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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 #563 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=\frac{{(2\cdot4^{(z^2)}cos{(x^2)}sin{(4y)})}}{9}\end{align*}$

Possible Answers:

${\frac{-{(32\cdot4^{(2z^2)}xcos{(x^2)}cos{(4y)}sin{(x^2)}sin{(4y)})}}{81}}$

${\frac{-{(16\cdot4^{(z^2)}xcos{(4y)}sin{(x^2)})}}{9}}$

${\frac{{(8\cdot4^{(z^2)}cos{(x^2)}cos{(4y)})}}{9} -\frac{ {(4\cdot4^{(z^2)}xsin{(x^2)}sin{(4y)})}}{9}}$

${\frac{{(64\cdot4^{(2z^2)}x^2cos{(4y)}sin{(x^2)}^2sin{(4y)})}}{81}}$

Correct answer:

${\frac{-{(16\cdot4^{(z^2)}xcos{(4y)}sin{(x^2)})}}{9}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(z^2)}cos{(x^2)}sin{(4y)})}}{9}\\&f_{x}=\frac{-{(4\cdot4^{(z^2)}xsin{(x^2)}sin{(4y)})}}{9}\\&f_{xy}=\frac{-{(16\cdot4^{(z^2)}xcos{(4y)}sin{(x^2)})}}{9}\end{align*}$

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Example Question #564 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xz}\\&\text{Where }f(x,y,z)=\frac{{(2^{(4y)}\cdot2^{(3z)}cos{(4x)})}}{2}\end{align*}$

Possible Answers:

${-3\cdot2^{(8y)}\cdot2^{(6z)}cos{(4x)}sin{(4x)}ln{(2)}}$

${-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}}$

${12\cdot2^{(8y)}\cdot2^{(6z)}sin{(4x)}^2ln{(2)}}$

${\frac{{(3\cdot2^{(4y)}\cdot2^{(3z)}cos{(4x)}ln{(2)})}}{2} - 2\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}}$

Correct answer:

${-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2^{(4y)}\cdot2^{(3z)}cos{(4x)})}}{2}\\&f_{x}=-2\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}\\&f_{xz}=-6\cdot2^{(4y)}\cdot2^{(3z)}sin{(4x)}ln{(2)}\end{align*}$

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Example Question #565 : Partial Derivatives

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

Possible Answers:

${\frac{{(18sin{(3z)})}}{{(xy^2)}}}$

${\frac{{(6cos{(3z)}ln{(3y)})}}{x} - 9ln{(x^2)}ln{(3y)}sin{(3z)} +\frac{ {(6cos{(3z)}ln{(x^2)})}}{y}}$

${\frac{-{(1458cos{(3z)}^2ln{(x^2)}^3sin{(3z)}^2)}}{{(xy^7)}}}$

${\frac{-{(486cos{(3z)}^3ln{(x^2)}^3ln{(3y)}^2sin{(3z)})}}{{(xy^2)}}}$

Correct answer:

${\frac{{(18sin{(3z)})}}{{(xy^2)}}}$

Explanation:

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

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Example Question #566 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zzz}\\&\text{Where }f(x,y,z)=\frac{{(2ln{(z^2)}e^{(x^2)}e^{(4y)})}}{9}\end{align*}$

Possible Answers:

${\frac{-{(128e^{(12y)}e^{(3x^2)})}}{{(729z^6)}}}$

${\frac{{(8e^{(x^2)}e^{(4y)})}}{{(9z^3)}}}$

${\frac{{(64e^{(12y)}e^{(3x^2)})}}{{(729z^3)}}}$

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

Correct answer:

${\frac{{(8e^{(x^2)}e^{(4y)})}}{{(9z^3)}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2ln{(z^2)}e^{(x^2)}e^{(4y)})}}{9}\\&f_{z}=\frac{{(4e^{(x^2)}e^{(4y)})}}{{(9z)}}\\&f_{zz}=\frac{-{(4e^{(x^2)}e^{(4y)})}}{{(9z^2)}}\\&f_{zzz}=\frac{{(8e^{(x^2)}e^{(4y)})}}{{(9z^3)}}\end{align*}$

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Example Question #567 : Partial Derivatives

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

Possible Answers:

${\frac{-{(4\cdot4^{(9x)}ln{(y^2)}ln{(3z)}^3ln{(4)})}}{{(9y^2)}}}$

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

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

${\frac{{(4\cdot4^{(9x)}ln{(y^2)}ln{(3z)}^3ln{(4)}^3)}}{y^3}}$

Correct answer:

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

Explanation:

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

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Example Question #568 : Partial Derivatives

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

Possible Answers:

${\frac{{(2y^2e^{(z^2)}e^{(x)})}}{9} +\frac{ {(4y^2ze^{(z^2)}e^{(x)})}}{9}}$

${\frac{{(8y^4ze^{(2x)}e^{(2z^2)})}}{81}}$

${\frac{{(4y^2ze^{(z^2)}e^{(x)})}}{9}}$

${\frac{{(16y^4z^2e^{(2x)}e^{(2z^2)})}}{81}}$

Correct answer:

${\frac{{(4y^2ze^{(z^2)}e^{(x)})}}{9}}$

Explanation:

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Example Question #569 : Partial Derivatives

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

Possible Answers:

${\frac{{(16x^2y^2z)}}{7}}$

${\frac{{(256x^8y^8z^4)}}{2401}}$

${\frac{{(16x^2y^2z)}}{49}}$

Correct answer:

Explanation:

$\text{The last derivation is, of course, trivial.}$

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Example Question #570 : Partial Derivatives

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

Possible Answers:

${2\cdot2^{(2z)}y^3e^{(2x^2)}ln{(2)}^2}$

${2\cdot2^{(2z)}y^3e^{(2x^2)}ln{(2)}}$

${-2\cdot2^zye^{(x^2)}ln{(2)}}$

${- 2\cdot2^zye^{(x^2)} - 2^zy^2e^{(x^2)}ln{(2)}}$

Correct answer:

${-2\cdot2^zye^{(x^2)}ln{(2)}}$

Explanation:

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Example Question #571 : Partial Derivatives

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

Possible Answers:

${-24cos{(y^2)}^3e^{(9x)}e^{(3z)}}$

${-6cos{(y^2)}e^{(3x)}e^{(z)}}$

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

Correct answer:

${-6cos{(y^2)}e^{(3x)}e^{(z)}}$

Explanation:

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

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Example Question #572 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zyzx}\\&\text{Where }f(x,y,z)=\frac{-{(2\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)})}}{9}\end{align*}$

Possible Answers:

${\frac{{(32\cdot2^{(12z)}\cdot3^{(4x^2)}xe^{(4y)}ln{(2)}^2ln{(3)})}}{729}}$

${-4\cdot2^{(3z)}\cdot3^{(x^2)}xe^{(y)}ln{(2)}^2ln{(3)}}$

${-\frac{ {(2\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)})}}{9} -\frac{ {(4\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)}ln{(2)})}}{3} -\frac{ {(4\cdot2^{(3z)}\cdot3^{(x^2)}xe^{(y)}ln{(3)})}}{9}}$

${\frac{{(32\cdot2^{(12z)}\cdot3^{(4x^2)}xe^{(4y)}ln{(2)}^6ln{(3)})}}{9}}$

Correct answer:

${-4\cdot2^{(3z)}\cdot3^{(x^2)}xe^{(y)}ln{(2)}^2ln{(3)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{-{(2\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)})}}{9}\\&f_{z}=\frac{-{(2\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)}ln{(2)})}}{3}\\&f_{zy}=\frac{-{(2\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)}ln{(2)})}}{3}\\&f_{zyz}=-2\cdot2^{(3z)}\cdot3^{(x^2)}e^{(y)}ln{(2)}^2\\&f_{zyzx}=-4\cdot2^{(3z)}\cdot3^{(x^2)}xe^{(y)}ln{(2)}^2ln{(3)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #563 : Partial Derivatives

Example Question #564 : Partial Derivatives

Example Question #565 : Partial Derivatives

Example Question #566 : Partial Derivatives

Example Question #567 : Partial Derivatives

Example Question #568 : Partial Derivatives

Example Question #569 : Partial Derivatives

Example Question #570 : Partial Derivatives

Example Question #571 : Partial Derivatives

Example Question #572 : Partial Derivatives

All Calculus 3 Resources

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