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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #2792 : Calculus 3

Given the function $f(x,y,z)=\sin(x^2+y^3+z^5)$ , find the partial derivative f_y .

Possible Answers:

$f_y=5y^4\cos(x^2+y^3+z^5)$

$f_y=3y^2\sin(x^2+y^3+z^5)$

$f_y=3y^2\cos(x^2+y^3+z^5)$

$f_y=2y\cos(x^2+y^3+z^5)$

Correct answer:

$f_y=3y^2\cos(x^2+y^3+z^5)$

Explanation:

Given the function

$f(x,y,z)=\sin(x^2+y^3+z^5)$ ,

we find f_y by taking the derivative of
$f(x,y,z)=\sin(x^2+y^3+z^5)$

with respect to , which means we hold x,z constant.

So we get (using the chain rule)

$\\f_y=\frac{\partial}{\partial y}f(x,y,z) \\f_y=\frac{\partial}{\partial y}\sin(x^2+y^3+z^5) \\f_y=\cos(x^2+y^3+z^5)\cdot \frac{\partial}{\partial y}(x^2+y^3+z^5)$

$=3y^2\cos(x^2+y^3+z^5)$

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Example Question #2793 : Calculus 3

Given the function $f(x,y,z)=\sin(x^2+y^3+z^5)$ , find the partial derivative f_z .

Possible Answers:

$f_z=3z^2\cos(x^2+y^3+z^5)$

$f_z=5y^4\cos(x^2+y^3+z^5)$

$f_z=5z^4\cos(x^2+y^3+z^5)$

$f_z=-5y^4\cos(x^2+y^3+z^5)$

Correct answer:

$f_z=5z^4\cos(x^2+y^3+z^5)$

Explanation:

Given the function

$f(x,y,z)=\sin(x^2+y^3+z^5)$ ,

we find f_z by taking the derivative of

$f(x,y,z)=\sin(x^2+y^3+z^5)$

with respect to , which means we hold x,y constant. So we get (using the chain rule)

$\\f_z=\frac{\partial}{\partial z}f(x,y,z) \\f_z=\frac{\partial}{\partial z}\sin(x^2+y^3+z^5) \\f_z=\cos(x^2+y^3+z^5)\cdot \frac{\partial}{\partial z}(x^2+y^3+z^5)$

$=5z^4\cos(x^2+y^3+z^5)$

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Example Question #2801 : Calculus 3

Given the function $f(x,y)=x^2\sin e^y$ , find the partial derivative f_x .

Possible Answers:

$f_x=2x\sin e^y$

$f_x=2xe^y\sin e^y$

$f_x=2y\sin e^x$

$f_x=2x\cos e^y$

Correct answer:

$f_x=2x\sin e^y$

Explanation:

To find the partial derivative f_x of $f(x,y)=x^2\sin e^y$ , we take its derivative with respect to while holding constant.

So we get

$\\f_x=\frac{\partial}{\partial x}f(x,y) \\f_x=\frac{\partial}{\partial x}x^2\sin e^y \\f_x=\sin e^y \cdot \frac{\partial}{\partial x}x^2 \\f_x=2x\sin e^y$

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Example Question #432 : Partial Derivatives

Given the function $f(x,y)=x^2\sin e^y$ , find the partial derivative f_y .

Possible Answers:

$f_y=x^2 \cos e^y$

$f_y=2xe^y \cos e^y$

$f_y=x^2e^y \cos e^y$

$f_y=2x \cos e^y$

Correct answer:

$f_y=x^2e^y \cos e^y$

Explanation:

To find the partial derivative f_y of $f(x,y)=x^2\sin e^y$ , we take its derivative with respect to while holding constant.

So we get

$\\f_y=\frac{\partial}{\partial y}f(x,y) \\ \\f_y=\frac{\partial}{\partial y}x^2\sin e^y \\ \\f_y=x^2 \cdot \frac{\partial}{\partial y}\sin e^y \\ \\ f_y=2x\cos e^y\cdot e^y \\ \\f_y=x^2e^y\cos e^y$

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Example Question #433 : Partial Derivatives

Given the function $f(x,y,z)=z^3e^{5x+y^4}$ , find the partial derivative f_x .

Possible Answers:

$f_x=5z^3e^{5x+y^4}$

$f_x=15z^2e^{5x+y^4}$

$f_x=20y^3z^3e^{5x+y^4}$

$f_x=3x^2e^{5x+y^4}$

Correct answer:

$f_x=5z^3e^{5x+y^4}$

Explanation:

Given the function $f(x,y,z)=z^3e^{5x+y^4}$ , we can find the partial derivative f_x by taking its derivative with respect to while holding y,z constant.

So we get

$\\f_x=\frac{\partial}{\partial x}f(x,y,z) \\ \\f_x=\frac{\partial}{\partial x}z^3e^{5x+y^4} \\ \\f_x=z^3\frac{\partial}{\partial x}e^{5x+y^4} \\ \\f_x=z^3e^{5x+y^4}\frac{\partial}{\partial x}(5x+y^4)$

$=5z^3e^{5x+y^4}$

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Example Question #434 : Partial Derivatives

Given the function $f(x,y,z)=z^3e^{5x+y^4}$ , find the partial derivative f_y .

Possible Answers:

$f_y=20y^3z^3e^{5x+y^4}$

$f_y=z^3e^{5+4y^3}$

$f_y=3z^2e^{5x+y^4}$

$f_y=4y^3z^3e^{5x+y^4}$

Correct answer:

$f_y=4y^3z^3e^{5x+y^4}$

Explanation:

Given the function $f(x,y,z)=z^3e^{5x+y^4}$ , we can find the partial derivative f_y by taking its derivative with respect to while holding x,z constant.

So we get

$\\f_y=\frac{\partial}{\partial y}f(x,y,z) \\ \\f_y=\frac{\partial}{\partial y}z^3e^{5x+y^4} \\ \\f_y=z^3\frac{\partial}{\partial y}e^{5x+y^4} \\ \\f_y=z^3e^{5x+y^4}\frac{\partial}{\partial y}(5x+y^4)$

$=4y^3z^3e^{5x+y^4}$

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Example Question #435 : Partial Derivatives

Given the function $f(x,y,z)=z^3e^{5x+y^4}$ , find the partial derivative f_z .

Possible Answers:

$f_z=3z^2e^{5x+y^4}$

$f_z=3z^2e^{5+4y^3}$

$f_z=z^3e^{5+4y^3}$

$f_z=z^3e^{5+y^4}$

Correct answer:

$f_z=3z^2e^{5x+y^4}$

Explanation:

Given the function $f(x,y,z)=z^3e^{5x+y^4}$ , we can find the partial derivative f_z by taking its derivative with respect to while holding x,y constant.

So we get

$\\f_z=\frac{\partial}{\partial z}f(x,y,z) \\ \\f_z=\frac{\partial}{\partial z}z^3e^{5x+y^4} \\ \\f_z=e^{5x+y^4}\frac{\partial}{\partial y}z^3 \\ \\ f_z=3z^2e^{5x+y^4}$

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Example Question #436 : Partial Derivatives

Find the partial derivative f_x of the function $f(x,y,z)=y\ln(x^2+z^5)$ .

Possible Answers:

$f_x=\ln(x^2+z^5)$

$f_x=\frac{2x}{x^2+z^5}$

$f_x=\frac{2xy}{x^2+z^5}$

$f_x=\frac{5yz^4}{x^2+z^5}$

Correct answer:

$f_x=\frac{2xy}{x^2+z^5}$

Explanation:

To find the partial derivative f_x of the function $f(x,y,z)=y\ln(x^2+z^5)$ , we take its derivative with respect to while holding y,z constant.

We use the chain rule to get

$\\f_x=\frac{\partial}{\partial x}f(x,y,z) \\ \\f_x=\frac{\partial}{\partial x}y\ln(x^2+z^5) \\ \\f_x=y\frac{\partial}{\partial x}\ln(x^2+z^5) \\ \\f_x=y\cdot \frac{\frac{\partial}{\partial x}(x^2+z^5)}{x^2+z^5}$

$=\frac{2xy}{x^2+z^5}$

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Example Question #437 : Partial Derivatives

Find the partial derivative f_y of the function $f(x,y,z)=y\ln(x^2+z^5)$ .

Possible Answers:

$f_y=\ln(x^2+z^5)$

$f_y=\frac{5yz^4}{x^2+z^5}$

$f_y=\frac{2xyz^5}{x^2+z^5}$

$f_y=\frac{2xy}{x^2+z^5}$

Correct answer:

$f_y=\ln(x^2+z^5)$

Explanation:

To find the partial derivative f_y of the function $f(x,y,z)=y\ln(x^2+z^5)$ , we take its derivative with respect to while holding x,z constant.

We get

$\\f_y=\frac{\partial}{\partial y}f(x,y,z) \\ \\f_y=\frac{\partial}{\partial y}y\ln(x^2+z^5) \\ \\f_y=\ln(x^2+z^5)\frac{\partial}{\partial y}y \\ \\f_y=\ln(x^2+z^5)$

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Example Question #438 : Partial Derivatives

Given the function $f(x,y,z)=y\ln(x^2+z^5)$ , find the partial derivative f_z .

Possible Answers:

$f_z=\frac{5yz^4}{x^2+z^5}$

$f_z=\ln(x^2+y^5)$

$f_z=\frac{2x}{x^2+z^5}$

$f_z=\frac{10xyz^4}{x^2+z^5}$

Correct answer:

$f_z=\frac{5yz^4}{x^2+z^5}$

Explanation:

To find the partial derivative f_z of the function $f(x,y,z)=y\ln(x^2+z^5)$ , we take its derivative with respect to while holding x,y constant.

We use the chain rule to get

$\\f_z=\frac{\partial}{\partial z}f(x,y,z) \\ \\f_z=\frac{\partial}{\partial z}y\ln(x^2+z^5) \\ \\f_z=y\frac{\partial}{\partial z}\ln(x^2+z^5) \\ \\f_z=y\cdot \frac{\frac{\partial}{\partial z}(x^2+z^5)}{x^2+z^5}$

$=\frac{5yz^4}{x^2+z^5}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #2792 : Calculus 3

Example Question #2793 : Calculus 3

Example Question #2801 : Calculus 3

Example Question #432 : Partial Derivatives

Example Question #433 : Partial Derivatives

Example Question #434 : Partial Derivatives

Example Question #435 : Partial Derivatives

Example Question #436 : Partial Derivatives

Example Question #437 : Partial Derivatives

Example Question #438 : Partial Derivatives

All Calculus 3 Resources

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