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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 #702 : Partial Derivatives

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

Possible Answers:

${\frac{(4cos(z^{2})sin(3y + y^{3}))}{(5x^{3})}-\frac{ (10e^{(z^{4})})}{(x^{3}y)}}$

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

${\frac{(10e^{(z^{4})})}{(x^{2}y)}-\frac{ (4cos(z^{2})sin(3y + y^{3}))}{(5x^{2})}}$

${-(\frac{(5e^{(z^{4})})}{(x^{2}y)}-\frac{ (2cos(z^{2})sin(3y + y^{3}))}{(5x^{2})})\cdot (\frac{(10e^{(z^{4})})}{(x^{3}y)}-\frac{ (4cos(z^{2})sin(3y + y^{3}))}{(5x^{3})})}$

Correct answer:

${\frac{(4cos(z^{2})sin(3y + y^{3}))}{(5x^{3})}-\frac{ (10e^{(z^{4})})}{(x^{3}y)}}$

Explanation:

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

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Example Question #704 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #705 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #706 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #707 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #703 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #704 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #705 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #711 : Partial Derivatives

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

Possible Answers:

${\frac{(cos(y^{3} - 3y)^{2}tan(11x + x^{3})^{2}\cdot (tan(11z + z^{3})^{2} + 1)^{2}\cdot (3z^{2} + 11)^{2})}{25}}$

${\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5}}$

${\frac{(cos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5})\cdot (3z^{2} + 11))}{5}}$

${\frac{(2cos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11))}{5}}$

Correct answer:

${\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5}}$

Explanation:

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

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Example Question #712 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #702 : Partial Derivatives

Example Question #704 : Partial Derivatives

Example Question #705 : Partial Derivatives

Example Question #706 : Partial Derivatives

Example Question #707 : Partial Derivatives

Example Question #703 : Partial Derivatives

Example Question #704 : Partial Derivatives

Example Question #705 : Partial Derivatives

Example Question #711 : Partial Derivatives

Example Question #712 : Partial Derivatives

All Calculus 3 Resources

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