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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 #693 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #694 : Partial Derivatives

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

Possible Answers:

${\frac{(8sin(y^{3}))}{(xz)}-\frac{ (4ze^{(4y)}sin(x))}{5}}$

${(\frac{(2e^{(4y)}sin(x))}{5}+\frac{ (4sin(y^{3}))}{(xz^{2})})\cdot (\frac{(2ze^{(4y)}sin(x))}{5}-\frac{ (4sin(y^{3}))}{(xz)})}$

${(\frac{(2ze^{(4y)}sin(x))}{5}-\frac{ (4sin(y^{3}))}{(xz)})^{2}}$

${-\frac{ (2e^{(4y)}sin(x))}{5}-\frac{ (4sin(y^{3}))}{(xz^{2})}}$

Correct answer:

${-\frac{ (2e^{(4y)}sin(x))}{5}-\frac{ (4sin(y^{3}))}{(xz^{2})}}$

Explanation:

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

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Example Question #695 : Partial Derivatives

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

Possible Answers:

${\frac{(3cos(z^{2}))}{(xy)}-\frac{ (27ln(z^{2})e^{(3y)}cos(x^{2} - 3x))}{2}-\frac{ (9e^{(3y)}cos(x^{2} - 3x))}{z}-\frac{ (6zsin(z^{2})ln(y))}{x}}$

${-\frac{ (27e^{(3y)}cos(x^{2} - 3x))}{z}-\frac{ (6zsin(z^{2}))}{(xy)}}$

${-(\frac{(3cos(z^{2}))}{(xy)}-\frac{ (27ln(z^{2})e^{(3y)}cos(x^{2} - 3x))}{2})\cdot (\frac{(9e^{(3y)}cos(x^{2} - 3x))}{z}+\frac{ (6zsin(z^{2})ln(y))}{x})}$

${-(\frac{(3cos(z^{2}))}{(xy)}-\frac{ (27ln(z^{2})e^{(3y)}cos(x^{2} - 3x))}{2})\cdot (\frac{(27e^{(3y)}cos(x^{2} - 3x))}{z}+\frac{ (6zsin(z^{2}))}{(xy)})}$

Correct answer:

${-\frac{ (27e^{(3y)}cos(x^{2} - 3x))}{z}-\frac{ (6zsin(z^{2}))}{(xy)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3cos(z^{2})ln(y))}{x}-\frac{ (9ln(z^{2})e^{(3y)}cos(x^{2} - 3x))}{2}\\&f_{y}=\frac{(3cos(z^{2}))}{(xy)}-\frac{ (27ln(z^{2})e^{(3y)}cos(x^{2} - 3x))}{2}\\&f_{yz}=-\frac{ (27e^{(3y)}cos(x^{2} - 3x))}{z}-\frac{ (6zsin(z^{2}))}{(xy)}\end{align*}$

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Example Question #696 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=\frac{(tan(y^{3})sin(4x + x^{4})tan(11z + z^{4}))}{3}- ln(x^{2})e^{(z^{3})}e^{(y)}\end{align*}$

Possible Answers:

${-(\frac{(tan(y^{3})sin(4x + x^{4})\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11))}{3}- 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)})\cdot (3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)} - y^{2}sin(4x + x^{4})\cdot (tan(y^{3})^{2} + 1)\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11))}$

${y^{2}sin(4x + x^{4})\cdot (tan(y^{3})^{2} + 1)\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11) - 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)}}$

${\frac{(tan(y^{3})sin(4x + x^{4})\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11))}{3}- ln(x^{2})e^{(z^{3})}e^{(y)} + y^{2}sin(4x + x^{4})tan(11z + z^{4})\cdot (tan(y^{3})^{2} + 1) - 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)}}$

${-(\frac{(tan(y^{3})sin(4x + x^{4})\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11))}{3}- 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)})\cdot (ln(x^{2})e^{(z^{3})}e^{(y)} - y^{2}sin(4x + x^{4})tan(11z + z^{4})\cdot (tan(y^{3})^{2} + 1))}$

Correct answer:

${y^{2}sin(4x + x^{4})\cdot (tan(y^{3})^{2} + 1)\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11) - 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(tan(y^{3})sin(4x + x^{4})tan(11z + z^{4}))}{3}- ln(x^{2})e^{(z^{3})}e^{(y)}\\&f_{z}=\frac{(tan(y^{3})sin(4x + x^{4})\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11))}{3}- 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)}\\&f_{zy}=y^{2}sin(4x + x^{4})\cdot (tan(y^{3})^{2} + 1)\cdot (tan(11z + z^{4})^{2} + 1)\cdot (4z^{3} + 11) - 3z^{2}ln(x^{2})e^{(z^{3})}e^{(y)}\end{align*}$

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Example Question #697 : Partial Derivatives

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

Possible Answers:

${36z^{4}ln(3y)^{2}ln(x)^{2}\cdot (tan(z^{3})^{2} + 1)^{2}}$

${12zln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1) + 36z^{4}ln(3y)tan(z^{3})ln(x)\cdot (tan(z^{3})^{2} + 1)}$

${12z^{2}ln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1)}$

${6z^{2}ln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1)\cdot (12zln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1) + 36z^{4}ln(3y)tan(z^{3})ln(x)\cdot (tan(z^{3})^{2} + 1))}$

Correct answer:

${12zln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1) + 36z^{4}ln(3y)tan(z^{3})ln(x)\cdot (tan(z^{3})^{2} + 1)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=2ln(3y)tan(z^{3})ln(x)\\&f_{z}=6z^{2}ln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1)\\&f_{zz}=12zln(3y)ln(x)\cdot (tan(z^{3})^{2} + 1) + 36z^{4}ln(3y)tan(z^{3})ln(x)\cdot (tan(z^{3})^{2} + 1)\end{align*}$

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Example Question #698 : Partial Derivatives

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

Possible Answers:

${\frac{(36\cdot 3^{(6x)}e^{(2z)}ln(3)^{2}cos(y^{3} - 4y)sin(y^{3} - 4y)\cdot (3y^{2} - 4))}{49}}$

${\frac{(12\cdot 3^{(6x)}e^{(2z)}ln(3)cos(y^{3} - 4y)sin(y^{3} - 4y)\cdot (3y^{2} - 4))}{49}}$

${-\frac{ (6\cdot 3^{(3x)}e^{(z)}ln(3)sin(y^{3} - 4y))}{7}-\frac{ (2\cdot 3^{(3x)}e^{(z)}cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{7}}$

${-\frac{(6\cdot 3^{(3x)}e^{(z)}ln(3)cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{7}}$

Correct answer:

${-\frac{(6\cdot 3^{(3x)}e^{(z)}ln(3)cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2\cdot 3^{(3x)}e^{(z)}sin(y^{3} - 4y))}{7}\\&f_{x}=-\frac{(6\cdot 3^{(3x)}e^{(z)}ln(3)sin(y^{3} - 4y))}{7}\\&f_{xy}=-\frac{(6\cdot 3^{(3x)}e^{(z)}ln(3)cos(y^{3} - 4y)\cdot (3y^{2} - 4))}{7}\end{align*}$

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Example Question #699 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{(2e^{(z^{2})}tan(11x + x^{4}))}{y}\end{align*}$

Possible Answers:

${\frac{(4e^{(2z^{2})}tan(11x + x^{4})\cdot (tan(11x + x^{4})^{2} + 1)\cdot (4x^{3} + 11))}{y^{4}}}$

${-\frac{(2e^{(z^{2})}\cdot (tan(11x + x^{4})^{2} + 1)\cdot (4x^{3} + 11))}{y^{2}}}$

${-\frac{(4e^{(2z^{2})}tan(11x + x^{4})\cdot (tan(11x + x^{4})^{2} + 1)\cdot (4x^{3} + 11))}{y^{3}}}$

${\frac{(2e^{(z^{2})}\cdot (tan(11x + x^{4})^{2} + 1)\cdot (4x^{3} + 11))}{y}-\frac{ (2e^{(z^{2})}tan(11x + x^{4}))}{y^{2}}}$

Correct answer:

${-\frac{(2e^{(z^{2})}\cdot (tan(11x + x^{4})^{2} + 1)\cdot (4x^{3} + 11))}{y^{2}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2e^{(z^{2})}tan(11x + x^{4}))}{y}\\&f_{y}=-\frac{(2e^{(z^{2})}tan(11x + x^{4}))}{y^{2}}\\&f_{yx}=-\frac{(2e^{(z^{2})}\cdot (tan(11x + x^{4})^{2} + 1)\cdot (4x^{3} + 11))}{y^{2}}\end{align*}$

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Example Question #700 : Partial Derivatives

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

Possible Answers:

${-(\frac{(48y^{3}e^{(z^{3})})}{x^{4}}- 3x^{2}yz^{2}sin(y^{2})\cdot (tan(x^{3})^{2} + 1))\cdot (\frac{(16y^{3}e^{(z^{3})})}{x^{3}}+ yz^{2}tan(x^{3})sin(y^{2}))}$

${\frac{(12y^{4}e^{(z^{3})})}{x^{4}}-\frac{ (16y^{3}e^{(z^{3})})}{x^{3}}- yz^{2}tan(x^{3})sin(y^{2}) +\frac{ (3x^{2}z^{2}cos(y^{2})\cdot (tan(x^{3})^{2} + 1))}{2}}$

${-(\frac{(12y^{4}e^{(z^{3})})}{x^{4}}+\frac{ (3x^{2}z^{2}cos(y^{2})\cdot (tan(x^{3})^{2} + 1))}{2})\cdot (\frac{(16y^{3}e^{(z^{3})})}{x^{3}}+ yz^{2}tan(x^{3})sin(y^{2}))}$

${\frac{(48y^{3}e^{(z^{3})})}{x^{4}}- 3x^{2}yz^{2}sin(y^{2})\cdot (tan(x^{3})^{2} + 1)}$

Correct answer:

${\frac{(48y^{3}e^{(z^{3})})}{x^{4}}- 3x^{2}yz^{2}sin(y^{2})\cdot (tan(x^{3})^{2} + 1)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(z^{2}cos(y^{2})tan(x^{3}))}{2}-\frac{ (4y^{4}e^{(z^{3})})}{x^{3}}\\&f_{y}=-\frac{ (16y^{3}e^{(z^{3})})}{x^{3}}- yz^{2}tan(x^{3})sin(y^{2})\\&f_{yx}=\frac{(48y^{3}e^{(z^{3})})}{x^{4}}- 3x^{2}yz^{2}sin(y^{2})\cdot (tan(x^{3})^{2} + 1)\end{align*}$

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Example Question #701 : Partial Derivatives

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

Possible Answers:

${\frac{(x^{4}e^{(2y)})}{(25z^{2})}}$

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

${-\frac{(x^{4}e^{(2y)})}{(25z^{3})}}$

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

Correct answer:

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

Explanation:

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Example Question #702 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yz}\\&\text{Where }f(x,y,z)=3cos(z^{3})sin(x^{3})e^{(y^{2})} +\frac{ (x^{2}z^{2}tan(12y + y^{2}))}{4}\end{align*}$

Possible Answers:

${\frac{(x^{2}ztan(12y + y^{2}))}{2}- 9z^{2}sin(x^{3})sin(z^{3})e^{(y^{2})} +\frac{ (x^{2}z^{2}\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{4}+ 6ycos(z^{3})sin(x^{3})e^{(y^{2})}}$

${\frac{(x^{2}z\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{2}- 18yz^{2}sin(x^{3})sin(z^{3})e^{(y^{2})}}$

${(\frac{(x^{2}z^{2}\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{4}+ 6ycos(z^{3})sin(x^{3})e^{(y^{2})})\cdot (\frac{(x^{2}ztan(12y + y^{2}))}{2}- 9z^{2}sin(x^{3})sin(z^{3})e^{(y^{2})})}$

${(\frac{(x^{2}z\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{2}- 18yz^{2}sin(x^{3})sin(z^{3})e^{(y^{2})})\cdot (\frac{(x^{2}z^{2}\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{4}+ 6ycos(z^{3})sin(x^{3})e^{(y^{2})})}$

Correct answer:

${\frac{(x^{2}z\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{2}- 18yz^{2}sin(x^{3})sin(z^{3})e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3cos(z^{3})sin(x^{3})e^{(y^{2})} +\frac{ (x^{2}z^{2}tan(12y + y^{2}))}{4}\\&f_{y}=\frac{(x^{2}z^{2}\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{4}+ 6ycos(z^{3})sin(x^{3})e^{(y^{2})}\\&f_{yz}=\frac{(x^{2}z\cdot (2y + 12)\cdot (tan(12y + y^{2})^{2} + 1))}{2}- 18yz^{2}sin(x^{3})sin(z^{3})e^{(y^{2})}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #693 : Partial Derivatives

Example Question #694 : Partial Derivatives

Example Question #695 : Partial Derivatives

Example Question #696 : Partial Derivatives

Example Question #697 : Partial Derivatives

Example Question #698 : Partial Derivatives

Example Question #699 : Partial Derivatives

Example Question #700 : Partial Derivatives

Example Question #701 : Partial Derivatives

Example Question #702 : Partial Derivatives

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