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Calculus 3 : Calculus 3

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

Example Questions

Example Question #433 : Partial Derivatives

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

Possible Answers:

$f_x=2x\cos e^y$

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

$f_x=2x\sin e^y$

$f_x=2y\sin e^x$

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 #2801 : Calculus 3

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

Possible Answers:

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

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

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

$f_y=2x \cos e^y$

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 #435 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #2801 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #437 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #438 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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 #2801 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #2802 : Calculus 3

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

Possible Answers:

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

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

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

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

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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Example Question #441 : Partial Derivatives

Find the value of f_x for f(x,y)=x^y at (2,3)

Possible Answers:

Correct answer:

Explanation:

Note that for this problem, we're told to take the derivative with respect to one particular variable. This is known as taking a partial derivative; often it is denoted with the Greek character delta, $\delta$ , or by the subscript of the variable being considered such as f_x or f_y .

For a problem like this, where we presume all variables are independent of each other, we need only consider the variable that we're taking the derivative of the function with respect to; all other variables can be treated as constants.

Taking the partial derivative of f(x,y)=x^y at (2,3)

We find:

$f_x= yx^{y-1}\rightarrow3(2^2)=12$

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Example Question #2803 : Calculus 3

Find the value of f_x for f(x,y)=x^3(y+5) at (2,1)

Possible Answers:

144

Correct answer:

Explanation:

Taking the partial derivative of f(x,y)=x^3(y+5) at (2,1)

We find:

$f_x= 3x^2(y+5)\rightarrow12(6)=72$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #433 : Partial Derivatives

Example Question #2801 : Calculus 3

Example Question #435 : Partial Derivatives

Example Question #2801 : Calculus 3

Example Question #437 : Partial Derivatives

Example Question #438 : Partial Derivatives

Example Question #2801 : Calculus 3

Example Question #2802 : Calculus 3

Example Question #441 : Partial Derivatives

Example Question #2803 : Calculus 3

All Calculus 3 Resources

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