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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #481 : Partial Derivatives

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

Possible Answers:

${\frac{{(5cos{(y^3)}e^{(z^3)})}}{x} + 45z^2cos{(y^3)}e^{(z^3)}ln{(x)}}$

${\frac{{(30cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(30zcos{(y^3)}e^{(z^3)})}}{x} + 30zcos{(y^3)}e^{(z^3)}ln{(x)} +\frac{ {(270z^3cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(45z^4cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(135z^6cos{(y^3)}e^{(z^3)})}}{x} + 15z^2cos{(y^3)}e^{(z^3)}ln{(x)} + 45z^4cos{(y^3)}e^{(z^3)}ln{(x)}}$

${\frac{{(30cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(270z^3cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(135z^6cos{(y^3)}e^{(z^3)})}}{x}}$

$\frac{{(16875z^6cos{(y^3)}^4e^{(4z^3)}ln{(x)}^3)}}{x}$

Correct answer:

${\frac{{(30cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(270z^3cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(135z^6cos{(y^3)}e^{(z^3)})}}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f_{z}=15z^2cos{(y^3)}e^{(z^3)}ln{(x)}\\&f_{zz}=30zcos{(y^3)}e^{(z^3)}ln{(x)} + 45z^4cos{(y^3)}e^{(z^3)}ln{(x)}\\&f_{zzx}=\frac{{(30zcos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(45z^4cos{(y^3)}e^{(z^3)})}}{x}\\&f_{zzxz}=\frac{{(30cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(270z^3cos{(y^3)}e^{(z^3)})}}{x} +\frac{ {(135z^6cos{(y^3)}e^{(z^3)})}}{x}\end{align*}$

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Example Question #482 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zyzz}\\&\text{Where }f(x,y,z)=-10cos{(x^4)}e^{(z^3)}cos{(y)}\end{align*}$

Possible Answers:

$-270000z^6cos{(x^4)}^4e^{(4z^3)}cos{(y)}^3sin{(y)}$

${60cos{(x^4)}e^{(z^3)}sin{(y)} + 60zcos{(x^4)}e^{(z^3)}sin{(y)} - 30z^2cos{(x^4)}e^{(z^3)}cos{(y)} + 30z^2cos{(x^4)}e^{(z^3)}sin{(y)} + 540z^3cos{(x^4)}e^{(z^3)}sin{(y)} + 90z^4cos{(x^4)}e^{(z^3)}sin{(y)} + 270z^6cos{(x^4)}e^{(z^3)}sin{(y)}}$

${10cos{(x^4)}e^{(z^3)}sin{(y)} - 90z^2cos{(x^4)}e^{(z^3)}cos{(y)}}$

${60cos{(x^4)}e^{(z^3)}sin{(y)} + 540z^3cos{(x^4)}e^{(z^3)}sin{(y)} + 270z^6cos{(x^4)}e^{(z^3)}sin{(y)}}$

Correct answer:

${60cos{(x^4)}e^{(z^3)}sin{(y)} + 540z^3cos{(x^4)}e^{(z^3)}sin{(y)} + 270z^6cos{(x^4)}e^{(z^3)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{z}=-30z^2cos{(x^4)}e^{(z^3)}cos{(y)}\\&f_{zy}=30z^2cos{(x^4)}e^{(z^3)}sin{(y)}\\&f_{zyz}=60zcos{(x^4)}e^{(z^3)}sin{(y)} + 90z^4cos{(x^4)}e^{(z^3)}sin{(y)}\\&f_{zyzz}=60cos{(x^4)}e^{(z^3)}sin{(y)} + 540z^3cos{(x^4)}e^{(z^3)}sin{(y)} + 270z^6cos{(x^4)}e^{(z^3)}sin{(y)}\end{align*}$

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Example Question #483 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=-7\cdot3^{(x^4)}cos{(y^4)}sin{(z^3)}\end{align*}$

Possible Answers:

${56\cdot3^{(x^4)}y^3sin{(y^4)}sin{(z^3)}}$

$784\cdot3^{(2x^4)}y^6sin{(y^4)}^2sin{(z^3)}^2$

${84\cdot3^{(x^4)}y^2sin{(y^4)}sin{(z^3)} + 112\cdot3^{(x^4)}y^6cos{(y^4)}sin{(z^3)}}$

${84\cdot3^{(x^4)}y^2sin{(y^4)}sin{(z^3)} + 28\cdot3^{(x^4)}y^3sin{(y^4)}sin{(z^3)} + 112\cdot3^{(x^4)}y^6cos{(y^4)}sin{(z^3)}}$

Correct answer:

${84\cdot3^{(x^4)}y^2sin{(y^4)}sin{(z^3)} + 112\cdot3^{(x^4)}y^6cos{(y^4)}sin{(z^3)}}$

Explanation:

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Example Question #484 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=-9\cdot3^zy^4cos{(x^2)}\end{align*}$

Possible Answers:

${-72\cdot3^zy^3cos{(x^2)}}$

${-108\cdot3^zy^2cos{(x^2)}}$

$1296\cdot3^{(2z)}y^6cos{(x^2)}^2$

${- 108\cdot3^zy^2cos{(x^2)} - 36\cdot3^zy^3cos{(x^2)}}$

Correct answer:

${-108\cdot3^zy^2cos{(x^2)}}$

Explanation:

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Example Question #485 : Partial Derivatives

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

Possible Answers:

${\frac{{(48\cdot4^{(y^3)}y^2z^2sin{(z^2)}ln{(4)})}}{x} - 8\cdot4^{(y^3)}zcos{(z^2)}ln{(x)} -\frac{ {(24\cdot4^{(y^3)}y^2cos{(z^2)}ln{(4)})}}{x} -\frac{ {(24\cdot4^{(y^3)}y^2zcos{(z^2)}ln{(4)})}}{x} -\frac{ {(8\cdot4^{(y^3)}zcos{(z^2)})}}{x}}$

${-\frac{ {(4\cdot4^{(y^3)}sin{(z^2)})}}{x} - 16\cdot4^{(y^3)}zcos{(z^2)}ln{(x)} - 12\cdot4^{(y^3)}y^2sin{(z^2)}ln{(4)}ln{(x)}}$

${\frac{{(48\cdot4^{(y^3)}y^2z^2sin{(z^2)}ln{(4)})}}{x} -\frac{ {(24\cdot4^{(y^3)}y^2cos{(z^2)}ln{(4)})}}{x}}$

$\frac{{(3072\cdot4^{(4y^3)}y^2z^2cos{(z^2)}^2sin{(z^2)}^2ln{(4)}ln{(x)}^3)}}{x}$

Correct answer:

${\frac{{(48\cdot4^{(y^3)}y^2z^2sin{(z^2)}ln{(4)})}}{x} -\frac{ {(24\cdot4^{(y^3)}y^2cos{(z^2)}ln{(4)})}}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{z}=-8\cdot4^{(y^3)}zcos{(z^2)}ln{(x)}\\&f_{zx}=\frac{-{(8\cdot4^{(y^3)}zcos{(z^2)})}}{x}\\&f_{zxy}=\frac{-{(24\cdot4^{(y^3)}y^2zcos{(z^2)}ln{(4)})}}{x}\\&f_{zxyz}=\frac{{(48\cdot4^{(y^3)}y^2z^2sin{(z^2)}ln{(4)})}}{x} -\frac{ {(24\cdot4^{(y^3)}y^2cos{(z^2)}ln{(4)})}}{x}\end{align*}$

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Example Question #486 : Partial Derivatives

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

Possible Answers:

$4\cdot4^{(2y)}cos{(x)}cos{(z)}sin{(x)}sin{(z)}$

${- 2\cdot4^ycos{(x)}sin{(z)} - 2\cdot4^ycos{(z)}sin{(x)}}$

${- 2\cdot4^ycos{(x)}sin{(z)} - 2\cdot4^ycos{(x)}cos{(z)}}$

${-2\cdot4^ycos{(x)}cos{(z)}}$

Correct answer:

${-2\cdot4^ycos{(x)}cos{(z)}}$

Explanation:

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Example Question #487 : Partial Derivatives

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

Possible Answers:

$-{4cos{(x^4)}e^{(z^4)}e^{(y)}}$

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

${4cos{(x^4)}e^{(z^4)}e^{(y)}}$

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

Correct answer:

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

Explanation:

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Example Question #488 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xyxz}\\&\text{Where }f(x,y,z)=9\cdot4^{(z^3)}cos{(x^3)}ln{(y)}\end{align*}$

Possible Answers:

${\frac{{(9\cdot4^{(z^3)}cos{(x^3)})}}{y} - 54\cdot4^{(z^3)}x^2sin{(x^3)}ln{(y)} + 27\cdot4^{(z^3)}z^2cos{(x^3)}ln{(4)}ln{(y)}}$

${-\frac{ {(54\cdot4^{(z^3)}xsin{(x^3)})}}{y} -\frac{ {(81\cdot4^{(z^3)}x^4cos{(x^3)})}}{y} -\frac{ {(27\cdot4^{(z^3)}x^2sin{(x^3)})}}{y} - 27\cdot4^{(z^3)}x^2sin{(x^3)}ln{(y)} -\frac{ {(162\cdot4^{(z^3)}xz^2sin{(x^3)}ln{(4)})}}{y} -\frac{ {(243\cdot4^{(z^3)}x^4z^2cos{(x^3)}ln{(4)})}}{y}}$

$\frac{{(177147\cdot4^{(4z^3)}x^4z^2cos{(x^3)}^2sin{(x^3)}^2ln{(4)}ln{(y)}^3)}}{y}$

${-\frac{ {(162\cdot4^{(z^3)}xz^2sin{(x^3)}ln{(4)})}}{y} -\frac{ {(243\cdot4^{(z^3)}x^4z^2cos{(x^3)}ln{(4)})}}{y}}$

Correct answer:

${-\frac{ {(162\cdot4^{(z^3)}xz^2sin{(x^3)}ln{(4)})}}{y} -\frac{ {(243\cdot4^{(z^3)}x^4z^2cos{(x^3)}ln{(4)})}}{y}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f_{x}=-27\cdot4^{(z^3)}x^2sin{(x^3)}ln{(y)}\\&f_{xy}=\frac{-{(27\cdot4^{(z^3)}x^2sin{(x^3)})}}{y}\\&f_{xyx}=-\frac{ {(54\cdot4^{(z^3)}xsin{(x^3)})}}{y} -\frac{ {(81\cdot4^{(z^3)}x^4cos{(x^3)})}}{y}\\&f_{xyxz}=-\frac{ {(162\cdot4^{(z^3)}xz^2sin{(x^3)}ln{(4)})}}{y} -\frac{ {(243\cdot4^{(z^3)}x^4z^2cos{(x^3)}ln{(4)})}}{y}\end{align*}$

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Example Question #489 : Partial Derivatives

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

Possible Answers:

${\frac{{(2x^2ln{(z^3)})}}{y^2} -\frac{ {(18x^2)}}{{(yz)}}}$

${\frac{{(12x^2)}}{{(y^2z^3)}}}$

${\frac{{(6x^2)}}{{(yz^2)}} -\frac{ {(6x^2)}}{{(yz)}} -\frac{ {(6x^2)}}{{(y^2z^2)}} +\frac{ {(12x^2)}}{{(y^2z^3)}}}$

$\frac{-{(432x^8ln{(z^3)})}}{{(y^5z^3)}}$

Correct answer:

${\frac{{(12x^2)}}{{(y^2z^3)}}}$

Explanation:

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Example Question #490 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xx}\\&\text{Where }f(x,y,z)=-8cos{(y^3)}ln{(x^3)}sin{(z)}\end{align*}$

Possible Answers:

${\frac{{(24cos{(y^3)}sin{(z)})}}{x^2} -\frac{ {(24cos{(y^3)}sin{(z)})}}{x}}$

$\frac{{(576cos{(y^3)}^2sin{(z)}^2)}}{x^2}$

${\frac{-{(48cos{(y^3)}sin{(z)})}}{x}}$

${\frac{{(24cos{(y^3)}sin{(z)})}}{x^2}}$

Correct answer:

${\frac{{(24cos{(y^3)}sin{(z)})}}{x^2}}$

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 #481 : Partial Derivatives

Example Question #482 : Partial Derivatives

Example Question #483 : Partial Derivatives

Example Question #484 : Partial Derivatives

Example Question #485 : Partial Derivatives

Example Question #486 : Partial Derivatives

Example Question #487 : Partial Derivatives

Example Question #488 : Partial Derivatives

Example Question #489 : Partial Derivatives

Example Question #490 : Partial Derivatives

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