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Calculus 3 : Calculus 3

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Example Questions

Example Question #3121 : Calculus 3

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

Possible Answers:

${\frac{(4\cdot 3^{(4z)}\cdot 4^x)}{(9y)}-\frac{ (6z^{3}e^{(y)})}{x}}$

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3122 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xx}\\&\text{Where }f(x,y,z)=\frac{(e^{(z)}sin(x^{4} - 3x)sin(y^{3} - 4y))}{2}-\frac{ 1}{(x^{2}y^{2})}\end{align*}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3123 : Calculus 3

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

Possible Answers:

${7y^{2}zcos(x^{3} - 3x)\cdot (3x^{2} - 3)}$

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

${7y^{2}zsin(x^{3} - 3x) +\frac{ (7y^{2}z^{2}cos(x^{3} - 3x)\cdot (3x^{2} - 3))}{2}}$

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

Correct answer:

${7y^{2}zcos(x^{3} - 3x)\cdot (3x^{2} - 3)}$

Explanation:

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

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Example Question #3124 : Calculus 3

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

Possible Answers:

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

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

${\frac{(4yln(3z)e^{(y^{2})}\cdot (\frac{(4ln(3z)e^{(y^{2})})}{x}+\frac{ (8y^{2}ln(3z)e^{(y^{2})})}{x}))}{x}}$

${\frac{(16y^{2}ln(3z)^{2}e^{(2y^{2})})}{x^{2}}}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=5z^{2}tan(x) +\frac{ (2ln(3z)e^{(y^{2})})}{x}\\&f_{y}=\frac{(4yln(3z)e^{(y^{2})})}{x}\\&f_{yy}=\frac{(4ln(3z)e^{(y^{2})})}{x}+\frac{ (8y^{2}ln(3z)e^{(y^{2})})}{x}\end{align*}$

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Example Question #3125 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3126 : Calculus 3

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

Possible Answers:

${\frac{(75\cdot 4^{(2z^{2})}e^{(6y)})}{(4x^{2})}}$

${\frac{(5\cdot 4^{(z^{2})}e^{(3y)}\cdot (\frac{(15\cdot 4^{(z^{2})}ln(4x)e^{(3y)})}{2}- 9y^{2}ln(z^{2})e^{(y^{3})}))}{(2x)}}$

${9y^{2}ln(z^{2})e^{(y^{3})} -\frac{ (5\cdot 4^{(z^{2})}e^{(3y)})}{(2x)}-\frac{ (15\cdot 4^{(z^{2})}ln(4x)e^{(3y)})}{2}}$

${-\frac{(15\cdot 4^{(z^{2})}e^{(3y)})}{(2x)}}$

Correct answer:

${-\frac{(15\cdot 4^{(z^{2})}e^{(3y)})}{(2x)}}$

Explanation:

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

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Example Question #3127 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3128 : Calculus 3

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

Possible Answers:

${\frac{(24x^{2}ycos(x^{3})e^{(y^{2})})}{z^{2}}-\frac{ z^{2}}{(2x^{2}y^{2})}}$

${(\frac{z^{2}}{(2xy^{2})}+\frac{ (8ysin(x^{3})e^{(y^{2})})}{z^{2}})\cdot (\frac{z^{2}}{(2x^{2}y)}+\frac{ (12x^{2}cos(x^{3})e^{(y^{2})})}{z^{2}})}$

${\frac{z^{2}}{(2xy^{2})}+\frac{ z^{2}}{(2x^{2}y)}+\frac{ (8ysin(x^{3})e^{(y^{2})})}{z^{2}}+\frac{ (12x^{2}cos(x^{3})e^{(y^{2})})}{z^{2}}}$

${-(\frac{z^{2}}{(2x^{2}y^{2})}-\frac{ (24x^{2}ycos(x^{3})e^{(y^{2})})}{z^{2}})\cdot (\frac{z^{2}}{(2x^{2}y)}+\frac{ (12x^{2}cos(x^{3})e^{(y^{2})})}{z^{2}})}$

Correct answer:

${\frac{(24x^{2}ycos(x^{3})e^{(y^{2})})}{z^{2}}-\frac{ z^{2}}{(2x^{2}y^{2})}}$

Explanation:

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

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Example Question #3129 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3130 : Calculus 3

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

Possible Answers:

${54y^{3}z^{2}ln(x^{2})^{2}tan(z^{3})\cdot (tan(z^{3})^{2} + 1)}$

${- 6yln(x^{2})tan(z^{3}) - 9y^{2}z^{2}ln(x^{2})\cdot (tan(z^{3})^{2} + 1)}$

${162y^{3}z^{4}ln(x^{2})^{2}\cdot (tan(z^{3})^{2} + 1)^{2}}$

${-18yz^{2}ln(x^{2})\cdot (tan(z^{3})^{2} + 1)}$

Correct answer:

${-18yz^{2}ln(x^{2})\cdot (tan(z^{3})^{2} + 1)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-3y^{2}ln(x^{2})tan(z^{3})\\&f_{z}=-9y^{2}z^{2}ln(x^{2})\cdot (tan(z^{3})^{2} + 1)\\&f_{zy}=-18yz^{2}ln(x^{2})\cdot (tan(z^{3})^{2} + 1)\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #3121 : Calculus 3

Example Question #3122 : Calculus 3

Example Question #3123 : Calculus 3

Example Question #3124 : Calculus 3

Example Question #3125 : Calculus 3

Example Question #3126 : Calculus 3

Example Question #3127 : Calculus 3

Example Question #3128 : Calculus 3

Example Question #3129 : Calculus 3

Example Question #3130 : Calculus 3

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