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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #471 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yz}\\&\text{Where }f(x,y,z)=-10y^2z^2ln{(x^4)}\end{align*}$

Possible Answers:

$400y^3z^3ln{(x^4)}^2$

${- 40yzln{(x^4)} - 20yz^2ln{(x^4)}}$

${-40yzln{(x^4)}}$

${- 20yz^2ln{(x^4)} - 20y^2zln{(x^4)}}$

Correct answer:

${-40yzln{(x^4)}}$

Explanation:

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Example Question #472 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{-{(6cos{(x^3)}ln{(z^4)})}}{y^3}\end{align*}$

Possible Answers:

$\frac{{(324x^2cos{(x^3)}ln{(z^4)}^2sin{(x^3)})}}{y^7}$

${\frac{{(18cos{(x^3)}ln{(z^4)})}}{y^4} -\frac{ {(54x^2ln{(z^4)}sin{(x^3)})}}{y^4}}$

${\frac{{(18cos{(x^3)}ln{(z^4)})}}{y^4} +\frac{ {(18x^2ln{(z^4)}sin{(x^3)})}}{y^3}}$

${\frac{-{(54x^2ln{(z^4)}sin{(x^3)})}}{y^4}}$

Correct answer:

${\frac{-{(54x^2ln{(z^4)}sin{(x^3)})}}{y^4}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&d[cos(u)]=-sin(u)du\\&f_{y}=\frac{{(18cos{(x^3)}ln{(z^4)})}}{y^4}\\&f_{yx}=\frac{-{(54x^2ln{(z^4)}sin{(x^3)})}}{y^4}\end{align*}$

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Example Question #472 : Partial Derivatives

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

Possible Answers:

${32x^3z^3cos{(y^2)}e^{(x^4)} - 64x^3y^2z^3sin{(y^2)}e^{(x^4)}}$

$64x^3y^2z^15cos{(y^2)}^2sin{(y^2)}^2e^{(4x^4)}$

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

${32x^3z^3cos{(y^2)}e^{(x^4)} + 4x^3z^4sin{(y^2)}e^{(x^4)} + 32x^3yz^3cos{(y^2)}e^{(x^4)} + 8x^3yz^4cos{(y^2)}e^{(x^4)} - 64x^3y^2z^3sin{(y^2)}e^{(x^4)}}$

Correct answer:

${32x^3z^3cos{(y^2)}e^{(x^4)} - 64x^3y^2z^3sin{(y^2)}e^{(x^4)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&f_{x}=4x^3z^4sin{(y^2)}e^{(x^4)}\\&f_{xy}=8x^3yz^4cos{(y^2)}e^{(x^4)}\\&f_{xyz}=32x^3yz^3cos{(y^2)}e^{(x^4)}\\&f_{xyzy}=32x^3z^3cos{(y^2)}e^{(x^4)} - 64x^3y^2z^3sin{(y^2)}e^{(x^4)}\end{align*}$

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Example Question #474 : Partial Derivatives

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

Possible Answers:

${32y^3e^{(x^4)}e^{(y^4)}cos{(z)} - 8e^{(x^4)}e^{(y^4)}sin{(z)}}$

$4096y^6e^{(4x^4)}e^{(4y^4)}cos{(z)}^2sin{(z)}^2$

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

${- 48y^2e^{(x^4)}e^{(y^4)}cos{(z)} - 64y^6e^{(x^4)}e^{(y^4)}cos{(z)}}$

Correct answer:

${- 48y^2e^{(x^4)}e^{(y^4)}cos{(z)} - 64y^6e^{(x^4)}e^{(y^4)}cos{(z)}}$

Explanation:

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

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Example Question #472 : Partial Derivatives

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

Possible Answers:

${3cos{(y^3)}ln{(x^4)}sin{(z)} -\frac{ {(12cos{(y^3)}cos{(z)})}}{x} + 18y^2ln{(x^4)}sin{(y^3)}cos{(z)}}$

${18yln{(x^4)}sin{(y^3)}cos{(z)} -\frac{ {(108y^4cos{(y^3)}sin{(z)})}}{x} - 18yln{(x^4)}sin{(y^3)}sin{(z)} + 27y^4cos{(y^3)}ln{(x^4)}cos{(z)} + 9y^2ln{(x^4)}sin{(y^3)}cos{(z)} - 27y^4cos{(y^3)}ln{(x^4)}sin{(z)} -\frac{ {(72ysin{(y^3)}sin{(z)})}}{x}}$

$\frac{-{(2916y^4cos{(y^3)}^2ln{(x^4)}^3sin{(y^3)}^2cos{(z)}^3sin{(z)})}}{x}$

${-\frac{ {(108y^4cos{(y^3)}sin{(z)})}}{x} -\frac{ {(72ysin{(y^3)}sin{(z)})}}{x}}$

Correct answer:

${-\frac{ {(108y^4cos{(y^3)}sin{(z)})}}{x} -\frac{ {(72ysin{(y^3)}sin{(z)})}}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f_{y}=9y^2ln{(x^4)}sin{(y^3)}cos{(z)}\\&f_{yy}=18yln{(x^4)}sin{(y^3)}cos{(z)} + 27y^4cos{(y^3)}ln{(x^4)}cos{(z)}\\&f_{yyz}=- 18yln{(x^4)}sin{(y^3)}sin{(z)} - 27y^4cos{(y^3)}ln{(x^4)}sin{(z)}\\&f_{yyzx}=-\frac{ {(108y^4cos{(y^3)}sin{(z)})}}{x} -\frac{ {(72ysin{(y^3)}sin{(z)})}}{x}\end{align*}$

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Example Question #472 : Partial Derivatives

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

Possible Answers:

$4\cdot3^{(2y)}z^3cos{(x^3)}^2cos{(z^4)}sin{(z^4)}ln{(3)}$

${4\cdot3^yz^3cos{(x^3)}cos{(z^4)} + 4\cdot3^yz^3cos{(x^3)}cos{(z^4)}ln{(3)}}$

${3^ycos{(x^3)}sin{(z^4)}ln{(3)} + 4\cdot3^yz^3cos{(x^3)}cos{(z^4)}}$

${4\cdot3^yz^3cos{(x^3)}cos{(z^4)}ln{(3)}}$

Correct answer:

${4\cdot3^yz^3cos{(x^3)}cos{(z^4)}ln{(3)}}$

Explanation:

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Example Question #477 : Partial Derivatives

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

Possible Answers:

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

${\frac{-{(8zsin{(y^3)}e^{(z^2)})}}{x^3}}$

$\frac{-{(16zsin{(y^3)}^2e^{(2z^2)})}}{x^5}$

${\frac{{(4zsin{(y^3)}e^{(z^2)})}}{x^2} -\frac{ {(8zsin{(y^3)}e^{(z^2)})}}{x^3}}$

Correct answer:

${\frac{-{(8zsin{(y^3)}e^{(z^2)})}}{x^3}}$

Explanation:

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Example Question #478 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xzy}\\&\text{Where }f(x,y,z)=6\cdot4^{(y^4)}z^2sin{(x)}\end{align*}$

Possible Answers:

$1728\cdot4^{(3y^4)}y^3z^5ln{(4)}cos{(x)}sin{(x)}^2$

${6\cdot4^{(y^4)}z^2cos{(x)} + 12\cdot4^{(y^4)}zcos{(x)} + 48\cdot4^{(y^4)}y^3zln{(4)}cos{(x)}}$

${6\cdot4^{(y^4)}z^2cos{(x)} + 12\cdot4^{(y^4)}zsin{(x)} + 24\cdot4^{(y^4)}y^3z^2ln{(4)}sin{(x)}}$

${48\cdot4^{(y^4)}y^3zln{(4)}cos{(x)}}$

Correct answer:

${48\cdot4^{(y^4)}y^3zln{(4)}cos{(x)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f_{x}=6\cdot4^{(y^4)}z^2cos{(x)}\\&f_{xz}=12\cdot4^{(y^4)}zcos{(x)}\\&f_{xzy}=48\cdot4^{(y^4)}y^3zln{(4)}cos{(x)}\end{align*}$

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Example Question #473 : Partial Derivatives

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

Possible Answers:

${- 8zsin{(z^2)}e^{(x^3)}e^{(y^3)} - 24y^2sin{(z^2)}e^{(x^3)}e^{(y^3)} - 24y^2zsin{(z^2)}e^{(x^3)}e^{(y^3)} - 48y^2z^2cos{(z^2)}e^{(x^3)}e^{(y^3)}}$

${12y^2cos{(z^2)}e^{(x^3)}e^{(y^3)} - 16zsin{(z^2)}e^{(x^3)}e^{(y^3)}}$

$768y^2z^2cos{(z^2)}sin{(z^2)}^2e^{(3x^3)}e^{(3y^3)}$

${- 24y^2sin{(z^2)}e^{(x^3)}e^{(y^3)} - 48y^2z^2cos{(z^2)}e^{(x^3)}e^{(y^3)}}$

Correct answer:

${- 24y^2sin{(z^2)}e^{(x^3)}e^{(y^3)} - 48y^2z^2cos{(z^2)}e^{(x^3)}e^{(y^3)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f_{z}=-8zsin{(z^2)}e^{(x^3)}e^{(y^3)}\\&f_{zy}=-24y^2zsin{(z^2)}e^{(x^3)}e^{(y^3)}\\&f_{zyz}=- 24y^2sin{(z^2)}e^{(x^3)}e^{(y^3)} - 48y^2z^2cos{(z^2)}e^{(x^3)}e^{(y^3)}\end{align*}$

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Example Question #473 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zyz}\\&\text{Where }f(x,y,z)=\frac{-{(6ln{(y^4)})}}{{(x^2z)}}\end{align*}$

Possible Answers:

${\frac{{(6ln{(y^4)})}}{{(x^2z^2)}} +\frac{ 24}{{(x^2yz^2)}} -\frac{ 48}{{(x^2yz^3)}}}$

$\frac{-{(864ln{(y^4)}^2)}}{{(x^6yz^5)}}$

${\frac{-48}{{(x^2yz^3)}}}$

${\frac{{(12ln{(y^4)})}}{{(x^2z^2)}} -\frac{ 24}{{(x^2yz)}}}$

Correct answer:

${\frac{-48}{{(x^2yz^3)}}}$

Explanation:

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #471 : Partial Derivatives

Example Question #472 : Partial Derivatives

Example Question #472 : Partial Derivatives

Example Question #474 : Partial Derivatives

Example Question #472 : Partial Derivatives

Example Question #472 : Partial Derivatives

Example Question #477 : Partial Derivatives

Example Question #478 : Partial Derivatives

Example Question #473 : Partial Derivatives

Example Question #473 : Partial Derivatives

All Calculus 3 Resources

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