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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #3039 : Calculus 3

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

Possible Answers:

${49ztan(x^{3})^{2}sin(y^{2})^{2}}$

${49z^{2}tan(x^{3})^{2}sin(y^{2})^{2}}$

${-14ztan(x^{3})sin(y^{2})}$

${-7tan(x^{3})sin(y^{2})}$

Correct answer:

${-7tan(x^{3})sin(y^{2})}$

Explanation:

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Example Question #3040 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #672 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #673 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #674 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #3041 : 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:

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

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

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

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

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 #676 : Partial Derivatives

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

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 #671 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #678 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #679 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3039 : Calculus 3

Example Question #3040 : Calculus 3

Example Question #672 : Partial Derivatives

Example Question #673 : Partial Derivatives

Example Question #674 : Partial Derivatives

Example Question #3041 : Calculus 3

Example Question #676 : Partial Derivatives

Example Question #671 : Partial Derivatives

Example Question #678 : Partial Derivatives

Example Question #679 : Partial Derivatives

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