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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 #2981 : Calculus 3

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

Possible Answers:

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

${\frac{(4e^{(3y)})}{(9x^{3}z)}}$

${\frac{(16e^{(6y)}ln(z))}{(81x^{6}z)}}$

${\frac{(4e^{(3y)}ln(z)\cdot(2zcos(y) -\frac{ (2e^{(3y)})}{(9x^{2}z)}))}{(9x^{3})}}$

Correct answer:

${\frac{(4e^{(3y)})}{(9x^{3}z)}}$

Explanation:

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Example Question #2982 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #2983 : Calculus 3

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

Possible Answers:

${-\frac{(cos(3y)ln(4z)sin(x)\cdot(\frac{(6sin(y))}{z}+\frac{ (cos(3y)cos(x))}{(4z)}))}{4}}$

${\frac{(6sin(y))}{z}+\frac{ (cos(3y)cos(x))}{(4z)}-\frac{ (cos(3y)ln(4z)sin(x))}{4}}$

${-\frac{(cos(3y)sin(x)\cdot(\frac{(6sin(y))}{z}+\frac{ (cos(3y)cos(x))}{(4z)}))}{(4z)}}$

${-\frac{(cos(3y)sin(x))}{(4z)}}$

Correct answer:

${-\frac{(cos(3y)sin(x))}{(4z)}}$

Explanation:

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

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Example Question #2984 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=4\cdot3^{(y^{2})}z^{2}ln(x) + 5ln(z^{2})sin(x^{3})e^{(y)}\end{align*}$

Possible Answers:

${\frac{(10sin(x^{3})e^{(y)})}{z}+ 5ln(z^{2})sin(x^{3})e^{(y)} + 8\cdot3^{(y^{2})}zln(x) + 8\cdot3^{(y^{2})}yz^{2}ln(3)ln(x)}$

${\frac{(10sin(x^{3})e^{(y)})}{z}+ 16\cdot3^{(y^{2})}yzln(3)ln(x)}$

${(\frac{(10sin(x^{3})e^{(y)})}{z}+ 16\cdot3^{(y^{2})}yzln(3)ln(x))\cdot(\frac{(10sin(x^{3})e^{(y)})}{z}+ 8\cdot3^{(y^{2})}zln(x))}$

${(5ln(z^{2})sin(x^{3})e^{(y)} + 8\cdot3^{(y^{2})}yz^{2}ln(3)ln(x))\cdot(\frac{(10sin(x^{3})e^{(y)})}{z}+ 8\cdot3^{(y^{2})}zln(x))}$

Correct answer:

${\frac{(10sin(x^{3})e^{(y)})}{z}+ 16\cdot3^{(y^{2})}yzln(3)ln(x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4\cdot3^{(y^{2})}z^{2}ln(x) + 5ln(z^{2})sin(x^{3})e^{(y)}\\&f_{z}=\frac{(10sin(x^{3})e^{(y)})}{z}+ 8\cdot3^{(y^{2})}zln(x)\\&f_{zy}=\frac{(10sin(x^{3})e^{(y)})}{z}+ 16\cdot3^{(y^{2})}yzln(3)ln(x)\end{align*}$

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Example Question #2985 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=5x^{2}sin(z^{2})ln(y) -\frac{ (3\cdot4^{(4x)}cos(y^{2})ln(4z))}{2}\end{align*}$

Possible Answers:

${10x^{2}cos(z^{2})ln(y) +\frac{ (3\cdot4^{(4x)}cos(y^{2}))}{(2z^{2})}- 20x^{2}z^{2}sin(z^{2})ln(y)}$

${-(\frac{(3\cdot4^{(4x)}cos(y^{2}))}{(2z)}- 10x^{2}zcos(z^{2})ln(y))\cdot(10x^{2}cos(z^{2})ln(y) +\frac{ (3\cdot4^{(4x)}cos(y^{2}))}{(2z^{2})}- 20x^{2}z^{2}sin(z^{2})ln(y))}$

${20x^{2}zcos(z^{2})ln(y) -\frac{ (3\cdot4^{(4x)}cos(y^{2}))}{z}}$

${(\frac{(3\cdot4^{(4x)}cos(y^{2}))}{(2z)}- 10x^{2}zcos(z^{2})ln(y))^{2}}$

Correct answer:

${10x^{2}cos(z^{2})ln(y) +\frac{ (3\cdot4^{(4x)}cos(y^{2}))}{(2z^{2})}- 20x^{2}z^{2}sin(z^{2})ln(y)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=5x^{2}sin(z^{2})ln(y) -\frac{ (3\cdot4^{(4x)}cos(y^{2})ln(4z))}{2}\\&f_{z}=10x^{2}zcos(z^{2})ln(y) -\frac{ (3\cdot4^{(4x)}cos(y^{2}))}{(2z)}\\&f_{zz}=10x^{2}cos(z^{2})ln(y) +\frac{ (3\cdot4^{(4x)}cos(y^{2}))}{(2z^{2})}- 20x^{2}z^{2}sin(z^{2})ln(y)\end{align*}$

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Example Question #612 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=4cos(y^{2})sin(z^{3})e^{(x)} -\frac{ (2cos(3y + y^{4} - 1)sin(z))}{x}\end{align*}$

Possible Answers:

${\frac{(24y^{2}sin(3y + y^{4} - 1)sin(z))}{x}- 8sin(y^{2})sin(z^{3})e^{(x)} +\frac{ (2cos(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3)^{2})}{x}- 16y^{2}cos(y^{2})sin(z^{3})e^{(x)}}$

${-(\frac{(2sin(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3))}{x}- 8ysin(y^{2})sin(z^{3})e^{(x)})\cdot(8sin(y^{2})sin(z^{3})e^{(x)} -\frac{ (24y^{2}sin(3y + y^{4} - 1)sin(z))}{x}-\frac{ (2cos(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3)^{2})}{x}+ 16y^{2}cos(y^{2})sin(z^{3})e^{(x)})}$

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

${\frac{(4sin(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3))}{x}- 16ysin(y^{2})sin(z^{3})e^{(x)}}$

Correct answer:

${\frac{(24y^{2}sin(3y + y^{4} - 1)sin(z))}{x}- 8sin(y^{2})sin(z^{3})e^{(x)} +\frac{ (2cos(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3)^{2})}{x}- 16y^{2}cos(y^{2})sin(z^{3})e^{(x)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=4cos(y^{2})sin(z^{3})e^{(x)} -\frac{ (2cos(3y + y^{4} - 1)sin(z))}{x}\\&f_{y}=\frac{(2sin(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3))}{x}- 8ysin(y^{2})sin(z^{3})e^{(x)}\\&f_{yy}=\frac{(24y^{2}sin(3y + y^{4} - 1)sin(z))}{x}- 8sin(y^{2})sin(z^{3})e^{(x)} +\frac{ (2cos(3y + y^{4} - 1)sin(z)\cdot(4y^{3} + 3)^{2})}{x}- 16y^{2}cos(y^{2})sin(z^{3})e^{(x)}\end{align*}$

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Example Question #2987 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{(3^{(x^{2})}\cdot3^ye^{(z)})}{3}+ 3xln(z^{2})sin(y^{3})\end{align*}$

Possible Answers:

${(9xy^{2}cos(y^{3})ln(z^{2}) +\frac{ (3^{(x^{2})}\cdot3^ye^{(z)}ln(3))}{3})\cdot(3ln(z^{2})sin(y^{3}) +\frac{ (2\cdot3^{(x^{2})}\cdot3^yxe^{(z)}ln(3))}{3})}$

${9y^{2}cos(y^{3})ln(z^{2}) +\frac{ (2\cdot3^{(x^{2})}\cdot3^yxe^{(z)}ln(3)^{2})}{3}}$

${(9xy^{2}cos(y^{3})ln(z^{2}) +\frac{ (3^{(x^{2})}\cdot3^ye^{(z)}ln(3))}{3})\cdot(9y^{2}cos(y^{3})ln(z^{2}) +\frac{ (2\cdot3^{(x^{2})}\cdot3^yxe^{(z)}ln(3)^{2})}{3})}$

${3ln(z^{2})sin(y^{3}) + 9xy^{2}cos(y^{3})ln(z^{2}) +\frac{ (3^{(x^{2})}\cdot3^ye^{(z)}ln(3))}{3}+\frac{ (2\cdot3^{(x^{2})}\cdot3^yxe^{(z)}ln(3))}{3}}$

Correct answer:

${9y^{2}cos(y^{3})ln(z^{2}) +\frac{ (2\cdot3^{(x^{2})}\cdot3^yxe^{(z)}ln(3)^{2})}{3}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3^{(x^{2})}\cdot3^ye^{(z)})}{3}+ 3xln(z^{2})sin(y^{3})\\&f_{y}=9xy^{2}cos(y^{3})ln(z^{2}) +\frac{ (3^{(x^{2})}\cdot3^ye^{(z)}ln(3))}{3}\\&f_{yx}=9y^{2}cos(y^{3})ln(z^{2}) +\frac{ (2\cdot3^{(x^{2})}\cdot3^yxe^{(z)}ln(3)^{2})}{3}\end{align*}$

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Example Question #2988 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xx}\\&\text{Where }f(x,y,z)=4x^{3}y^{3}z +\frac{ (2\cdot4^{(3z)}ln(x^{2})ln(y))}{3}\end{align*}$

Possible Answers:

${\frac{(8\cdot4^{(3z)}ln(y))}{(3x)}+ 24x^{2}y^{3}z}$

${24xy^{3}z -\frac{ (4\cdot4^{(3z)}ln(y))}{(3x^{2})}}$

${(\frac{(4\cdot4^{(3z)}ln(y))}{(3x)}+ 12x^{2}y^{3}z)\cdot(24xy^{3}z -\frac{ (4\cdot4^{(3z)}ln(y))}{(3x^{2})})}$

${(\frac{(4\cdot4^{(3z)}ln(y))}{(3x)}+ 12x^{2}y^{3}z)^{2}}$

Correct answer:

${24xy^{3}z -\frac{ (4\cdot4^{(3z)}ln(y))}{(3x^{2})}}$

Explanation:

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

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Example Question #2989 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=2ln(z^{2})e^{(y^{2})}ln(x) -\frac{ (4^xcos(4y + y^{2} - 1)cos(3z + z^{2} + 1))}{4}\end{align*}$

Possible Answers:

${(4yln(z^{2})e^{(y^{2})}ln(x) +\frac{ (4^xcos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4})\cdot(\frac{(2ln(z^{2})e^{(y^{2})})}{x}-\frac{ (4^xln(4)cos(4y + y^{2} - 1)cos(3z + z^{2} + 1))}{4})}$

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

${\frac{(2ln(z^{2})e^{(y^{2})})}{x}+ 4yln(z^{2})e^{(y^{2})}ln(x) +\frac{ (4^xcos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4}-\frac{ (4^xln(4)cos(4y + y^{2} - 1)cos(3z + z^{2} + 1))}{4}}$

${(4yln(z^{2})e^{(y^{2})}ln(x) +\frac{ (4^xcos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4})\cdot(\frac{(4yln(z^{2})e^{(y^{2})})}{x}+\frac{ (4^xln(4)cos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4})}$

Correct answer:

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

Explanation:

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

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Example Question #2990 : Calculus 3

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

Possible Answers:

${\frac{(30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}}-\frac{ (ln(z)sin(x))}{(4y)}}$

${\frac{(cos(x)ln(z))}{(4y)}-\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}}-\frac{ (ln(4y)ln(z)sin(x))}{4}-\frac{ (15y^{2}cos(z^{3})sin(y^{3}))}{x^{2}}}$

${(\frac{(ln(z)sin(x))}{(4y)}-\frac{ (30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}})\cdot(\frac{(ln(4y)ln(z)sin(x))}{4}+\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}})}$

${-(\frac{(ln(4y)ln(z)sin(x))}{4}+\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}})\cdot(\frac{(cos(x)ln(z))}{(4y)}-\frac{ (15y^{2}cos(z^{3})sin(y^{3}))}{x^{2}})}$

Correct answer:

${\frac{(30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}}-\frac{ (ln(z)sin(x))}{(4y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(ln(4y)cos(x)ln(z))}{4}+\frac{ (5cos(y^{3})cos(z^{3}))}{x^{2}}\\&f_{x}=-\frac{ (ln(4y)ln(z)sin(x))}{4}-\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}}\\&f_{xy}=\frac{(30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}}-\frac{ (ln(z)sin(x))}{(4y)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #2981 : Calculus 3

Example Question #2982 : Calculus 3

Example Question #2983 : Calculus 3

Example Question #2984 : Calculus 3

Example Question #2985 : Calculus 3

Example Question #612 : Partial Derivatives

Example Question #2987 : Calculus 3

Example Question #2988 : Calculus 3

Example Question #2989 : Calculus 3

Example Question #2990 : Calculus 3

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