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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3041 : Calculus 3

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

Possible Answers:

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

${-\frac{ (2^yztan(x^{4}))}{2}-\frac{ (2^yz^{2}tan(x^{4})ln(2))}{4}}$

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

${-\frac{(2^yztan(x^{4})ln(2))}{2}}$

Correct answer:

${-\frac{(2^yztan(x^{4})ln(2))}{2}}$

Explanation:

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

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Example Question #3042 : Calculus 3

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

Possible Answers:

${(\frac{(12x^{3}y^{3}e^{(y^{4})})}{z}-\frac{ (4\cdot 3^{(y^{2})}ye^{(4x)}ln(3)sin(z))}{7})\cdot (\frac{(9x^{2}e^{(y^{4})})}{z}-\frac{ (8\cdot 3^{(y^{2})}e^{(4x)}sin(z))}{7})}$

${\frac{(36x^{2}y^{3}e^{(y^{4})})}{z}-\frac{ (16\cdot 3^{(y^{2})}ye^{(4x)}ln(3)sin(z))}{7}}$

${\frac{(9x^{2}e^{(y^{4})})}{z}-\frac{ (8\cdot 3^{(y^{2})}e^{(4x)}sin(z))}{7}+\frac{ (12x^{3}y^{3}e^{(y^{4})})}{z}-\frac{ (4\cdot 3^{(y^{2})}ye^{(4x)}ln(3)sin(z))}{7}}$

${(\frac{(36x^{2}y^{3}e^{(y^{4})})}{z}-\frac{ (16\cdot 3^{(y^{2})}ye^{(4x)}ln(3)sin(z))}{7})\cdot (\frac{(9x^{2}e^{(y^{4})})}{z}-\frac{ (8\cdot 3^{(y^{2})}e^{(4x)}sin(z))}{7})}$

Correct answer:

${\frac{(36x^{2}y^{3}e^{(y^{4})})}{z}-\frac{ (16\cdot 3^{(y^{2})}ye^{(4x)}ln(3)sin(z))}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3x^{3}e^{(y^{4})})}{z}-\frac{ (2\cdot 3^{(y^{2})}e^{(4x)}sin(z))}{7}\\&f_{x}=\frac{(9x^{2}e^{(y^{4})})}{z}-\frac{ (8\cdot 3^{(y^{2})}e^{(4x)}sin(z))}{7}\\&f_{xy}=\frac{(36x^{2}y^{3}e^{(y^{4})})}{z}-\frac{ (16\cdot 3^{(y^{2})}ye^{(4x)}ln(3)sin(z))}{7}\end{align*}$

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Example Question #3043 : Calculus 3

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

Possible Answers:

${\frac{(4^{(y^{2})}z^{2}\cdot (\frac{3}{y}+\frac{ (2\cdot 4^{(y^{2})}zln(4x))}{3}))}{(3x)}}$

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

${-\frac{ 3}{y}-\frac{ (4^{(y^{2})}z^{2})}{(3x)}-\frac{ (2\cdot 4^{(y^{2})}zln(4x))}{3}}$

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

Correct answer:

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

Explanation:

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

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Example Question #3044 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yz}\\&\text{Where }f(x,y,z)=\frac{(4z^{3})}{y^{2}}+\frac{ (2e^{(z^{2})}sin(3x + x^{4})cos(y))}{7}\end{align*}$

Possible Answers:

${\frac{(12z^{2})}{y^{2}}-\frac{ (8z^{3})}{y^{3}}-\frac{ (2e^{(z^{2})}sin(3x + x^{4})sin(y))}{7}+\frac{ (4ze^{(z^{2})}sin(3x + x^{4})cos(y))}{7}}$

${(\frac{(8z^{3})}{y^{3}}+\frac{ (2e^{(z^{2})}sin(3x + x^{4})sin(y))}{7})\cdot (\frac{(24z^{2})}{y^{3}}+\frac{ (4ze^{(z^{2})}sin(3x + x^{4})sin(y))}{7})}$

${-(\frac{(8z^{3})}{y^{3}}+\frac{ (2e^{(z^{2})}sin(3x + x^{4})sin(y))}{7})\cdot (\frac{(12z^{2})}{y^{2}}+\frac{ (4ze^{(z^{2})}sin(3x + x^{4})cos(y))}{7})}$

${-\frac{ (24z^{2})}{y^{3}}-\frac{ (4ze^{(z^{2})}sin(3x + x^{4})sin(y))}{7}}$

Correct answer:

${-\frac{ (24z^{2})}{y^{3}}-\frac{ (4ze^{(z^{2})}sin(3x + x^{4})sin(y))}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4z^{3})}{y^{2}}+\frac{ (2e^{(z^{2})}sin(3x + x^{4})cos(y))}{7}\\&f_{y}=-\frac{ (8z^{3})}{y^{3}}-\frac{ (2e^{(z^{2})}sin(3x + x^{4})sin(y))}{7}\\&f_{yz}=-\frac{ (24z^{2})}{y^{3}}-\frac{ (4ze^{(z^{2})}sin(3x + x^{4})sin(y))}{7}\end{align*}$

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Example Question #3045 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yz}\\&\text{Where }f(x,y,z)=\frac{(2\cdot 4^ytan(4z)cos(x^{2} - 3x))}{9}\end{align*}$

Possible Answers:

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

${\frac{(2\cdot 4^ycos(x^{2} - 3x)\cdot (4tan(4z)^{2} + 4))}{9}+\frac{ (2\cdot 4^ytan(4z)ln(4)cos(x^{2} - 3x))}{9}}$

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 4^ytan(4z)cos(x^{2} - 3x))}{9}\\&f_{y}=\frac{(2\cdot 4^ytan(4z)ln(4)cos(x^{2} - 3x))}{9}\\&f_{yz}=\frac{(2\cdot 4^yln(4)cos(x^{2} - 3x)\cdot (4tan(4z)^{2} + 4))}{9}\end{align*}$

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Example Question #3046 : Calculus 3

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

Possible Answers:

${(\frac{(2\cdot 2^{(3z)}ysin(x))}{3}+\frac{ (12xcos(x^{2}))}{(yz)})\cdot (\frac{(6sin(x^{2}))}{(yz)}-\frac{ (2\cdot 2^{(3z)}ycos(x))}{3})}$

${(\frac{(6sin(x^{2}))}{(yz)}-\frac{ (2\cdot 2^{(3z)}ycos(x))}{3})\cdot (\frac{(2^{(3z)}y^{2}sin(x))}{3}+\frac{ (6xcos(x^{2})ln(y^{2}))}{z})}$

${\frac{(2\cdot 2^{(3z)}ycos(x))}{3}-\frac{ (6sin(x^{2}))}{(yz)}-\frac{ (2^{(3z)}y^{2}sin(x))}{3}-\frac{ (6xcos(x^{2})ln(y^{2}))}{z}}$

${-\frac{ (2\cdot 2^{(3z)}ysin(x))}{3}-\frac{ (12xcos(x^{2}))}{(yz)}}$

Correct answer:

${-\frac{ (2\cdot 2^{(3z)}ysin(x))}{3}-\frac{ (12xcos(x^{2}))}{(yz)}}$

Explanation:

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

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Example Question #3047 : Calculus 3

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

Possible Answers:

${6\cdot 3^{(4y)}xln(z^{2}) -\frac{ (2e^{(y^{2})}e^{(z^{4})})}{x^{2}}+ 12\cdot 3^{(4y)}x^{2}ln(z^{2})ln(3) +\frac{ (4ye^{(y^{2})}e^{(z^{4})})}{x}}$

${(6\cdot 3^{(4y)}xln(z^{2}) -\frac{ (2e^{(y^{2})}e^{(z^{4})})}{x^{2}})\cdot (12\cdot 3^{(4y)}x^{2}ln(z^{2})ln(3) +\frac{ (4ye^{(y^{2})}e^{(z^{4})})}{x})}$

${24\cdot 3^{(4y)}xln(z^{2})ln(3) -\frac{ (4ye^{(y^{2})}e^{(z^{4})})}{x^{2}}}$

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

Correct answer:

${24\cdot 3^{(4y)}xln(z^{2})ln(3) -\frac{ (4ye^{(y^{2})}e^{(z^{4})})}{x^{2}}}$

Explanation:

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

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Example Question #3048 : Calculus 3

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

Possible Answers:

${\frac{(xsin(y))}{z^{3}}}$

${-\frac{(x^{2}sin(y)^{2})}{(2z^{5})}}$

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

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

Correct answer:

${\frac{(xsin(y))}{z^{3}}}$

Explanation:

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

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Example Question #3049 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3050 : Calculus 3

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

Possible Answers:

${6x^{3}zsin(y) - 40ze^{(4y)}e^{(z^{2})}ln(x)}$

${-(20e^{(4y)}e^{(z^{2})}ln(x) - 3x^{3}z^{2}sin(y))\cdot (6x^{3}zsin(y) - 40ze^{(4y)}e^{(z^{2})}ln(x))}$

${3x^{3}z^{2}sin(y) - 20e^{(4y)}e^{(z^{2})}ln(x) - 6x^{3}zcos(y) - 10ze^{(4y)}e^{(z^{2})}ln(x)}$

${(20e^{(4y)}e^{(z^{2})}ln(x) - 3x^{3}z^{2}sin(y))\cdot (6x^{3}zcos(y) + 10ze^{(4y)}e^{(z^{2})}ln(x))}$

Correct answer:

${6x^{3}zsin(y) - 40ze^{(4y)}e^{(z^{2})}ln(x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=- 5e^{(4y)}e^{(z^{2})}ln(x) - 3x^{3}z^{2}cos(y)\\&f_{y}=3x^{3}z^{2}sin(y) - 20e^{(4y)}e^{(z^{2})}ln(x)\\&f_{yz}=6x^{3}zsin(y) - 40ze^{(4y)}e^{(z^{2})}ln(x)\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3041 : Calculus 3

Example Question #3042 : Calculus 3

Example Question #3043 : Calculus 3

Example Question #3044 : Calculus 3

Example Question #3045 : Calculus 3

Example Question #3046 : Calculus 3

Example Question #3047 : Calculus 3

Example Question #3048 : Calculus 3

Example Question #3049 : Calculus 3

Example Question #3050 : Calculus 3

All Calculus 3 Resources

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