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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 #651 : Partial Derivatives

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #652 : Partial Derivatives

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

Possible Answers:

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

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

${-\frac{(ycos(y^{2})sin(y^{2})e^{(2z)}sin(x)^{2})}{8}}$

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

Correct answer:

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

Explanation:

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

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Example Question #3023 : Calculus 3

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

Possible Answers:

${(\frac{(20z^{3}ln(x^{2}))}{y^{2}}-\frac{ (sin(3y + y^{2})cos(x)\cdot (2y + 3))}{z})\cdot (\frac{(5z^{4}ln(x^{2}))}{y^{2}}- sin(3y + y^{2})cos(x)ln(z)\cdot (2y + 3))}$

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

${-(\frac{(20z^{3}ln(x^{2}))}{y}-\frac{ (cos(3y + y^{2})cos(x))}{z})\cdot (\frac{(5z^{4}ln(x^{2}))}{y^{2}}- sin(3y + y^{2})cos(x)ln(z)\cdot (2y + 3))}$

${\frac{(5z^{4}ln(x^{2}))}{y^{2}}-\frac{ (20z^{3}ln(x^{2}))}{y}+\frac{ (cos(3y + y^{2})cos(x))}{z}- sin(3y + y^{2})cos(x)ln(z)\cdot (2y + 3)}$

Correct answer:

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

Explanation:

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

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Example Question #653 : Partial Derivatives

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

Possible Answers:

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

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

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

${-(\frac{(8y^{3})}{(x^{3}z^{2})}+ 6xzcos(x^{2})ln(4y)sin(z^{2}))\cdot (\frac{(8y^{3})}{(x^{3}z)}+ 3xcos(x^{2})cos(z^{2})ln(4y))}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3cos(z^{2})ln(4y)sin(x^{2}))}{2}-\frac{ (4y^{3})}{(x^{2}z)}\\&f_{x}=\frac{(8y^{3})}{(x^{3}z)}+ 3xcos(x^{2})cos(z^{2})ln(4y)\\&f_{xz}=-\frac{ (8y^{3})}{(x^{3}z^{2})}- 6xzcos(x^{2})ln(4y)sin(z^{2})\end{align*}$

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Example Question #3025 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #654 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3027 : Calculus 3

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

Possible Answers:

${(\frac{(3\cdot 3^{(4x)}e^{(3y)}sin(4z + z^{3}))}{5}+\frac{ (15y^{2}sin(y^{3})sin(x))}{z^{2}})\cdot (\frac{(15y^{2}sin(y^{3})cos(x))}{z^{2}}+\frac{ (12\cdot 3^{(4x)}e^{(3y)}ln(3)sin(4z + z^{3}))}{5})}$

${-(\frac{(3\cdot 3^{(4x)}e^{(3y)}sin(4z + z^{3}))}{5}+\frac{ (15y^{2}sin(y^{3})sin(x))}{z^{2}})\cdot (\frac{(5cos(y^{3})cos(x))}{z^{2}}-\frac{ (4\cdot 3^{(4x)}e^{(3y)}ln(3)sin(4z + z^{3}))}{5})}$

${\frac{(3\cdot 3^{(4x)}e^{(3y)}sin(4z + z^{3}))}{5}-\frac{ (5cos(y^{3})cos(x))}{z^{2}}+\frac{ (15y^{2}sin(y^{3})sin(x))}{z^{2}}+\frac{ (4\cdot 3^{(4x)}e^{(3y)}ln(3)sin(4z + z^{3}))}{5}}$

${\frac{(15y^{2}sin(y^{3})cos(x))}{z^{2}}+\frac{ (12\cdot 3^{(4x)}e^{(3y)}ln(3)sin(4z + z^{3}))}{5}}$

Correct answer:

${\frac{(15y^{2}sin(y^{3})cos(x))}{z^{2}}+\frac{ (12\cdot 3^{(4x)}e^{(3y)}ln(3)sin(4z + z^{3}))}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3^{(4x)}e^{(3y)}sin(4z + z^{3}))}{5}-\frac{ (5cos(y^{3})sin(x))}{z^{2}}\\&f_{y}=\frac{(3\cdot 3^{(4x)}e^{(3y)}sin(4z + z^{3}))}{5}+\frac{ (15y^{2}sin(y^{3})sin(x))}{z^{2}}\\&f_{yx}=\frac{(15y^{2}sin(y^{3})cos(x))}{z^{2}}+\frac{ (12\cdot 3^{(4x)}e^{(3y)}ln(3)sin(4z + z^{3}))}{5}\end{align*}$

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Example Question #655 : Partial Derivatives

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

Possible Answers:

${-\frac{ (6e^{(x^{2})})}{z}-\frac{ (8\cdot 3^{(3z)}\cdot 3^yln(3)sin(x^{3} - 3x))}{7}}$

${\frac{(12\cdot 3^{(2y)}\cdot 3^{(6z)}ln(3)^{3}sin(x^{3} - 3x)^{2})}{49}}$

${\frac{(2\cdot 3^{(3z)}\cdot 3^yln(3)sin(x^{3} - 3x)\cdot (\frac{(6e^{(x^{2})})}{z}+\frac{ (6\cdot 3^{(3z)}\cdot 3^yln(3)sin(x^{3} - 3x))}{7}))}{7}}$

${-\frac{(6\cdot 3^{(3z)}\cdot 3^yln(3)^{2}sin(x^{3} - 3x))}{7}}$

Correct answer:

${-\frac{(6\cdot 3^{(3z)}\cdot 3^yln(3)^{2}sin(x^{3} - 3x))}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=- 3ln(z^{2})e^{(x^{2})} -\frac{ (2\cdot 3^{(3z)}\cdot 3^ysin(x^{3} - 3x))}{7}\\&f_{y}=-\frac{(2\cdot 3^{(3z)}\cdot 3^yln(3)sin(x^{3} - 3x))}{7}\\&f_{yz}=-\frac{(6\cdot 3^{(3z)}\cdot 3^yln(3)^{2}sin(x^{3} - 3x))}{7}\end{align*}$

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Example Question #661 : Partial Derivatives

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

Possible Answers:

${-(\frac{(9ln(y^{2}))}{z^{4}}+ 2cos(x^{2})e^{(z)}sin(3y + y^{3}))\cdot (\frac{6}{(yz^{3})}- 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3))}$

${\frac{(9ln(y^{2}))}{z^{4}}-\frac{ 6}{(yz^{3})}+ 2cos(x^{2})e^{(z)}sin(3y + y^{3}) + 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3)}$

${\frac{18}{(yz^{4})}+ 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3)}$

${-(\frac{6}{(yz^{3})}- 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3))\cdot (\frac{18}{(yz^{4})}+ 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3))}$

Correct answer:

${\frac{18}{(yz^{4})}+ 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=2cos(x^{2})e^{(z)}sin(3y + y^{3}) -\frac{ (3ln(y^{2}))}{z^{3}}\\&f_{y}=2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3) -\frac{ 6}{(yz^{3})}\\&f_{yz}=\frac{18}{(yz^{4})}+ 2cos(x^{2})e^{(z)}cos(3y + y^{3})\cdot (3y^{2} + 3)\end{align*}$

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Example Question #662 : Partial Derivatives

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

Possible Answers:

${\frac{(49\cdot 2^{(2z)}y^{3}ln(4x))}{(2x)}}$

${\frac{(49\cdot 2^{(2z)}y^{2}ln(4x))}{x}}$

${-\frac{(7\cdot 2^zy)}{x}}$

${-\frac{ (7\cdot 2^zy^{2})}{(2x)}- 7\cdot 2^zyln(4x)}$

Correct answer:

${-\frac{(7\cdot 2^zy)}{x}}$

Explanation:

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #651 : Partial Derivatives

Example Question #652 : Partial Derivatives

Example Question #3023 : Calculus 3

Example Question #653 : Partial Derivatives

Example Question #3025 : Calculus 3

Example Question #654 : Partial Derivatives

Example Question #3027 : Calculus 3

Example Question #655 : Partial Derivatives

Example Question #661 : Partial Derivatives

Example Question #662 : Partial Derivatives

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