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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #3109 : Calculus 3

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

Possible Answers:

${(\frac{(6y^{2}cos(y^{3})sin(z))}{x^{2}}+\frac{ (2^ysin(z^{2})ln(2)sin(x))}{3})\cdot (\frac{(2sin(y^{3})cos(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})sin(x))}{3})}$

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

${\frac{(2sin(y^{3})cos(z))}{x^{2}}+\frac{ (6y^{2}cos(y^{3})sin(z))}{x^{2}}+\frac{ (2\cdot 2^yzcos(z^{2})sin(x))}{3}+\frac{ (2^ysin(z^{2})ln(2)sin(x))}{3}}$

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

Correct answer:

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

Explanation:

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

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Example Question #3110 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yz}\\&\text{Where }f(x,y,z)=\frac{(5cos(z^{2})ln(y))}{x}+\frac{ (y^{2}e^{(z)}tan(11x + x^{3}))}{5}\end{align*}$

Possible Answers:

${\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)}+\frac{ (y^{2}e^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2})ln(y))}{x}}$

${(\frac{(y^{2}e^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2})ln(y))}{x})\cdot (\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)})}$

${(\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)})\cdot (\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)})}$

${\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)}}$

Correct answer:

${\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(5cos(z^{2})ln(y))}{x}+\frac{ (y^{2}e^{(z)}tan(11x + x^{3}))}{5}\\&f_{y}=\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}+\frac{ (5cos(z^{2}))}{(xy)}\\&f_{yz}=\frac{(2ye^{(z)}tan(11x + x^{3}))}{5}-\frac{ (10zsin(z^{2}))}{(xy)}\end{align*}$

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Example Question #3111 : Calculus 3

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

Possible Answers:

${\frac{(4\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{3})}-\frac{ (6\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{2})}}$

${\frac{(48\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{6})}}$

${\frac{(12\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{3})}}$

${-\frac{(24\cdot 4^{(2z^{2})}x^{2}tan(x^{3})\cdot (tan(x^{3})^{2} + 1))}{(49y^{5})}}$

Correct answer:

${\frac{(12\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{3})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{2})}\\&f_{y}=\frac{(4\cdot 4^{(z^{2})}tan(x^{3}))}{(7y^{3})}\\&f_{yx}=\frac{(12\cdot 4^{(z^{2})}x^{2}\cdot (tan(x^{3})^{2} + 1))}{(7y^{3})}\end{align*}$

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Example Question #3112 : Calculus 3

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

Possible Answers:

${-\frac{(32\cdot 2^{(8x)}zcos(z^{2})ln(4y)sin(z^{2}))}{y}}$

${\frac{(64\cdot 2^{(8x)}z^{2}ln(4y)sin(z^{2})^{2})}{y}}$

${\frac{(8\cdot 2^{(4x)}zsin(z^{2}))}{y}}$

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

Correct answer:

${\frac{(8\cdot 2^{(4x)}zsin(z^{2}))}{y}}$

Explanation:

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

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Example Question #3113 : Calculus 3

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

Possible Answers:

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

${tan(\frac{y^{3})^{2}}{(25x^{2}z^{2})}}$

${\frac{tan(y^{3})}{(5x^{2}z)}}$

${-\frac{(2tan(y^{3}))}{(5xz)}}$

Correct answer:

${\frac{tan(y^{3})}{(5x^{2}z)}}$

Explanation:

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

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Example Question #3114 : Calculus 3

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

Possible Answers:

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

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

${\frac{(2x^{3}ze^{(z^{2})})}{y^{2}}+\frac{ (4zcos(z^{2})tan(x^{4})sin(4y + y^{2})\cdot (2y + 4))}{5}}$

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

Correct answer:

${\frac{(2x^{3}ze^{(z^{2})})}{y^{2}}+\frac{ (4zcos(z^{2})tan(x^{4})sin(4y + y^{2})\cdot (2y + 4))}{5}}$

Explanation:

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

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Example Question #3115 : Calculus 3

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

Possible Answers:

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

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

${\frac{(9\cdot 4^{(2y)}e^{(6z)})}{(16x^{2})}}$

${\frac{(27\cdot 4^{(2y)}e^{(6z)})}{(16x^{2})}}$

Correct answer:

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

Explanation:

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

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Example Question #3116 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{(2y^{2}sin(x^{4} - 4x)tan(12z + z^{4}))}{9}\end{align*}$

Possible Answers:

${-\frac{ (4ysin(x^{4} - 4x)tan(12z + z^{4}))}{9}-\frac{ (2y^{2}cos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$

${-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$

${\frac{(8y^{3}cos(x^{4} - 4x)sin(x^{4} - 4x)tan(12z + z^{4})^{2}\cdot (4x^{3} - 4))}{81}}$

${\frac{(16y^{2}cos(x^{4} - 4x)sin(x^{4} - 4x)tan(12z + z^{4})^{2}\cdot (4x^{3} - 4))}{81}}$

Correct answer:

${-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2y^{2}sin(x^{4} - 4x)tan(12z + z^{4}))}{9}\\&f_{y}=-\frac{(4ysin(x^{4} - 4x)tan(12z + z^{4}))}{9}\\&f_{yx}=-\frac{(4ycos(x^{4} - 4x)tan(12z + z^{4})\cdot (4x^{3} - 4))}{9}\end{align*}$

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Example Question #3117 : Calculus 3

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

Possible Answers:

${2^{(z^{2})}xzln(2)ln(y)}$

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

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

${\frac{(2^{(2z^{2})}x^{2}zln(2)ln(y)^{2})}{2}}$

Correct answer:

${2^{(z^{2})}xzln(2)ln(y)}$

Explanation:

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

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Example Question #3118 : Calculus 3

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

Possible Answers:

${-(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))\cdot (2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x))}$

${32y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 4\cdot 4^{(3x)}cos(4z + z^{4})cos(y)}$

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

${2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)}$

Correct answer:

${2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})sin(y)\\&f_{y}=16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})cos(y)\\&f_{yy}=2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3109 : Calculus 3

Example Question #3110 : Calculus 3

Example Question #3111 : Calculus 3

Example Question #3112 : Calculus 3

Example Question #3113 : Calculus 3

Example Question #3114 : Calculus 3

Example Question #3115 : Calculus 3

Example Question #3116 : Calculus 3

Example Question #3117 : Calculus 3

Example Question #3118 : Calculus 3

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