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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #3251 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(tan(y^{3})e^{(4x)})}{(3z)})\widehat{i}+(\frac{(4cos(y^{2})e^{(z^{2})})}{x^{3}})\widehat{j}+(\frac{(e^{(x^{2})}e^{(z)})}{y^{3}})\widehat{k}\\&\text{at the point }(-3.07,1.54,-0.67)\end{align*}$

Possible Answers:

-8684.18

434.21

-289.47

1736.84

Correct answer:

1736.84

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(tan(y^{3})e^{(4x)})}{(3z)})\widehat{i}+(\frac{(4cos(y^{2})e^{(z^{2})})}{x^{3}})\widehat{j}+(\frac{(e^{(x^{2})}e^{(z)})}{y^{3}})\widehat{k}\\&\text{at the point }(-3.07,1.54,-0.67)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(\frac{(4tan(y^{3})e^{(4x)})}{(3z)})+(-\frac{(8ysin(y^{2})e^{(z^{2})})}{x^{3}})+(\frac{(e^{(x^{2})}e^{(z)})}{y^{3}})\\&divF(-3.07,1.54,-0.67)=(-5.177\cdot10^{-6})+(0.464)+(1736.37)=1736.84\end{align*}$

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Example Question #3252 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(tan(y^{3})cos(x^{3} - 4x))}{(3z)})\widehat{i}+(2y^{2}sin(x^{2})sin(z^{3}))\widehat{j}+(2xycos(z))\widehat{k}\\&\text{at the point }(3.04,-3.34,-2.37)\end{align*}$

Possible Answers:

2.14

51.44

-12.86

-3.22

Correct answer:

-12.86

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(tan(y^{3})cos(x^{3} - 4x))}{(3z)})\widehat{i}+(2y^{2}sin(x^{2})sin(z^{3}))\widehat{j}+(2xycos(z))\widehat{k}\\&\text{at the point }(3.04,-3.34,-2.37)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&divF(x,y,z)=(-\frac{(tan(y^{3})sin(x^{3} - 4x)\cdot (3x^{2} - 4))}{(3z)})+(4ysin(x^{2})sin(z^{3}))+(-2xysin(z))\\&divF(3.04,-3.34,-2.37)=(-0.352)+(1.651)+(-14.16)=-12.86\end{align*}$

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Example Question #3261 : Calculus 3

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

Possible Answers:

${\frac{(8\cdot 3^{(8x)}z^{3}e^{(8y)}ln(3))}{9}}$

${\frac{(8\cdot 3^{(4x)}ze^{(4y)}ln(3))}{3}}$

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

${\frac{(32\cdot 3^{(8x)}z^{3}e^{(8y)}ln(3)^{2})}{9}}$

Correct answer:

${\frac{(8\cdot 3^{(4x)}ze^{(4y)}ln(3))}{3}}$

Explanation:

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Example Question #3262 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #3263 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xzx}\\&\text{Where }f(x,y,z)=\frac{(4cos(y^{2}))}{z^{2}}+\frac{ (2\cdot 3^{(3x)}sin(3y + y^{3})tan(11z + z^{3}))}{9}\end{align*}$

Possible Answers:

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

${\frac{(8\cdot 3^{(9x)}ln(3)^{4}sin(3y + y^{3})^{3}tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)^{2}\cdot (3z^{2} + 11)^{2})}{9}}$

${\frac{(4\cdot 3^{(3x)}ln(3)sin(3y + y^{3})tan(11z + z^{3}))}{3}-\frac{ (8cos(y^{2}))}{z^{3}}+\frac{ (2\cdot 3^{(3x)}sin(3y + y^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11))}{9}}$

${-\frac{(4\cdot 3^{(6x)}ln(3)^{2}sin(3y + y^{3})^{2}tan(11z + z^{3})^{2}\cdot (\frac{(8cos(y^{2}))}{z^{3}}-\frac{ (2\cdot 3^{(3x)}sin(3y + y^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11))}{9}))}{9}}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4cos(y^{2}))}{z^{2}}+\frac{ (2\cdot 3^{(3x)}sin(3y + y^{3})tan(11z + z^{3}))}{9}\\&f_{x}=\frac{(2\cdot 3^{(3x)}ln(3)sin(3y + y^{3})tan(11z + z^{3}))}{3}\\&f_{xz}=\frac{(2\cdot 3^{(3x)}ln(3)sin(3y + y^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11))}{3}\\&f_{xzx}=2\cdot 3^{(3x)}ln(3)^{2}sin(3y + y^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)\end{align*}$

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Example Question #3264 : Calculus 3

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

Possible Answers:

${2\cdot 4^xycos(y^{2})ln(4)sin(4z + z^{3})}$

${2\cdot 4^xycos(y^{2})sin(4z + z^{3}) + 4^xsin(y^{2})ln(4)sin(4z + z^{3})}$

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

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

Correct answer:

${2\cdot 4^xycos(y^{2})ln(4)sin(4z + z^{3})}$

Explanation:

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Example Question #3265 : Calculus 3

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

Possible Answers:

${-\frac{(32768\cdot 4^{(8y)}ln(3z)^{3}e^{(32x)}ln(4)^{22})}{(6561z^{12})}}$

${-\frac{(16\cdot 4^ye^{(4x)}ln(4)^{4})}{(3z^{3})}}$

${\frac{(1024\cdot 4^{(8y)}ln(3z)^{5}e^{(32x)}ln(4)^{4})}{(6561z^{3})}}$

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

Correct answer:

${-\frac{(16\cdot 4^ye^{(4x)}ln(4)^{4})}{(3z^{3})}}$

Explanation:

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

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Example Question #3266 : Calculus 3

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

Possible Answers:

${\frac{(8x^{2})}{(y^{2}z)}-\frac{ (15e^{(y)}cos(x)sin(z))}{2}-\frac{ (24x^{2}ln(z))}{y^{3}}- 5e^{(y)}cos(x)cos(z)}$

${-(\frac{(4x^{2})}{(y^{2}z)}-\frac{ (5e^{(y)}cos(x)cos(z))}{2})\cdot (\frac{(8x^{2})}{(y^{3}z)}+\frac{ (5e^{(y)}cos(x)cos(z))}{2})\cdot (\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (8x^{2})}{(y^{3}z^{2})})\cdot (\frac{(5e^{(y)}cos(x)sin(z))}{2}-\frac{ (24x^{2})}{(y^{4}z^{2})})\cdot (\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (96x^{2})}{(y^{5}z^{2})})}$

${-(\frac{(4x^{2})}{(y^{2}z)}-\frac{ (5e^{(y)}cos(x)cos(z))}{2})^{2}\cdot (\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (8x^{2}ln(z))}{y^{3}})^{3}}$

${\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (96x^{2})}{(y^{5}z^{2})}}$

Correct answer:

${\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (96x^{2})}{(y^{5}z^{2})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4x^{2}ln(z))}{y^{2}}-\frac{ (5e^{(y)}cos(x)sin(z))}{2}\\&f_{z}=\frac{(4x^{2})}{(y^{2}z)}-\frac{ (5e^{(y)}cos(x)cos(z))}{2}\\&f_{zy}=-\frac{ (8x^{2})}{(y^{3}z)}-\frac{ (5e^{(y)}cos(x)cos(z))}{2}\\&f_{zyz}=\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (8x^{2})}{(y^{3}z^{2})}\\&f_{zyzy}=\frac{(5e^{(y)}cos(x)sin(z))}{2}-\frac{ (24x^{2})}{(y^{4}z^{2})}\\&f_{zyzyy}=\frac{(5e^{(y)}cos(x)sin(z))}{2}+\frac{ (96x^{2})}{(y^{5}z^{2})}\end{align*}$

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Example Question #3267 : Calculus 3

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

Possible Answers:

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

${-\frac{(xcos(y^{2})sin(z^{4} - 4z)\cdot (5cos(y^{3}) -\frac{ (x^{2}cos(y^{2})cos(z^{4} - 4z)\cdot (4z^{3} - 4))}{4})\cdot (15y^{2}zsin(y^{3}) -\frac{ (x^{2}ysin(y^{2})sin(z^{4} - 4z))}{2}))}{2}}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #3268 : Calculus 3

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

Possible Answers:

${(60\cdot 4^{(3x)}y^{3}e^{(y^{2})}ln(4)sin(z) + 90\cdot 4^{(3x)}ye^{(y^{2})}ln(4)sin(z))\cdot (4x^{3}z^{2} + 15\cdot 4^{(3x)}ye^{(y^{2})}ln(4)sin(z))\cdot (4x^{3}yz^{2} +\frac{ (15\cdot 4^{(3x)}e^{(y^{2})}ln(4)sin(z))}{2})\cdot (15\cdot 4^{(3x)}e^{(y^{2})}ln(4)sin(z) + 30\cdot 4^{(3x)}y^{2}e^{(y^{2})}ln(4)sin(z))}$

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

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

${60\cdot 4^{(3x)}y^{3}e^{(y^{2})}ln(4)sin(z) + 90\cdot 4^{(3x)}ye^{(y^{2})}ln(4)sin(z)}$

Correct answer:

${60\cdot 4^{(3x)}y^{3}e^{(y^{2})}ln(4)sin(z) + 90\cdot 4^{(3x)}ye^{(y^{2})}ln(4)sin(z)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=x^{4}yz^{2} +\frac{ (5\cdot 4^{(3x)}e^{(y^{2})}sin(z))}{2}\\&f_{x}=4x^{3}yz^{2} +\frac{ (15\cdot 4^{(3x)}e^{(y^{2})}ln(4)sin(z))}{2}\\&f_{xy}=4x^{3}z^{2} + 15\cdot 4^{(3x)}ye^{(y^{2})}ln(4)sin(z)\\&f_{xyy}=15\cdot 4^{(3x)}e^{(y^{2})}ln(4)sin(z) + 30\cdot 4^{(3x)}y^{2}e^{(y^{2})}ln(4)sin(z)\\&f_{xyyy}=60\cdot 4^{(3x)}y^{3}e^{(y^{2})}ln(4)sin(z) + 90\cdot 4^{(3x)}ye^{(y^{2})}ln(4)sin(z)\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3251 : Calculus 3

Example Question #3252 : Calculus 3

Example Question #3261 : Calculus 3

Example Question #3262 : Calculus 3

Example Question #3263 : Calculus 3

Example Question #3264 : Calculus 3

Example Question #3265 : Calculus 3

Example Question #3266 : Calculus 3

Example Question #3267 : Calculus 3

Example Question #3268 : Calculus 3

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