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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1051 : Partial Derivatives

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

Possible Answers:

$ye^y-\sin(y)$

$e^y-\sin(y)$

$-\sin(y)$

$\sin(y)$

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 #3420 : Calculus 3

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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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Example Question #1052 : Partial Derivatives

Calculate the partial derivative with respect to of f(x,y,z)=x^2y^4z^5 at the point (3,1,-2) .

Possible Answers:

864

1620

720

Correct answer:

720

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 are given f(x,y,z)=x^2y^4z^5 , and its partial derivative with respect to can be calculated by treating and and constants and differentiating z^5 :

$\frac{\partial f}{\partial z}=5x^2y^4z^4$ .

Now substitute the values for the point (3,1,-2) into $\frac{\partial f}{\partial z}$ to calculate its value at that point:

$F_z(x,y,z)=5(3)^2(1)^4(-2)^4=5\times9\times1\times16=720$ .

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Example Question #1053 : Partial Derivatives

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

$f(x,y,z)=7e^x^z\ln(y)$

Possible Answers:

$\frac{\partial f}{\partial y}=\frac{7e^x^z}{\ln(y)}$

$\frac{\partial f}{\partial y}=7xze^x^z\ln(y)$

$\frac{\partial f}{\partial y}=\frac{7e^x^z}{y}$

$\frac{\partial f}{\partial y}=\frac{7xze^x^z}{y}$

$\frac{\partial f}{\partial y}=7ye^x^z\ln(y)$

Correct answer:

$\frac{\partial f}{\partial y}=\frac{7e^x^z}{y}$

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(y) of a single variable, and treat the other variables as constants. In this case, we have:

$f(x,y,z)=7e^x^z\ln(y)$

We are being asked to differentiate f(x,y,z) with respect to , so we treat the variables and as constants, recognize that the term 7e^x^z is now just a constant, and apply the rule of differentiation for the natural logarithm to find the partial derivative $\frac{\partial f}{\partial y}$ , as shown:

$\frac{\partial f}{\partial y}=7e^x^z(\ln(y))'$

$=7e^x^z(\frac{1}{y})$

$=\frac{7e^x^z}{y}$

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Example Question #1052 : Partial Derivatives

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

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

Possible Answers:

$2xy^2e^{x^2y^2}+\tan(y)$

$4xye^{x^2y^2}+4x^3y^3e^{x^2y^2}+\sec^2(y)$

$4xye^{x^2y^2}+4x^3y^3e^{x^2y^2}+\sec(y)$

$4xye^{x^2y^2}+4x^2y^2e^{x^2y^2}+\sec^2(y)$

Correct answer:

$4xye^{x^2y^2}+4x^3y^3e^{x^2y^2}+\sec^2(y)$

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

$f_x=2xy^2e^{x^2y^2}+\tan(y)$

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

$f_{xy}=4xye^{x^2y^2}+4x^3y^3e^{x^2y^2}+\sec^2(y)$

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Example Question #1052 : Partial Derivatives

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

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

Possible Answers:

$y^4\cos(xy)$

$-y^4\cos(xy)$

$-y^3\sin(xy)$

Correct answer:

$-y^4\cos(xy)$

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

$f_x=-y^3\sin(xy)$

Next, we take the partial derivative of this function with respect to x:

$f_{xx}=-y^4\cos(xy)$

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Example Question #1052 : Partial Derivatives

Find f_x of the function $f(x,y,z)=xy+z\cos(x)$

Possible Answers:

$x-z\sin(x)$

$y-z\cos(y)$

$y-z\sin(x)$

$y+z\sin(x)$

Correct answer:

$y-z\sin(x)$

Explanation:

To find f_x of the function, you take the partial derivative of the function with respect to x

$\frac{\partial }{\partial x}(xy+z\cos(x))=y-z\sin(x)$

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Example Question #1055 : Partial Derivatives

Find $f_{yy}$ of the function $f(x,y,z)=e^{2xy}+2xy^3$

Possible Answers:

$4x^2e^{2xy}+12x^2$

$4x^2e^{2xy}+12xy$

$x^3e^{2xy}+12xy$

$e^{5xy}+10xy$

Correct answer:

$4x^2e^{2xy}+12xy$

Explanation:

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

$\frac{\partial }{\partial y}(e^{2xy}+2xy^3)=2xe^{2xy}+6xy^2$

$\frac{\partial }{\partial y}(2xe^{2xy}+6xy^2)=4x^2e^{2xy}+12xy$

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Example Question #1053 : Partial Derivatives

Find $(f_{xy})^2$ of the following function:

f(x,y)=y^2x+x^2e^y

Possible Answers:

$4y^2+4x^2e^{2y}$

2y+2xe^y

$y^2+2xye^y+x^2e^{2y}$

$4y^2+8xye^y+4x^2e^{2y}$

Correct answer:

$4y^2+8xye^y+4x^2e^{2y}$

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 take the partial derivative of the function with respect to x:

f_x=y^2+2xe^y

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

$f_{xy}=2y+2xe^y$

Finally, we square this:

$(f_{xy})^2=(2y+2xe^y)^2=4y^2+8xye^y+4x^2e^{2y}$

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Example Question #1054 : Partial Derivatives

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

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

Possible Answers:

$\frac{2}{y^2}-\frac{1}{x^2}$

$\frac{2x}{y^2}+\frac{y}{x}$

$\frac{2}{y^2}-\frac{y}{x^2}$

Correct answer:

$\frac{2}{y^2}-\frac{y}{x^2}$

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 take the partial derivative of the function with respect to x:

$f_x=\frac{2x}{y^2}+\frac{y}{x}$

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

$f_{xx}=\frac{2}{y^2}-\frac{y}{x^2}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1051 : Partial Derivatives

Example Question #3420 : Calculus 3

Example Question #1052 : Partial Derivatives

Example Question #1053 : Partial Derivatives

Example Question #1052 : Partial Derivatives

Example Question #1052 : Partial Derivatives

Example Question #1052 : Partial Derivatives

Example Question #1055 : Partial Derivatives

Example Question #1053 : Partial Derivatives

Example Question #1054 : Partial Derivatives

All Calculus 3 Resources

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