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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #991 : Partial Derivatives

Calculate the partial derivative with respect to x, f_x(x, y), of the following function:

$f(x, y)=xy+3x^2y+\sin(xy)$

Possible Answers:

$x[1+3x+\cos(xy)]$

$y[1+6x+\cos(xy)]$

$x+3x^2+\cos(xy)$

$y+6xy+\cos(xy)$

Correct answer:

$y[1+6x+\cos(xy)]$

Explanation:

The partial derivative of f(x,y) is calculated by treating y as a constant. The third term requires an application of the chain rule:

$\frac{\partial f(x,y)}{\partial x} =y+6xy +\cos(xy)y=y[1+6x+\cos(xy)]$

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Example Question #992 : Partial Derivatives

Calculate the partial derivative with respect to y, fy(x, y), of the following function:

$f(x, y)=xy+3x^2y+\sin(xy)$

Possible Answers:

$x[1+3x+\cos(xy)]$

$y[1+6x+\cos(xy)]$

$y+6xy+\cos(xy)$

$x+3x^2+\cos(xy)$

Correct answer:

$x[1+3x+\cos(xy)]$

Explanation:

The partial derivative of f(x,y) is calculated by treating x as a constant. The third term requires an application of the chain rule:

$\frac{\partial f(x,y)}{\partial y} =x+3x^2+\cos(xy)x=x[1+3x+\cos(xy)]$

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Example Question #995 : Partial Derivatives

Find $f_{xyy}$ of the function 3xy^3+xy^2z

Possible Answers:

18y+2z

9y^2+2yz

11y

y+3z

Correct answer:

18y+2z

Explanation:

To find $f_{xyy}$ of the function, we take three consecutive partial derivatives:

$\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial y}$ . Applying to the function given, we get:

$\frac{\partial }{\partial x}(3xy^3+xy^2z)=3y^3+y^2z$

$\frac{\partial }{\partial y}(3y^3+y^2z)=9y^2+2yz$

$\frac{\partial }{\partial y}(9y^2+2yz)=18y+2z$

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Example Question #996 : Partial Derivatives

Find $f_{xyy}$ of the function x^3y^5+xyz

Possible Answers:

60xyz

15x^2y^4+z

30x^4y

60x^2y^3

Correct answer:

60x^2y^3

Explanation:

o find $f_{xyy}$ of the function, we take three consecutive partial derivatives:

$\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial y}$ . Applying to the function given, we get:

$\frac{\partial }{\partial x}(x^3y^5+xyz)=3x^2y^5+yz$

$\frac{\partial }{\partial y}(3x^2y^5+yz)=15x^2y^4+z$

$\frac{\partial }{\partial y}(15x^2y^4+z)=60x^2y^3$

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Example Question #997 : Partial Derivatives

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

f(x,y,z)=12xyz

Possible Answers:

12yz

12z

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.

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

f_x=12yz

The derivative was found using the following rule:

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

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

$f_{xy}=12z$

The rule above was used.

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

$f_{xyy}=0$

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Example Question #998 : Partial Derivatives

Find $f_{yzz}$ of the function xy^2z+2yz^3

Possible Answers:

2xy

2z^3

12z

Correct answer:

12z

Explanation:

To find $f_{yzz}$ , you take three consecutive partial derivatives: $\frac{\partial }{\partial y},\frac{\partial }{\partial z},\frac{\partial }{\partial z}$

$\frac{\partial }{\partial y}(xy^2z+2yz^3)=2xy+2z^3$

$\frac{\partial }{\partial z}(2xyz+2z^3)=2xy+6z^2$

$\frac{\partial }{\partial z}(2xy+6z^2)=12z$

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Example Question #999 : Partial Derivatives

Find $f_{yzz}$ of the function $2y\sin(z)+xy^2z^2$

Possible Answers:

$2\cos(z)$

$2\sin(z)+2xyz^2$

$-2\cos(z)+4xyz$

$-2\sin(z)+4xy$

Correct answer:

$-2\sin(z)+4xy$

Explanation:

To find $f_{yzz}$ , you take three consecutive partial derivatives: $\frac{\partial }{\partial y},\frac{\partial }{\partial z},\frac{\partial }{\partial z}$

$\frac{\partial }{\partial y}(2y\sin(z)+xy^2z^2)=2\sin(z)+2xyz^2$

$\frac{\partial }{\partial z}(2\sin(z)+2xyz^2)=2\cos(z)+4xyz$

$\frac{\partial }{\partial z}(2\cos(z)+4xyz)=-2\sin(z)+4xy$

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Example Question #1000 : Partial Derivatives

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

$f(x, y, z)=e^{yz}+x^3$

Possible Answers:

$ye^{yz}$

$e^{yz}$

$e^{yz}(1+y)$

Correct answer:

$e^{yz}(1+y)$

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 y:

$f_y=ze^{yz}$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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

$f_{yz}=e^{yz}+ye^{yz}=e^{yz}(y+1)$

The derivative was found using the above rules as well as

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

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Example Question #1001 : Partial Derivatives

Find f_y of the following function:

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

Possible Answers:

$-\sin(y)+3y$

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

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

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

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:

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

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

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

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

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #991 : Partial Derivatives

Example Question #992 : Partial Derivatives

Example Question #995 : Partial Derivatives

Example Question #996 : Partial Derivatives

Example Question #997 : Partial Derivatives

Example Question #998 : Partial Derivatives

Example Question #999 : Partial Derivatives

Example Question #1000 : Partial Derivatives

Example Question #1001 : Partial Derivatives

Example Question #1002 : Partial Derivatives

All Calculus 3 Resources

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