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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #971 : Partial Derivatives

Find the partial derivative $\small \small \small \small f_y$ of the function $\small \small \small \small f(x,y,z)=x^4(y+z)^5$ .

Possible Answers:

$\small \small \small f_y=\frac{1}{6}x^4(y+z)^6$

$\small \small \small f_y=4x^3(y+z)^4$

$\small \small f_y=5x^4(y+z)^4$

$\small \small \small f_y=20x^4(y+z)^4$

Correct answer:

$\small \small f_y=5x^4(y+z)^4$

Explanation:

To find the partial derivative $\small \small \small \small f_y$ of the function $\small \small \small \small f(x,y,z)=x^4(y+z)^5$ , we take the derivative with respect to $\small y$ while holding $\small \small \small \small \small x,z$ constant. So we use the power rule to get

$\small \small \small f_y=\frac{\partial}{\partial y}f(x,y,z)=\frac{\partial}{\partial y}x^4(y+z)^5=x^4\frac{\partial}{\partial y}(y+z)^5=x^4\cdot(5(y+z)^4)$

$\small =5x^4(y+z)^4$

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Example Question #972 : Partial Derivatives

Find the partial derivative $\small \small \small \small f_z$ of the function $\small \small \small \small f(x,y,z)=x^4(y+z)^5$ .

Possible Answers:

$\small \small f_z=4x^3(y+z)^4$

$\small \small f_z=x^3(y+z)^4$

$\small \small f_z=20x^4(y+z)^4$

$\small \small f_z=5x^4(y+z)^4$

Correct answer:

$\small \small f_z=5x^4(y+z)^4$

Explanation:

To find the partial derivative $\small \small \small \small f_z$ of the function $\small \small \small \small f(x,y,z)=x^4(y+z)^5$ , we take the derivative with respect to while holding $\small \small \small \small \small x,y$ constant. So we use the power rule to get

$\small \small \small \small f_z=\frac{\partial}{\partial z}f(x,y,z)=\frac{\partial}{\partial z}x^4(y+z)^5=x^4\frac{\partial}{\partial z}(y+z)^5=x^4\cdot(5(y+z)^4)$

$\small =5x^4(y+z)^4$

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Example Question #973 : Partial Derivatives

Find $f_{xzy}$ of the function x^3yz+z^5+xy^5

Possible Answers:

xy^2+5x

3x^2z

3x^2

Correct answer:

3x^2

Explanation:

To find $f_{xzy}$ , you must take $\frac{\partial }{\partial x},\frac{\partial }{\partial z},\frac{\partial }{\partial y}$ consecutively of the function. In doing so, we get $\frac{\partial }{\partial x}(x^3yz+z^5+xy^5)=3x^2yz+y^5$ , $\frac{\partial }{\partial z}(3x^2yz+y^5)=3x^2y$ , and finally $\frac{\partial }{\partial y}(3x^2y)=3x^2$

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Example Question #974 : Partial Derivatives

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

$f(x, z)=x^2\cos(z^2)$

Possible Answers:

$-4xz\sin(z^2)$

$4z\sin(z^2)$

$-4z\sin(z^2)$

$-2z\sin(z^2)$

Correct answer:

$-4z\sin(z^2)$

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:

$f_x=2x\cos(z^2)$

The derivative was found using the following rules:

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

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

$f_{xz}=-4xz\sin(z^2)$

The derivative was found using rules above as well as the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Finally, we take the partial derivative of the above function with respect to x:

$f_{xzx}=-4z\sin(z^2)$

The rule used to find the partial derivative is above.

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Example Question #975 : Partial Derivatives

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

$f(x,y,z)= (\cos x)y^3e^\tan z$

Possible Answers:

$f_{xxy}=-3y^2e^ {tan z}(\cos x)(\sec^2 z)$

$f_{xxy}=y^2e^ {tan z}(\sin x)$

$f_{xxy}=-3y^2e^ {tan z}(\cos x)$

$f_{xxy}=-3y(\cos x)$

Correct answer:

$f_{xxy}=-3y^2e^ {tan z}(\cos x)$

Explanation:

The first thing to do is to take one partial derivative with respect to only one variable while everything else behaves as a constant. In this case, you can start by taking the partial derivative of one , which would be:

$f_{x}=-y^3e^{\tan z}(\sin x)$

The derivative was found by using the following rule:

$\frac{d}{dx}\cos x=-\sin x$

Then, you can take the partial derivative with respect to again because is expressed twice. This partial derivative is:

$f_{xx}=-y^3e^{\tan z}(\cos x)$

The derivative was found by using the following rule:

$\frac{d}{dx}\sin x=\cos x$

Finally, you must take the partial derivative with respect to which is:

$f_{xxy}=-3y^2e^{\tan z}(\cos x)$

The derivative was taken by using the following rule:

$\frac{d}{dx}x^n=nx^{n-1}$

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Example Question #976 : Partial Derivatives

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

f(x,y,z)=5x^4y^3e^z

Possible Answers:

$f_{xyz}=20x^3y^2e^z$

$f_{xyz}=60x^3y^2e^z$

$f_{xyz}=360x^3ye^z$

$f_{xyz}=60x^3y^2ze^z$

Correct answer:

$f_{xyz}=60x^3y^2e^z$

Explanation:

$f_{x}=20x^3y^3e^z$

The derivative was found by using the following rule:

$\frac{d}{dx}x^n=nx^{n-1}$

Then, you can take the partial derivative with respect to . This partial derivative is:

$f_{xy}=60x^3y^2e^z$

The derivative was found by using the following rule:

$\frac{d}{dx}x^n=nx^{n-1}$

Finally, you must take the partial derivative with respect to which is:

$f_{xyz}=60x^3y^2e^z$

The derivative was taken by using the following rule:

$\frac{d}{dx}e^x=e^x\cdot x'$

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Example Question #977 : Partial Derivatives

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

$f(x,y,z)=\cos xe^{2y}\sqrt{z}$

Possible Answers:

$\frac{-2\sin xe^{2y}}{\sqrt{z}}$

$\frac{4\cos xe^{2y}}{\sqrt{z}}$

${2\cos xe^{2y}}{\sqrt{z}}$

$\frac{2\cos xe^{2y}}{\sqrt{z}}$

Correct answer:

$\frac{2\cos xe^{2y}}{\sqrt{z}}$

Explanation:

$f_{y}=2\cos xe^{2y}\sqrt{z}$

The derivative was found by using the following rule:

$\frac{d}{dx} e^x=e^x\cdot x'$

Then, you can take the partial derivative with respect to again because is expressed twice. This partial derivative is:

$f_{yy}=4\cos xe^{2y}\sqrt{z}$

The derivative was found by using the following rule:

$\frac{d}{dx}e^x=e^x\cdot x'$

Finally, you must take the partial derivative with respect to which is:

$f_{yyz}=4\cos xe^{2y}(\frac{1}{2}z^{\frac{-1}{2}})=\frac{2\cos xe^{2y}}{\sqrt{z}}$

The derivative was taken by using the following rule:

$\frac{d}{dx}x^n=nx^{n-1}$

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Example Question #978 : Partial Derivatives

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

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

Possible Answers:

$f_{xzy}=-2ye^{\tan z}\sec^2 z\sin x$

$f_{xzy}=-2ye^{\tan z} \sin x$

$f_{xzy}=-2y^2e^{\tan z}\sec^2 z\sin x$

$f_{xzy}=-2ye^{\tan z}\sec^2 z\cos x$

Correct answer:

$f_{xzy}=-2ye^{\tan z}\sec^2 z\sin x$

Explanation:

The first thing to do is to take the partial derivative with respect to one variable while everything else behaves as a constant. In this case, you can take the partial derivative with respect to first which would be:

$f_{x}=-y^2e^{tan z}\sin x$

Then, you can take the partial derivative of which would be:

$f_{xz}=-y^2e^{tan z}\sec^2 z\sin x$

Finally, you must take the partial derivative with respect to :

$f_{xzy}=-2ye^{tan z}\sec^2 z\sin x$

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Example Question #979 : Partial Derivatives

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

$f(x,y,z)=5x^3y^4\sqrt{z}$

Possible Answers:

${60xy^4}{\sqrt{z}}$

$\frac{60xy^3}{\sqrt{z}}$

$\frac{120xy^3}{\sqrt{z}}$

${60xy^3}{\sqrt{z}}$

Correct answer:

$\frac{60xy^3}{\sqrt{z}}$

Explanation:

The first thing to do is to find the partial derivative with respect to one variable while everything else behaves as a constant. In this case, you can start by taking the partial derivative with respect to which is:

$15x^2y^4\sqrt{z}$

Then, you can take the partial derivative with respect to again because is expressed twice in the problem:

$30xy^4\sqrt{z}$

Then, you can take the partial derivative with respect to which is:

$120xy^3\sqrt{z}$

Finally, you must take the partial derivative with respect to which is:

$120xy^3(\frac{1}{2}z^\frac{-1}{2})=\frac{60xy^3}{\sqrt{z}}$

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Example Question #980 : Partial Derivatives

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

$f(x,y,z)=z^3x\ln y$

Possible Answers:

$f_{yyz}=\frac{3z^2x}{y^2}$

$f_{yyz}=\frac{-3z^2x}{y^2}$

$f_{yyz}=\frac{-3z^2x}{y}$

$f_{yyz}=\frac{-3zx}{y^2}$

Correct answer:

$f_{yyz}=\frac{-3z^2x}{y^2}$

Explanation:

The first thing to do is to take the partial derivative of one variable while everything else behaves as a constant. In this case, you can take the partial derivative with respect to which is:

$f_y=\frac{z^3x}{y}$

Then, you can take the partial derivative with respect to again because the problem expresses twice. This would be:

$f_{yy}=\frac{-z^3x}{y^2}$

Finally, you must take the partial derivative with respect to which would be:

$f_{yyz}=\frac{-3z^2x}{y^2}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #971 : Partial Derivatives

Example Question #972 : Partial Derivatives

Example Question #973 : Partial Derivatives

Example Question #974 : Partial Derivatives

Example Question #975 : Partial Derivatives

Example Question #976 : Partial Derivatives

Example Question #977 : Partial Derivatives

Example Question #978 : Partial Derivatives

Example Question #979 : Partial Derivatives

Example Question #980 : Partial Derivatives

All Calculus 3 Resources

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