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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #771 : Partial Derivatives

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

Possible Answers:

${-(2zsin(x^{2})sin(y) +\frac{ (4z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}})^{2}\cdot (2xz^{2}cos(x^{2})sin(y) +\frac{ (2xe^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}})}$

${-(2zsin(x^{2})sin(y) +\frac{ (4z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}})\cdot (4xcos(x^{2})sin(y) +\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}+\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}})\cdot (2sin(x^{2})sin(y) +\frac{ (12z^{2}tan(x^{2})e^{(z^{4})})}{y^{3}}+\frac{ (16z^{6}tan(x^{2})e^{(z^{4})})}{y^{3}})}$

${- 4xcos(x^{2})sin(y) -\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}-\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}}$

${- 4zsin(x^{2})sin(y) -\frac{ (8z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}}- 2xz^{2}cos(x^{2})sin(y) -\frac{ (2xe^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}}$

Correct answer:

${- 4xcos(x^{2})sin(y) -\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}-\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=- z^{2}sin(x^{2})sin(y) -\frac{ (tan(x^{2})e^{(z^{4})})}{y^{3}}\\&f_{z}=- 2zsin(x^{2})sin(y) -\frac{ (4z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}}\\&f_{zz}=- 2sin(x^{2})sin(y) -\frac{ (12z^{2}tan(x^{2})e^{(z^{4})})}{y^{3}}-\frac{ (16z^{6}tan(x^{2})e^{(z^{4})})}{y^{3}}\\&f_{zzx}=- 4xcos(x^{2})sin(y) -\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}-\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}\end{align*}$

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Example Question #772 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yzz}\\&\text{Where }f(x,y,z)=5x^{4}y^{4}z^{2} - 3z^{2}e^{(3y)}tan(11x + x^{4})\end{align*}$

Possible Answers:

${20x^{4}y^{4}z - 12ze^{(3y)}tan(11x + x^{4}) + 20x^{4}y^{3}z^{2} - 9z^{2}e^{(3y)}tan(11x + x^{4})}$

${40x^{4}y^{3} - 18e^{(3y)}tan(11x + x^{4})}$

${(20x^{4}y^{3}z^{2} - 9z^{2}e^{(3y)}tan(11x + x^{4}))\cdot (40x^{4}y^{3}z - 18ze^{(3y)}tan(11x + x^{4}))\cdot (40x^{4}y^{3} - 18e^{(3y)}tan(11x + x^{4}))}$

${(20x^{4}y^{3}z^{2} - 9z^{2}e^{(3y)}tan(11x + x^{4}))\cdot (10x^{4}y^{4}z - 6ze^{(3y)}tan(11x + x^{4}))^{2}}$

Correct answer:

${40x^{4}y^{3} - 18e^{(3y)}tan(11x + x^{4})}$

Explanation:

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Example Question #773 : Partial Derivatives

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

Possible Answers:

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

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

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

${\frac{(4sin(4z + z^{2})^{2}cos(y)^{2}sin(x)^{2}\cdot (2z + 4)^{2})}{81}}$

Correct answer:

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

Explanation:

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

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Example Question #774 : Partial Derivatives

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

Possible Answers:

${-\frac{(256z^{4}ln(x^{2})^{3})}{y^{4}}}$

${\frac{(128z^{5}ln(x^{2})^{3}ln(4y))}{y^{2}}}$

${-\frac{(8zln(x^{2}))}{y^{2}}}$

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

Correct answer:

${-\frac{(8zln(x^{2}))}{y^{2}}}$

Explanation:

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Example Question #775 : Partial Derivatives

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

Possible Answers:

${-\frac{ (4sin(z))}{y^{3}}-\frac{ (6e^{(x^{2})}e^{(4y)}e^{(3z)})}{5}-\frac{ (4xe^{(x^{2})}e^{(4y)}e^{(3z)})}{5}}$

${\frac{(4xe^{(x^{2})}e^{(4y)}e^{(3z)}\cdot (\frac{(4sin(z))}{y^{3}}+\frac{ (6e^{(x^{2})}e^{(4y)}e^{(3z)})}{5}))}{5}}$

${\frac{(48x^{2}e^{(8y)}e^{(6z)}e^{(2x^{2})})}{25}}$

${-\frac{(12xe^{(x^{2})}e^{(4y)}e^{(3z)})}{5}}$

Correct answer:

${-\frac{(12xe^{(x^{2})}e^{(4y)}e^{(3z)})}{5}}$

Explanation:

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Example Question #776 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=- 2x^{2}tan(11y + y^{3})tan(z^{2} - 12z) -\frac{ (5x^{4}cos(y^{2}))}{z^{2}}\end{align*}$

Possible Answers:

${\frac{(40x^{3}ysin(y^{2}))}{z^{2}}- 4xtan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11)}$

${\frac{(10x^{4}ysin(y^{2}))}{z^{2}}- 4xtan(11y + y^{3})tan(z^{2} - 12z) - 2x^{2}tan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11) -\frac{ (20x^{3}cos(y^{2}))}{z^{2}}}$

${(\frac{(20x^{3}cos(y^{2}))}{z^{2}}+ 4xtan(11y + y^{3})tan(z^{2} - 12z))\cdot (2x^{2}tan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11) -\frac{ (10x^{4}ysin(y^{2}))}{z^{2}})}$

${(4xtan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11) -\frac{ (40x^{3}ysin(y^{2}))}{z^{2}})\cdot (2x^{2}tan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11) -\frac{ (10x^{4}ysin(y^{2}))}{z^{2}})}$

Correct answer:

${\frac{(40x^{3}ysin(y^{2}))}{z^{2}}- 4xtan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=- 2x^{2}tan(11y + y^{3})tan(z^{2} - 12z) -\frac{ (5x^{4}cos(y^{2}))}{z^{2}}\\&f_{y}=\frac{(10x^{4}ysin(y^{2}))}{z^{2}}- 2x^{2}tan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11)\\&f_{yx}=\frac{(40x^{3}ysin(y^{2}))}{z^{2}}- 4xtan(z^{2} - 12z)\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11)\end{align*}$

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Example Question #777 : Partial Derivatives

Find $\frac{\partial F}{\partial y}$ of the following function:

f(x, y)=x^2y+y^2x

Possible Answers:

2xy+y^2

x^2+2xy

4xy

Correct answer:

x^2+2xy

Explanation:

To find the partial derivative of the function, we must treat the other variable(s) as constants.

Doing this, we get

$\frac{\partial F}{\partial y}f(x, y)=x^2+2xy$

We used the following rules:

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

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Example Question #778 : Partial Derivatives

Determine $f_{yx}$ of the following function:

$f(x, y)=x\cos(y)+e^{2y}x^2$

Possible Answers:

$-x\sin(y)+2x^2e^{2y}$

$-\sin(y)+4xe^{2y}$

$-\sin(y)+2xe^{2y}$

$\sin(y)+4xe^{2y}$

Correct answer:

$-\sin(y)+4xe^{2y}$

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.

For us, this means first finding the partial derivative with respect to y:

$f_y=-x\sin(y)+2xe^{2y}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ .

Next, we find the partial derivative of the function above, this time with respect to x:

$f_{yx}=-\sin(y)+4xe^{2y}$

The derivatives were found using rules above as well as

$\frac{\mathrm{d} }{\mathrm{d} x}ax=a$

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Example Question #779 : Partial Derivatives

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

$f(x, y)=y^3\tan(x)+\cos(xy)$

Possible Answers:

$2y^3\sec(x)(\sec(x)\tan(x))-y^2\cos(xy)$

$2y^3\sec(x)(\sec(x)\tan(x))+y^2\cos(xy)$

$2y^3(\sec(x)\tan(x))-y^2\cos(xy)$

$2y^2\sec(x)(\sec(x)\tan(x))-y^2\cos(xy)$

Correct answer:

$2y^3\sec(x)(\sec(x)\tan(x))-y^2\cos(xy)$

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 with respect to x of the given function:

$f_x=y^3\sec^2(x)-y\sin(xy)$

The following rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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

$f_{xx}=2y^3\sec(x)(\sec(x)\tan(x))-y^2\cos(xy)$

We used the rules above, along with the following:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Example Question #780 : Partial Derivatives

Find $f_{xz}$ for the function

$f(x, y, z)=2e^{x^2}+xyz+y\sec(z)$

Possible Answers:

$4xe^{x^2}+y$

$4xe^{x^2}+yz$

Correct answer:

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 partial derivative with respect to x of the function:

$f_x=4xe^{x^2}+yz$

which was found using the following rules:

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

Next, we must take the partial derivative of the above function with respect to z:

$f_{xz}=y$

using the last rule above.

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #771 : Partial Derivatives

Example Question #772 : Partial Derivatives

Example Question #773 : Partial Derivatives

Example Question #774 : Partial Derivatives

Example Question #775 : Partial Derivatives

Example Question #776 : Partial Derivatives

Example Question #777 : Partial Derivatives

Example Question #778 : Partial Derivatives

Example Question #779 : Partial Derivatives

Example Question #780 : Partial Derivatives

All Calculus 3 Resources

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