2x^{2}y^{2} - 6zcos(y^{3})sin(z^{3})e^{(x)} - 9z^{4}cos(y^{3})cos(z^{3})e^{(x)}}$

Explanation:

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

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Example Question #612 : Partial Derivatives

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

Possible Answers:

${2z^{2}e^{(x)} - 3\cdot2^{(3z)}ln(4x)e^{(3y)}}$

${-(2^{(3z)}ln(4x)e^{(3y)} - 2yz^{2}e^{(x)})\cdot(2z^{2}e^{(x)} - 3\cdot2^{(3z)}ln(4x)e^{(3y)})}$

${(2^{(3z)}ln(4x)e^{(3y)} - 2yz^{2}e^{(x)})^{2}}$

${4yz^{2}e^{(x)} - 2\cdot2^{(3z)}ln(4x)e^{(3y)}}$

Correct answer:

${2z^{2}e^{(x)} - 3\cdot2^{(3z)}ln(4x)e^{(3y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=y^{2}z^{2}e^{(x)} -\frac{ (2^{(3z)}ln(4x)e^{(3y)})}{3}\\&f_{y}=2yz^{2}e^{(x)} - 2^{(3z)}ln(4x)e^{(3y)}\\&f_{yy}=2z^{2}e^{(x)} - 3\cdot2^{(3z)}ln(4x)e^{(3y)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #601 : Partial Derivatives

Example Question #604 : Partial Derivatives

Example Question #605 : Partial Derivatives

Example Question #606 : Partial Derivatives

Example Question #607 : Partial Derivatives

Example Question #608 : Partial Derivatives

Example Question #609 : Partial Derivatives

Example Question #610 : Partial Derivatives

Example Question #611 : Partial Derivatives

Example Question #612 : Partial Derivatives

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