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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1021 : Partial Derivatives

Find f_x $\displaystyle f_x$ of the following function:

$f(x, y, z)=\cos(xz)e^{xyz}$ $\displaystyle f(x, y, z)=\cos(xz)e^{xyz}$

Possible Answers:

$-z\sin(xz)+yz\cos(xz)$ $\displaystyle -z\sin(xz)+yz\cos(xz)$

$e^{xyz}(-z\sin(xz)+\cos(xz))$ $\displaystyle e^{xyz}(-z\sin(xz)+\cos(xz))$

$e^{xyz}(-z\sin(xz)+yz\cos(xz))$ $\displaystyle e^{xyz}(-z\sin(xz)+yz\cos(xz))$

$e^{xyz}(z\sin(xz)+yz\cos(xz))$ $\displaystyle e^{xyz}(z\sin(xz)+yz\cos(xz))$

Correct answer:

$e^{xyz}(-z\sin(xz)+yz\cos(xz))$ $\displaystyle e^{xyz}(-z\sin(xz)+yz\cos(xz))$

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 x is

$f_x=e^{xyz}(-z\sin(xz)+yz\cos(xz))$ $\displaystyle f_x=e^{xyz}(-z\sin(xz)+yz\cos(xz))$

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Example Question #3385 : Calculus 3

Find $f_{yx}$ $\displaystyle f_{yx}$ of the following function:

$f(x, y, z)=xyz^2+3y^3e^{xy}$ $\displaystyle f(x, y, z)=xyz^2+3y^3e^{xy}$

Possible Answers:

$z^2+15y^3e^{xy}$ $\displaystyle z^2+15y^3e^{xy}$

$\displaystyle 0$

$z^2+e^{xy}(12y^3+3y^4)$ $\displaystyle z^2+e^{xy}(12y^3+3y^4)$

$z^2+3xy^3e^{xy}$ $\displaystyle z^2+3xy^3e^{xy}$

$z^2+e^{xy}(12y^3+3xy^4)$ $\displaystyle z^2+e^{xy}(12y^3+3xy^4)$

Correct answer:

$z^2+e^{xy}(12y^3+3xy^4)$ $\displaystyle z^2+e^{xy}(12y^3+3xy^4)$

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 must find the partial derivative of the function with respect to y:

$f_y=xz^2+9y^2e^{xy}+3xy^3e^{xy}$ $\displaystyle f_y=xz^2+9y^2e^{xy}+3xy^3e^{xy}$

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

$f_{yx}=z^2+e^{xy}(12y^3+3xy^4)$ $\displaystyle f_{yx}=z^2+e^{xy}(12y^3+3xy^4)$

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Example Question #3391 : Calculus 3

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

$\displaystyle f(x, y, z)=x^2+2xy^2+3yz^$

Possible Answers:

2+4y $\displaystyle 2+4y$

$\displaystyle 6x$

2x+4xy $\displaystyle 2x+4xy$

$\displaystyle 0$

Correct answer:

2+4y $\displaystyle 2+4y$

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:

$\displaystyle f_x=2x+4xy$

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

$f_{xx}=2+4y$ $\displaystyle f_{xx}=2+4y$

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Example Question #3392 : Calculus 3

Find $f_{xy}$ $\displaystyle f_{xy}$ for the following function:

$f(x,y,z)=z^3\sin(xyz)+x^2y^2z$ $\displaystyle f(x,y,z)=z^3\sin(xyz)+x^2y^2z$

Possible Answers:

$z^4\cos(xyz)+xyz^5\sin(xyz)+4xyz$ $\displaystyle z^4\cos(xyz)+xyz^5\sin(xyz)+4xyz$

$z^4\cos(xyz)-xyz^5\sin(xyz)+4xyz$ $\displaystyle z^4\cos(xyz)-xyz^5\sin(xyz)+4xyz$

$\displaystyle 0$

$z^4\cos(xyz)-yz^4\sin(xyz)+4xyz$ $\displaystyle z^4\cos(xyz)-yz^4\sin(xyz)+4xyz$

Correct answer:

$z^4\cos(xyz)-xyz^5\sin(xyz)+4xyz$ $\displaystyle z^4\cos(xyz)-xyz^5\sin(xyz)+4xyz$

Explanation:

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

We must find the partial derivative of the function with respect to x:

$f_x=yz^4\cos(xyz)+2xy^2z$ $\displaystyle f_x=yz^4\cos(xyz)+2xy^2z$

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

$f_{xy}=z^4\cos(xyz)-xyz^5\sin(xyz)+4xyz$ $\displaystyle f_{xy}=z^4\cos(xyz)-xyz^5\sin(xyz)+4xyz$

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Example Question #3393 : Calculus 3

Find $f_{xy}$ $\displaystyle f_{xy}$ of the function $\displaystyle xy^2+3xyz$

Possible Answers:

2y^2+3y $\displaystyle 2y^2+3y$

y^2+3z $\displaystyle y^2+3z$

2y+3z $\displaystyle 2y+3z$

y+3 $\displaystyle y+3$

Correct answer:

2y+3z $\displaystyle 2y+3z$

Explanation:

To find $f_{xy}$ $\displaystyle f_{xy}$ of the function, you must take two consecutive partial derivatives:

$\frac{\partial }{\partial x},\frac{\partial }{\partial y}$ $\displaystyle \frac{\partial }{\partial x},\frac{\partial }{\partial y}$

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

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

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Example Question #3394 : Calculus 3

Find $f_{xyy}$ $\displaystyle f_{xyy}$ of the following function: $x\cos(y)+4x^2y^3$ $\displaystyle x\cos(y)+4x^2y^3$

Possible Answers:

$-\cos(y)+48xy$ $\displaystyle -\cos(y)+48xy$

$-\cos(y)+4xy$ $\displaystyle -\cos(y)+4xy$

$-\sin(y)+48xy$ $\displaystyle -\sin(y)+48xy$

$-\cos(y)+16xyz$ $\displaystyle -\cos(y)+16xyz$

Correct answer:

$-\cos(y)+48xy$ $\displaystyle -\cos(y)+48xy$

Explanation:

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

$\frac{\partial }{\partial x}(x\cos(y)+4x^2y^3)=\cos(y)+8xy^3$ $\displaystyle \frac{\partial }{\partial x}(x\cos(y)+4x^2y^3)=\cos(y)+8xy^3$

$\frac{\partial }{\partial y}(\cos(y)+8xy^3)=-\sin(y)+24xy^2$ $\displaystyle \frac{\partial }{\partial y}(\cos(y)+8xy^3)=-\sin(y)+24xy^2$

$\frac{\partial }{\partial y}(-\sin(y)+24xy^2)=-\cos(y)+48xy$ $\displaystyle \frac{\partial }{\partial y}(-\sin(y)+24xy^2)=-\cos(y)+48xy$

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Example Question #3395 : Calculus 3

Find $f_{zyz}$ $\displaystyle f_{zyz}$ for the following function:

$\displaystyle f(x, y, z)=8z^2e^zy$

Possible Answers:

$\displaystyle e^zy(8z^2+32z+16)$

$\displaystyle e^z(z+8z^2)$

$\displaystyle 8z^2+32z+16$

$\displaystyle e^z(8z^2+32z+16)$

Correct answer:

$\displaystyle e^z(8z^2+32z+16)$

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 must find the partial derivative of the function with respect to z:

$\displaystyle f_z=y(16ze^z+8z^2e^z)$

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

$f_{zy}=16ze^z+8z^2e^z$ $\displaystyle f_{zy}=16ze^z+8z^2e^z$

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

$f_{zyz}=16e^z+16ze^z+16ze^z+8z^2e^z=e^z(16+32z+8z^2)$ $\displaystyle f_{zyz}=16e^z+16ze^z+16ze^z+8z^2e^z=e^z(16+32z+8z^2)$

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Example Question #3396 : Calculus 3

Find $f_{yy}$ $\displaystyle f_{yy}$ for the following function:

$\displaystyle f(x, y)=5x+4xy^2$

Possible Answers:

$\displaystyle 8x$

5+4y^2 $\displaystyle 5+4y^2$

8xy $\displaystyle 8xy$

$\displaystyle 0$

$\displaystyle 8y$

Correct answer:

$\displaystyle 8x$

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=8xy $\displaystyle f_y=8xy$

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

$f_{yy}=8x$ $\displaystyle f_{yy}=8x$

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Example Question #3397 : Calculus 3

Find $f_{xy}$ $\displaystyle f_{xy}$ for the following function:

$\displaystyle f(x, y)=x^2+5xy^2+20xy+4y$

Possible Answers:

10y+24 $\displaystyle 10y+24$

10y+20 $\displaystyle 10y+20$

$\displaystyle 0$

$\displaystyle 2x+10y+20$

Correct answer:

10y+20 $\displaystyle 10y+20$

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:

$\displaystyle f_x=2x+5y^2+20y$

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

$f_{xy}=10y+20$ $\displaystyle f_{xy}=10y+20$

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Example Question #1022 : Partial Derivatives

Find f_x $\displaystyle f_x$ for the following function:

$f(x, y, z)=x^2y+\sqrt{xz}$ $\displaystyle f(x, y, z)=x^2y+\sqrt{xz}$

Possible Answers:

$2xy+\frac{z}{2}(xz)^{\frac{1}{2}}$ $\displaystyle 2xy+\frac{z}{2}(xz)^{\frac{1}{2}}$

$2xy+\frac{z}{2}(xz)^{-\frac{1}{2}}$ $\displaystyle 2xy+\frac{z}{2}(xz)^{-\frac{1}{2}}$

$2xy+\frac{(xz)^{-\frac{1}{2}}}{2}$ $\displaystyle 2xy+\frac{(xz)^{-\frac{1}{2}}}{2}$

$2xy+\sqrt{z}$ $\displaystyle 2xy+\sqrt{z}$

Correct answer:

$2xy+\frac{z}{2}(xz)^{-\frac{1}{2}}$ $\displaystyle 2xy+\frac{z}{2}(xz)^{-\frac{1}{2}}$

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 x is

$f_x=2xy+\frac{z}{2}(xz)^{-\frac{1}{2}}$ $\displaystyle f_x=2xy+\frac{z}{2}(xz)^{-\frac{1}{2}}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1021 : Partial Derivatives

Example Question #3385 : Calculus 3

Example Question #3391 : Calculus 3

Example Question #3392 : Calculus 3

Example Question #3393 : Calculus 3

Example Question #3394 : Calculus 3

Example Question #3395 : Calculus 3

Example Question #3396 : Calculus 3

Example Question #3397 : Calculus 3

Example Question #1022 : Partial Derivatives

All Calculus 3 Resources

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