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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #3169 : Calculus 3

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

Possible Answers:

${\frac{(32x^{3}e^{(2z^{2})})}{(81y^{6})}}$

${-\frac{(16x^{3}e^{(2z^{2})})}{(81y^{5})}}$

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

${-\frac{(8xe^{(z^{2})})}{(9y^{3})}}$

Correct answer:

${-\frac{(8xe^{(z^{2})})}{(9y^{3})}}$

Explanation:

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Example Question #801 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=\frac{(ln(4y)e^{(4x)}tan(z^{2} - 12z))}{5}\end{align*}$

Possible Answers:

${\frac{(e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(5y)}}$

${\frac{(ln(4y)e^{(8x)}\cdot (2z - 12)^{2}\cdot (tan(z^{2} - 12z)^{2} + 1)^{2})}{(25y)}}$

${\frac{(e^{(4x)}tan(z^{2} - 12z))}{(5y)}+\frac{ (ln(4y)e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{5}}$

${\frac{(ln(4y)e^{(8x)}tan(z^{2} - 12z)\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(25y)}}$

Correct answer:

${\frac{(e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(5y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(ln(4y)e^{(4x)}tan(z^{2} - 12z))}{5}\\&f_{z}=\frac{(ln(4y)e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{5}\\&f_{zy}=\frac{(e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(5y)}\end{align*}$

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Example Question #803 : Partial Derivatives

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

Possible Answers:

${(4x^{4}cos(z) +\frac{ (6\cdot 3^{(3z)}y^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5})\cdot (16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5})}$

${(16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5})\cdot (16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5})}$

${16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}}$

${4x^{4}cos(z) + 16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5}+\frac{ (6\cdot 3^{(3z)}y^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}}$

Correct answer:

${16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 3^{(3z)}tan(y^{3})e^{(x^{2})})}{5}+ 4x^{4}ycos(z)\\&f_{x}=16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5}\\&f_{xy}=16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}\end{align*}$

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Example Question #801 : Partial Derivatives

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

Possible Answers:

${\frac{(4xsin(z))}{y^{3}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5}}$

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

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

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

Correct answer:

${\frac{(4xsin(z))}{y^{3}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5}}$

Explanation:

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

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Example Question #3172 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3171 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #807 : Partial Derivatives

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

Possible Answers:

${12ysin(z^{3})sin(x) - 2\cdot 3^{(y^{2})}cos(z)sin(x) + 9y^{2}z^{2}cos(z^{3})sin(x) - 8\cdot 3^{(y^{2})}yln(3)sin(x)sin(z)}$

${-(18yz^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)cos(z)sin(x))\cdot (6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z))\cdot (4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 18z^{2}cos(z^{3})sin(x) + 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x))}$

${18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)}$

${-(6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z))^{2}\cdot (2\cdot 3^{(y^{2})}cos(z)sin(x) - 9y^{2}z^{2}cos(z^{3})sin(x))}$

Correct answer:

${18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3y^{2}sin(z^{3})sin(x) - 2\cdot 3^{(y^{2})}sin(x)sin(z)\\&f_{y}=6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z)\\&f_{yz}=18yz^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)cos(z)sin(x)\\&f_{yzy}=18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)\end{align*}$

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Example Question #3171 : Calculus 3

Find $f_{xyz}$ of the following function:

$f(x, y, z)=z^3x^2y+\cos(z)xy^2+xyz$

Possible Answers:

$6xz^2-2y\sin(z)$

$6xz^2-2y\sin(z)+1$

$2xz^3+2y\cos(z)+z$

$6xz^2+2y\sin(z)+1$

Correct answer:

$6xz^2-2y\sin(z)+1$

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must find the partial derivative of the function with respect to x:

$f_x=2xz^3y+\cos(z)y^2+yz$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we find the partial derivative of the above function with respect to y:

$f_{xy}=2xz^3+2y\cos(z)+z$

We used rules stated above.

Finally, we find the derivative of the above function with respect to z:

$f_{xyz}=6xz^2-2y\sin(z)+1$

We used all of the rules used previously along with

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #3173 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})\widehat{i}+(\frac{(4sin(y^{2}))}{z})\widehat{j}+(5e^{(x^{2})}e^{(z)}sin(y))\widehat{k}\\&\text{at the point }(0.98,1.11,2.39)\end{align*}$

Possible Answers:

-283.32

849.95

-70.83

141.66

Correct answer:

141.66

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})\widehat{i}+(\frac{(4sin(y^{2}))}{z})\widehat{j}+(5e^{(x^{2})}e^{(z)}sin(y))\widehat{k}\\&\text{at the point }(0.98,1.11,2.39)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})+(\frac{(8ycos(y^{2}))}{z})+(5e^{(x^{2})}e^{(z)}sin(y))\\&divF(0.98,1.11,2.39)=(12.72)+(1.23)+(127.7)=141.66\end{align*}$

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Example Question #3172 : Calculus 3

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(5e^{(y^{2})}e^{(4z)})}{x})\widehat{i}+(\frac{(3ln(x^{2})sin(z^{2}))}{y^{3}})\widehat{j}+(2x^{3}ze^{(y^{4})})\widehat{k}\\&\text{at the point }(1.57,0.68,1.71)\end{align*}$

Possible Answers:

3008.58

-3008.58

501.43

-1504.29

Correct answer:

-3008.58

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(5e^{(y^{2})}e^{(4z)})}{x})\widehat{i}+(\frac{(3ln(x^{2})sin(z^{2}))}{y^{3}})\widehat{j}+(2x^{3}ze^{(y^{4})})\widehat{k}\\&\text{at the point }(1.57,0.68,1.71)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&divF(x,y,z)=(-\frac{(5e^{(y^{2})}e^{(4z)})}{x^{2}})+(-\frac{(9ln(x^{2})sin(z^{2}))}{y^{4}})+(2x^{3}e^{(y^{4})})\\&divF(1.57,0.68,1.71)=(-3009.97)+(-8.19)+(9.58)=-3008.58\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3169 : Calculus 3

Example Question #801 : Partial Derivatives

Example Question #803 : Partial Derivatives

Example Question #801 : Partial Derivatives

Example Question #3172 : Calculus 3

Example Question #3171 : Calculus 3

Example Question #807 : Partial Derivatives

Example Question #3171 : Calculus 3

Example Question #3173 : Calculus 3

Example Question #3172 : Calculus 3

All Calculus 3 Resources

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