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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 #1041 : Partial Derivatives

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

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

Possible Answers:

$3x^2+2y^3\sec(xy)\tan(xy)$

$6x+y^2\sec(xy)\tan(xy)$

$6x+2y^3\sec(xy)\tan(xy)$

$6x+y^3\sec(xy)\tan(xy)$

$6x+2y^2\sec(xy)\tan(xy)$

Correct answer:

$6x+2y^3\sec(xy)\tan(xy)$

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 x:

$f_x=3x^2+y^2\sec^2(xy)$

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

$f_{xx}=6x+2y^3\sec(xy)\tan(xy)$

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Example Question #1042 : Partial Derivatives

Find $f_{xxz}$ of the function $x^2z^2+x^3\cos(z)$

Possible Answers:

$4z+6x\sin(z)$

$4z-x\cos(z)$

$4z-6x\sin(z)$

$4z^2-6x\sin(z)$

Correct answer:

$4z-6x\sin(z)$

Explanation:

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

$\frac{\partial }{\partial x}(x^2z^2+x^3\cos(z))=2xz^2+3x^2\cos(z)$

$\frac{\partial }{\partial x}(2xz^2+3x^2\cos(z))=2z^2+6x\cos(z)$

$\frac{\partial }{\partial z}(2z^2+6x\cos(z))=4z-6x\sin(z)$

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Example Question #1043 : Partial Derivatives

Find $f_{xxz}$ of the function $5x^2z^5+ze^{5x}$

Possible Answers:

$50z^4-25e^{10x}$

$50z^3+20e^{5x}$

$50z^4+25e^{5x}$

$50z^5+20e^{4x}$

Correct answer:

$50z^4+25e^{5x}$

Explanation:

To find $f_{xxz}$ of the function, you must take three consecutive partial derivatives:

$\frac{\partial }{\partial x}(5x^2z^5+ze^{5x})=10xz^5+5ze^{5x}$

$\frac{\partial }{\partial x}(10xz^5+5ze^{5x})=10z^5+25ze^{5x}$

$\frac{\partial }{\partial z}(10z^5+25ze^{5x})=50z^4+25e^{5x}$

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Example Question #1044 : Partial Derivatives

Find $\frac{\partial F}{\partial y}$ given

$F(x,y)=e^x+e^y+cos\,x+sin\,y$

Possible Answers:

$\frac{\partial F}{\partial y}=e^{xy}+cos\,y-sin\,x$

$\frac{\partial F}{\partial y}=e^x-sin\,x$

$\frac{\partial F}{\partial y}=e^y+cos\,y$

Correct answer:

$\frac{\partial F}{\partial y}=e^y+cos\,y$

Explanation:

To find $\frac{\partial F}{\partial y}$ we take the derivative of the function F(x,y) with respect to and treat the other variables as constants.

As such,

$\frac{\partial }{\partial y}[F(x,y)]=\frac{\partial }{\partial y}[e^x+e^y+cos\,x+sin\,y]$

yields

$\frac{\partial F}{\partial y}=e^y+cos\,y$

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Example Question #1045 : Partial Derivatives

Find $\frac{\partial F}{\partial x}$ given

$F(x,y)=tan\,x-cot\,y+y^2$

Possible Answers:

$\frac{\partial F}{\partial x}=csc^2\,y+2y$

$\frac{\partial F}{\partial x}=0$

$\frac{\partial F}{\partial x}=sec^2\,x$

Correct answer:

$\frac{\partial F}{\partial x}=sec^2\,x$

Explanation:

To find $\frac{\partial F}{\partial x}$ we take the derivative of the function F(x,y) with respect to and treat the other variables as constants.

As such,

$\frac{\partial }{\partial x}[F(x,y)]=\frac{\partial }{\partial x}[tan\,x-cot\,y+y^2]$

yields

$\frac{\partial F}{\partial x}=sec^2\,x$

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Example Question #1042 : Partial Derivatives

Find $\frac{\partial F}{\partial x}$ given

$F(x,y)=sec\,x+csc\,y$

Possible Answers:

$\frac{\partial F}{\partial x}=-csc(x)cot(x)$

$\frac{\partial F}{\partial x}=sec(x)tan(x)$

$\frac{\partial F}{\partial x}=sec^2\,x$

Correct answer:

$\frac{\partial F}{\partial x}=sec(x)tan(x)$

Explanation:

To find $\frac{\partial F}{\partial x}$ we take the derivative of the function F(x,y) with respect to and treat the other variables as constants.

As such,

$\frac{\partial }{\partial x}[F(x,y)]=\frac{\partial }{\partial x}[sec\,x+csc\,y]$

yields

$\frac{\partial F}{\partial x}=sec(x)tan(x)$

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Example Question #1047 : Partial Derivatives

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

f(x,y,z)=z^3y^4+xz^2y

Possible Answers:

12y^3z^2+2xz

36y^2z

72y^2z

36y^2z^2

Correct answer:

72y^2z

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=4y^3z^3+xz^2

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

$f_{yz}=12y^3z^2+2xz$

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

$f_{yzy}=36y^2z^2$

Finally, we take the partial derivative of this function with respect to z:

$f_{yzyz}=72y^2z$

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Example Question #1048 : Partial Derivatives

Find $f_{xy}$ of the function $x^3\sin(y)+xyz$

Possible Answers:

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

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

$3x^2\cos(y)-z$

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #3411 : Calculus 3

Find $f_{xy}$ of the function $e^x+x\cos(y)$

Possible Answers:

$\sin(y)$

$-\sin(y)$

$e^y-\sin(y)$

$ye^y-\sin(y)$

Correct answer:

$-\sin(y)$

Explanation:

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

$\frac{\partial }{\partial x}(e^x+x\cos(y))=e^x+\cos(y)$

$\frac{\partial }{\partial y}(e^x+\cos(y))=-\sin(y)$

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Example Question #1052 : Partial Derivatives

Calculate the partial derivative with respect to of the following function:

$f(x,y)=x^2y^2+\frac{x}{x+y}$

Possible Answers:

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

$\frac{\partial f}{\partial x}=2xy^2+\frac{1}{(x+y)^2}$

$\frac{\partial f}{\partial x}=x(2xy-\frac{1}{(x+y)^2})$

$\frac{\partial f}{\partial x}=2xy-\frac{y}{(x+y)^2}$

$\frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})$

Correct answer:

$\frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})$

Explanation:

When calculating the partial derivative with respect to the variable of a function of more than one variable, apply the standard rules for differentiating a function f(x) of a single variable, and treat the other variables as constants. In this case, we have:

$f(x,y)=x^2y^2+\frac{x}{x+y}$

We are being asked to differentiate f(x,y) with respect to , so we treat the variable as a constant and apply the sum and quotient rules of differentiation to find the partial derivative $\frac{\partial f}{\partial x}$ , as shown:

$\frac{\partial f}{\partial x}=2x(y^2)+\frac{(1)(x+y)-(1)(x)}{(x+y)^2}$

$=2xy^2+\frac{x+y-x}{(x+y)^2}$

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

Since is treated as a constant, we can factor out from the sum of these two terms and simplify the expression:

$\frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1041 : Partial Derivatives

Example Question #1042 : Partial Derivatives

Example Question #1043 : Partial Derivatives

Example Question #1044 : Partial Derivatives

Example Question #1045 : Partial Derivatives

Example Question #1042 : Partial Derivatives

Example Question #1047 : Partial Derivatives

Example Question #1048 : Partial Derivatives

Example Question #3411 : Calculus 3

Example Question #1052 : Partial Derivatives

All Calculus 3 Resources

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