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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 #3031 : Calculus 3

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

Possible Answers:

${\frac{(24x^{3}sin(y^{2})^{2}e^{(6z)})}{49}}$

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

${\frac{(72x^{3}sin(y^{2})^{2}e^{(6z)})}{49}}$

${-\frac{(12xsin(y^{2})e^{(3z)})}{7}}$

Correct answer:

${-\frac{(12xsin(y^{2})e^{(3z)})}{7}}$

Explanation:

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Example Question #3031 : Calculus 3

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

Possible Answers:

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

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

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

${\frac{(tan(x^{4})^{2}e^{(8z)}sin(y^{4} - 3y)^{2}\cdot (4y^{3} - 3)^{2})}{16}}$

Correct answer:

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

Explanation:

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

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Example Question #3032 : Calculus 3

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

Possible Answers:

${(\frac{(2e^{(3y)}e^{(x)}sin(z^{3} - 3z))}{5}-\frac{ (8xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{2}})\cdot (\frac{(8tan(x^{2})e^{(y^{3})})}{z^{3}}+\frac{ (2e^{(3y)}e^{(x)}cos(z^{3} - 3z)\cdot (3z^{2} - 3))}{5})}$

${-\frac{ (2e^{(3y)}e^{(x)}cos(z^{3} - 3z)\cdot (3z^{2} - 3))}{5}-\frac{ (16xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{3}}}$

${\frac{(8xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{2}}-\frac{ (8tan(x^{2})e^{(y^{3})})}{z^{3}}-\frac{ (2e^{(3y)}e^{(x)}cos(z^{3} - 3z)\cdot (3z^{2} - 3))}{5}-\frac{ (2e^{(3y)}e^{(x)}sin(z^{3} - 3z))}{5}}$

${(\frac{(2e^{(3y)}e^{(x)}sin(z^{3} - 3z))}{5}-\frac{ (8xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{2}})\cdot (\frac{(2e^{(3y)}e^{(x)}cos(z^{3} - 3z)\cdot (3z^{2} - 3))}{5}+\frac{ (16xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{3}})}$

Correct answer:

${-\frac{ (2e^{(3y)}e^{(x)}cos(z^{3} - 3z)\cdot (3z^{2} - 3))}{5}-\frac{ (16xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{3}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=du(tan^2(u)+1)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4tan(x^{2})e^{(y^{3})})}{z^{2}}-\frac{ (2e^{(3y)}e^{(x)}sin(z^{3} - 3z))}{5}\\&f_{x}=\frac{(8xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{2}}-\frac{ (2e^{(3y)}e^{(x)}sin(z^{3} - 3z))}{5}\\&f_{xz}=-\frac{ (2e^{(3y)}e^{(x)}cos(z^{3} - 3z)\cdot (3z^{2} - 3))}{5}-\frac{ (16xe^{(y^{3})}\cdot (tan(x^{2})^{2} + 1))}{z^{3}}\end{align*}$

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Example Question #666 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #667 : Partial Derivatives

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

Possible Answers:

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

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

${\frac{(yz^{4}cos(y^{2})sin(y^{2})sin(3x + x^{3})^{2}\cdot (3x^{2} + 3)^{2})}{2}}$

${-yz^{2}cos(y^{2})sin(3x + x^{3})\cdot (3x^{2} + 3)}$

Correct answer:

${-yz^{2}cos(y^{2})sin(3x + x^{3})\cdot (3x^{2} + 3)}$

Explanation:

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

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Example Question #3033 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=\frac{(2cos(y^{2})ln(x^{2})ln(4z))}{9}\end{align*}$

Possible Answers:

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

${-\frac{(16ycos(y^{2})ln(x^{2})ln(4z)^{2}sin(y^{2}))}{(81x)}}$

${-\frac{(32ycos(y^{2})ln(4z)^{2}sin(y^{2}))}{(81x^{2})}}$

${-\frac{(8yln(4z)sin(y^{2}))}{(9x)}}$

Correct answer:

${-\frac{(8yln(4z)sin(y^{2}))}{(9x)}}$

Explanation:

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Example Question #3032 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yz}\\&\text{Where }f(x,y,z)=\frac{(2e^{(z^{2})}e^{(y)}cos(x))}{7}- 5e^{(y)}ln(x)\end{align*}$

Possible Answers:

${\frac{(2e^{(z^{2})}e^{(y)}cos(x))}{7}- 5e^{(y)}ln(x) +\frac{ (4ze^{(z^{2})}e^{(y)}cos(x))}{7}}$

${-\frac{(4ze^{(z^{2})}e^{(y)}cos(x)\cdot (5e^{(y)}ln(x) -\frac{ (2e^{(z^{2})}e^{(y)}cos(x))}{7}))}{7}}$

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

${-\frac{(4ze^{(z^{2})}e^{(y)}cos(x)\cdot (5e^{(y)}ln(x) -\frac{ (2e^{(z^{2})}e^{(y)}cos(x))}{7}))}{7}}$

Correct answer:

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

Explanation:

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Example Question #3033 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=\frac{(2x^{2}y^{2}cos(z^{4} - 3z))}{9}-\frac{ (3yln(z))}{x^{2}}\end{align*}$

Possible Answers:

${(\frac{(3ln(z))}{x^{2}}-\frac{ (4x^{2}ycos(z^{4} - 3z))}{9})^{2}}$

${-\frac{(4x^{2}cos(z^{4} - 3z)\cdot (\frac{(3ln(z))}{x^{2}}-\frac{ (4x^{2}ycos(z^{4} - 3z))}{9}))}{9}}$

${\frac{(8x^{2}ycos(z^{4} - 3z))}{9}-\frac{ (6ln(z))}{x^{2}}}$

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

Correct answer:

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

Explanation:

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Example Question #3034 : Calculus 3

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

Possible Answers:

${-7tan(x^{3})sin(y^{2})}$

${49z^{2}tan(x^{3})^{2}sin(y^{2})^{2}}$

${49ztan(x^{3})^{2}sin(y^{2})^{2}}$

${-14ztan(x^{3})sin(y^{2})}$

Correct answer:

${-7tan(x^{3})sin(y^{2})}$

Explanation:

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Example Question #3035 : Calculus 3

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

Possible Answers:

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

${(4ytan(x) +\frac{ (tan(3z)e^{(4x)})}{(2y^{3})})^{2}}$

${(4tan(x) -\frac{ (3tan(3z)e^{(4x)})}{(2y^{4})})\cdot (4ytan(x) +\frac{ (tan(3z)e^{(4x)})}{(2y^{3})})}$

${8ytan(x) +\frac{ (tan(3z)e^{(4x)})}{y^{3}}}$

Correct answer:

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

Explanation:

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3031 : Calculus 3

Example Question #3031 : Calculus 3

Example Question #3032 : Calculus 3

Example Question #666 : Partial Derivatives

Example Question #667 : Partial Derivatives

Example Question #3033 : Calculus 3

Example Question #3032 : Calculus 3

Example Question #3033 : Calculus 3

Example Question #3034 : Calculus 3

Example Question #3035 : Calculus 3

All Calculus 3 Resources

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