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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 #733 : Partial Derivatives

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

Possible Answers:

${-\frac{(25tan(z^{3})^{2}cos(4y + y^{3})^{2})}{x^{3}}}$

${\frac{(25tan(z^{3})^{2}cos(4y + y^{3})^{2})}{x^{2}}}$

${-\frac{(10tan(z^{3})cos(4y + y^{3}))}{x}}$

${\frac{(5tan(z^{3})cos(4y + y^{3}))}{x^{2}}}$

Correct answer:

${\frac{(5tan(z^{3})cos(4y + y^{3}))}{x^{2}}}$

Explanation:

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Example Question #734 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #735 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3101 : Calculus 3

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

Possible Answers:

${\frac{(2cos(x^{2} - 3x)sin(4y + y^{4})sin(3z + z^{3})\cdot (3z^{2} + 3)^{2})}{3}- 4zcos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4}) -\frac{ (45z^{4}tan(x^{2})e^{(z^{3})})}{y^{2}}-\frac{ (30ztan(x^{2})e^{(z^{3})})}{y^{2}}}$

${(\frac{(2cos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4})\cdot (3z^{2} + 3))}{3}+\frac{ (15z^{2}tan(x^{2})e^{(z^{3})})}{y^{2}})\cdot (\frac{(30ztan(x^{2})e^{(z^{3})})}{y^{2}}+ 4zcos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4}) +\frac{ (45z^{4}tan(x^{2})e^{(z^{3})})}{y^{2}}-\frac{ (2cos(x^{2} - 3x)sin(4y + y^{4})sin(3z + z^{3})\cdot (3z^{2} + 3)^{2})}{3})}$

${-\frac{ (4cos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4})\cdot (3z^{2} + 3))}{3}-\frac{ (30z^{2}tan(x^{2})e^{(z^{3})})}{y^{2}}}$

${(\frac{(2cos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4})\cdot (3z^{2} + 3))}{3}+\frac{ (15z^{2}tan(x^{2})e^{(z^{3})})}{y^{2}})^{2}}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ (5tan(x^{2})e^{(z^{3})})}{y^{2}}-\frac{ (2cos(x^{2} - 3x)sin(4y + y^{4})sin(3z + z^{3}))}{3}\\&f_{z}=-\frac{ (2cos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4})\cdot (3z^{2} + 3))}{3}-\frac{ (15z^{2}tan(x^{2})e^{(z^{3})})}{y^{2}}\\&f_{zz}=\frac{(2cos(x^{2} - 3x)sin(4y + y^{4})sin(3z + z^{3})\cdot (3z^{2} + 3)^{2})}{3}- 4zcos(x^{2} - 3x)cos(3z + z^{3})sin(4y + y^{4}) -\frac{ (45z^{4}tan(x^{2})e^{(z^{3})})}{y^{2}}-\frac{ (30ztan(x^{2})e^{(z^{3})})}{y^{2}}\end{align*}$

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Example Question #735 : Partial Derivatives

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

Possible Answers:

${-\frac{(12\cdot 3^{(3x)}zsin(z^{2})ln(3)cos(y))}{7}}$

${-\frac{(72\cdot 3^{(6x)}zcos(z^{2})sin(z^{2})ln(3)^{2}cos(y)^{2})}{49}}$

${\frac{(6\cdot 3^{(3x)}cos(z^{2})ln(3)cos(y))}{7}-\frac{ (4\cdot 3^{(3x)}zsin(z^{2})cos(y))}{7}}$

${-\frac{(24\cdot 3^{(6x)}zcos(z^{2})sin(z^{2})ln(3)cos(y)^{2})}{49}}$

Correct answer:

${-\frac{(12\cdot 3^{(3x)}zsin(z^{2})ln(3)cos(y))}{7}}$

Explanation:

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Example Question #3102 : Calculus 3

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

Possible Answers:

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

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

${-(\frac{(2y^{3})}{(xz)}+ 2\cdot 3^{(y^{2})}z^{3}tan(x^{4})\cdot (tan(z^{4})^{2} + 1))\cdot (\frac{(2y^{3})}{(x^{2}z)}- 8\cdot 3^{(y^{2})}x^{3}z^{3}\cdot (tan(x^{4})^{2} + 1)\cdot (tan(z^{4})^{2} + 1))}$

${8\cdot 3^{(y^{2})}x^{3}z^{3}\cdot (tan(x^{4})^{2} + 1)\cdot (tan(z^{4})^{2} + 1) -\frac{ (2y^{3})}{(x^{2}z)}}$

Correct answer:

${8\cdot 3^{(y^{2})}x^{3}z^{3}\cdot (tan(x^{4})^{2} + 1)\cdot (tan(z^{4})^{2} + 1) -\frac{ (2y^{3})}{(x^{2}z)}}$

Explanation:

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

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Example Question #736 : Partial Derivatives

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

Possible Answers:

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

${\frac{(16e^{(8z)}tan(12y + y^{4})^{2})}{x^{2}}}$

${-\frac{(8e^{(4z)}tan(12y + y^{4}))}{x}}$

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

Correct answer:

${-\frac{(8e^{(4z)}tan(12y + y^{4}))}{x}}$

Explanation:

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Example Question #737 : Partial Derivatives

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

Possible Answers:

${16z^{2}tan(y^{3})^{2}e^{(8x)}}$

${-8ztan(y^{3})e^{(4x)}}$

${8z^{3}tan(y^{3})^{2}e^{(8x)}}$

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

Correct answer:

${-8ztan(y^{3})e^{(4x)}}$

Explanation:

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Example Question #3103 : Calculus 3

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

Possible Answers:

${\frac{(6y^{2}cos(y^{3})cos(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})ln(2)sin(x))}{3}}$

${\frac{(2sin(y^{3})cos(z))}{x^{2}}+\frac{ (6y^{2}cos(y^{3})sin(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})sin(x))}{3}+\frac{ (2^ysin(z^{2})ln(2)sin(x))}{3}}$

${(\frac{(6y^{2}cos(y^{3})sin(z))}{x^{2}}+\frac{ (2^ysin(z^{2})ln(2)sin(x))}{3})\cdot (\frac{(2sin(y^{3})cos(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})sin(x))}{3})}$

${(\frac{(2sin(y^{3})cos(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})sin(x))}{3})\cdot (\frac{(6y^{2}cos(y^{3})cos(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})ln(2)sin(x))}{3})}$

Correct answer:

${\frac{(6y^{2}cos(y^{3})cos(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})ln(2)sin(x))}{3}}$

Explanation:

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

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Example Question #3104 : Calculus 3

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

Possible Answers:

${(\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)})\cdot (\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)})}$

${\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)}}$

${\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)}+\frac{ (y^{2}e^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2})ln(y))}{x}}$

${(\frac{(y^{2}e^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2})ln(y))}{x})\cdot (\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)})}$

Correct answer:

${\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(5cos(z^{2})ln(y))}{x}+\frac{ (y^{2}e^{(z)}tan(11x + x^{3}))}{5}\\&f_{y}=\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)}\\&f_{yz}=\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #733 : Partial Derivatives

Example Question #734 : Partial Derivatives

Example Question #735 : Partial Derivatives

Example Question #3101 : Calculus 3

Example Question #735 : Partial Derivatives

Example Question #3102 : Calculus 3

Example Question #736 : Partial Derivatives

Example Question #737 : Partial Derivatives

Example Question #3103 : Calculus 3

Example Question #3104 : Calculus 3

All Calculus 3 Resources

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