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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1001 : Partial Derivatives

Find f_y of the following function:

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

Possible Answers:

$-\sin(y)+y+2yz$

$-\sin(y)+3y$

$-\sin(y)+x+2yz$

$\sin(y)+x+2yz$

Correct answer:

$-\sin(y)+x+2yz$

Explanation:

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

The partial derivative of the function with respect to y is

$f_y=-\sin(y)+x+2yz$

The derivative was found using the following rules:

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

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Example Question #1002 : Partial Derivatives

Find $f_{xxx}$ of the function $x^5yz+\sin(x)$

Possible Answers:

$60x^2yz+\cos(x)$

$50x^2yz-\cos(x)$

$60x^2yz-\cos(x)$

$20x^3yx+\sin(x)$

Correct answer:

$60x^2yz-\cos(x)$

Explanation:

To find $f_{xxx}$ , we take three consecutive partial derivatives: $\frac{\partial }{\partial x},\frac{\partial }{\partial x},\frac{\partial }{\partial x}$

$\frac{\partial }{\partial x}(x^5yz+\sin(x))=5x^4yz+\cos(x)$

$\frac{\partial }{\partial x}\left ( 5x^4yz+\cos(x) \right )=20x^3yz-\sin(x)$

$\frac{\partial }{\partial x}(20x^3yz-\sin(x))=60x^2yz-\cos(x)$

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Example Question #1001 : Partial Derivatives

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

$f(x,y,z)=5x^3e^{5y}\tan z$

Possible Answers:

$f_{xxy}=30xe^{5y}\tan z$

$f_{xxy}=150xe^{5y}\tan z$

$f_{xxy}=150xe^{5y}\sec^2 z$

$f_{xxy}=-150xe^{5y}\tan z$

Correct answer:

$f_{xxy}=150xe^{5y}\tan z$

Explanation:

The way to solve this problem is by taking consecutive partial derivatives while everything else in the function behaves as a constant and is therefore not differentiated. For example, you can start by taking the partial derivative of the function with respect to x:

$f_{x}=15x^2e^{5y}\tan z$

Then, you can take the partial derivative with respect to x again because x is stated twice:

$f_{xx}=30xe^{5y}\tan z$

Finally, you must take the partial derivative of the function with respect to y to get the final answer:

$f_{xxy}=150xe^{5y}\tan z$

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Example Question #1003 : Partial Derivatives

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

$f(x,y,z)=14x^4y^3\tan z$

Possible Answers:

$f_{yyz}=84x^4y\sec^2 z$

$f_{yyz}=-84x^4y\sec^2 z$

$f_{yyz}=42x^4y^2\sec^2 z$

$f_{yyz}=84x^4y^2\sec^2 z\tan z$

Correct answer:

$f_{yyz}=84x^4y\sec^2 z$

Explanation:

$f_{y}=42x^4y^2\tan z$

Then, you can take the partial derivative with respect to y again because y is stated twice:

$f_{yy}=84x^4y\tan z$

Finally, you must take the partial derivative of the function with respect to z to get the final answer:

$f_{yyz}=84x^4y\sec^2 z$

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Example Question #3371 : Calculus 3

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

$f(x,y,z)=e^{5x}y^3\csc z$

Possible Answers:

$f_{xyz}=3e^{5x}y^2\csc z\cot z$

$f_{xyz}=15e^{5x}y^2\csc z\cot z$

$f_{xyz}=-15e^{5x}y^2\csc z\cot z$

$f_{xyz}=15e^{5x}y^2\cot^2 z$

Correct answer:

$f_{xyz}=-15e^{5x}y^2\csc z\cot z$

Explanation:

$f_{x}=5e^{5x}y^3\csc z$

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

$f_{xy}=15e^{5x}y^2\csc z$

Finally, you must take the partial derivative of the function with respect to z to get the final answer:

$f_{xyz}=-15e^{5x}y^2\csc z\cot z$

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Example Question #3371 : Calculus 3

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

$f(x,y,z)=\arctan(5x)ye^{2z}$

Possible Answers:

$f_{xzz}=\frac{20ye^{2z}}{\sqrt{1-25x^2}}$

$f_{xzz}=\frac{20ye^{2z}}{1+25x^2}$

$f_{xzz}=-\frac{20ye^{2z}}{1+25x^2}$

$f_{xzz}=\frac{10ye^{2z}}{1+25x^2}$

Correct answer:

$f_{xzz}=\frac{20ye^{2z}}{1+25x^2}$

Explanation:

$f_{x}=\frac{5ye^{2z}}{1+25x^2}$

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

$f_{xz}=\frac{10ye^{2z}}{1+25x^2}$

Finally, you must take the partial derivative of the function with respect to z to get the final answer because z is expressed twice in the question:

$f_{xzz}=\frac{20ye^{2z}}{1+25x^2}$

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Example Question #3371 : Calculus 3

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

$f(x,y,z)=\ln(2x)y^2\arcsin z$

Possible Answers:

$f_{xxz}=-\frac{y^2}{x^2\sqrt{1-z^2}}$

$f_{xxz}=-\frac{y^2}{x^2(1+z^2)}$

$f_{xxz}=-\frac{y^2}{2x^2\sqrt{1-z^2}}$

$f_{xxz}=\frac{y^2}{x^2\sqrt{1-z^2}}$

Correct answer:

$f_{xxz}=-\frac{y^2}{x^2\sqrt{1-z^2}}$

Explanation:

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

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

$f_{xx}=-\frac{y^2\arcsin z}{x^2}$

Finally, you must take the partial derivative of the function with respect to z to get the final answer:

$f_{xxz}=-\frac{y^2}{x^2\sqrt{1-z^2}}$

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Example Question #3372 : Calculus 3

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

$f(x,y,z)=6^{x}\tan (y) z$

Possible Answers:

$f_{xyy}=2\ln6(6^x)(\sec^2 y\tan y)z$

$f_{xyy}=12(6^x)(\sec^2 y\tan y)z$

$f_{xyy}=2\ln6(6^x)(\sec^2 y\tan y)$

$f_{xyy}=-2\ln6(6^x)(\sec^2 y\tan y)z$

Correct answer:

$f_{xyy}=2\ln6(6^x)(\sec^2 y\tan y)z$

Explanation:

$f_{x}=\ln 6(6^x)\tan(y)z$

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

$f_{xy}=\ln 6(6^x)\sec^2(y)z$

Finally, you must take the partial derivative of the function with respect to y to get the final answer again because y is expressed twice in the question:

$f_{xyy}=2\ln 6(6^x)(\sec^2 y\tan y)z$

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Example Question #3372 : Calculus 3

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

$f(x,y,z)=6xy^4\sin z$

Possible Answers:

$f_{yyz}=12xy^2\cos z$

$f_{yyz}=72xy^3\cos z$

$f_{yyz}=-72xy^2\cos z$

$f_{yyz}=72xy^2\cos z$

Correct answer:

$f_{yyz}=72xy^2\cos z$

Explanation:

$f_{y}=24xy^3\sin z$

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

$f_{yy}=72xy^2\sin z$

Finally, you must take the partial derivative of the function with respect to z to get the final answer:

$f_{yyz}=72xy^2\cos z$

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Example Question #3373 : Calculus 3

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

$f(x,y,z)=3x\cot(y)z$

Possible Answers:

$f_{yyz}=-6x(\csc^2 y\cot y)$

$f_{yyz}=6x(\csc^2 y\cot y)$

$f_{yyz}=6x(\csc^2 y\cot y)z$

$f_{yyz}=3x(\csc^2 y\cot y)$

Correct answer:

$f_{yyz}=6x(\csc^2 y\cot y)$

Explanation:

$f_{y}=-3x\csc^2 (y)z$

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

$f_{yy}=6x(\csc^2 y\cot y)z$

Finally, you must take the partial derivative of the function with respect to z to get the final answer:

$f_{yyz}=6x(\csc^2 y\cot y)$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1001 : Partial Derivatives

Example Question #1002 : Partial Derivatives

Example Question #1001 : Partial Derivatives

Example Question #1003 : Partial Derivatives

Example Question #3371 : Calculus 3

Example Question #3371 : Calculus 3

Example Question #3371 : Calculus 3

Example Question #3372 : Calculus 3

Example Question #3372 : Calculus 3

Example Question #3373 : Calculus 3

All Calculus 3 Resources

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