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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #941 : Partial Derivatives

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

$f(x, y)=\sqrt{\cos(xy)+y^2}$

Possible Answers:

$\frac{1}{2}(\cos(xy)+y^2)^{-\frac{1}{2}}(-x\sin(xy)+2y)$

$\frac{1}{2}(\cos(xy)+y^2)^{-\frac{1}{2}}(-\sin(xy)+2y)$

$\frac{1}{2}(\cos(xy)+y^2)^{-\frac{1}{2}}(-x\sin(xy)+y^2)$

$\frac{1}{2}(\cos(xy)+y^2)^{\frac{1}{2}}(-x\sin(xy)+2y)$

Correct answer:

$\frac{1}{2}(\cos(xy)+y^2)^{-\frac{1}{2}}(-x\sin(xy)+2y)$

Explanation:

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

The partial derivative is

$f_y=\frac{1}{2}(\cos(xy)+y^2)^{-\frac{1}{2}}(-x\sin(xy)+2y)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #942 : Partial Derivatives

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

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

Possible Answers:

6xz+2e^y

3xz^2+2ze^y

6xz

6z^2x+2e^y

Correct answer:

6xz+2e^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.

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

f_z=3xz^2+2ze^y

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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

$f_{zz}=6xz+2e^y$

The same rules above were used.

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Example Question #943 : Partial Derivatives

Find $\nabla f$ for the given function:

$f(x, y, z)=\sin(xyz)$

Possible Answers:

$\left \langle yz\cos(xyz), xz\cos(xyz), xy\cos(xyz)\right \rangle$

$\left \langle x\cos(xyz), y\cos(xyz), z\cos(xyz)\right \rangle$

$yz\cos(xyz)+ xz\cos(xyz) + xy\cos(xyz)$

$\left \langle -yz\cos(xyz), -xz\cos(xyz), -xy\cos(xyz)\right \rangle$

Correct answer:

$\left \langle yz\cos(xyz), xz\cos(xyz), xy\cos(xyz)\right \rangle$

Explanation:

The gradient of a function is given by

$\nabla f =\left \langle f_x, f_y, f_z\right \rangle$

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

The partial derivatives are

$f_x=yz\cos(xyz)$

$f_y=xz\cos(xyz)$

$f_z=xy\cos(xyz)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #944 : Partial Derivatives

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

$f(x, y, z)=y\sin(e^{xz^2})$

Possible Answers:

$z^4\sin(e^{xz^2})$

$z^2\cos(e^{xz^2})$

$-z^4\sin(e^{xz^2})$

Correct answer:

$-z^4\sin(e^{xz^2})$

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.

To start, we must find the partial derivative of the function with respect to x:

$f_x=z^2\cos(e^{xz^2})$

The following rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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

$f_{xy}=z^2\cos(e^{xz^2})$

The rule used is stated above.

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

$f_{xyx}=-z^4\sin(e^{xz^2})$

The rules used are stated above, along with

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #945 : Partial Derivatives

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

$f(x,y,z)=y^3z^4\cos(x)$

Possible Answers:

$24z^3y\cos(x)$

$24z^3y\sin(x)$

$12z^3y^2\cos(x)$

$-24z^3y\sin(x)$

Correct answer:

$-24z^3y\sin(x)$

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 z:

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

The rules used were

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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

$f_{zy}=12z^3y^2\cos(x)$

The same rules above were used.

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

$f_{zyy}=24z^3y\cos(x)$

The same rules above were used.

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

$f_{zyyx}=-24z^3y\sin(x)$

The same rules above were used along with

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #946 : Partial Derivatives

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

$f(x,y,z)=\cos(z)e^zy$

Possible Answers:

$-\sin(z)e^z+\cos(z)e^z$

$\sin(z)e^z+\cos(z)e^z$

$y(-\sin(z)e^z+\cos(z)e^z)$

$\sin(z)e^z$

Correct answer:

$-\sin(z)e^z+\cos(z)e^z$

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 z:

$f_z=(-\sin(z)e^z+\cos(z)e^z)y$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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

$f_{zy}=-\sin(z)e^z+\cos(z)e^z$

The rule used is stated above.

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Example Question #947 : Partial Derivatives

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

$f(x, y, z)=y^3\cos(z)$

Possible Answers:

$6y\cos(z)$

$3y^2\cos(z)$

Correct answer:

$6y\cos(z)$

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:

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

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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

$f_{yy}=6y\cos(z)$

The derivative was found using the same rules as above.

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Example Question #948 : Partial Derivatives

Find $f_{xyx}$ of the function x^2y+3xz^3 .

Possible Answers:

3z^3

Correct answer:

Explanation:

To find $f_{xyx}$ , you must perform three partial derivatives consecutively:

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

The first derivative, using

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ ,

and treating all other variables as constants, we get

2xy+3z^3 .

Taking $\frac{\partial }{\partial y}$ of that gets you .

Finally, taking $\frac{\partial }{\partial x}$ of the previous expression gets you .

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Example Question #949 : Partial Derivatives

Compute $f_{xxy}$ of the function 3x^3y+xyz+xz^3 .

Possible Answers:

18x

18y

9x^2y

9x^2+xyz

Correct answer:

18x

Explanation:

In order to solve, you must take a total of three derivatives: the first is $\frac{\partial }{\partial x}$ , then again ${\frac{\partial }{\partial x}}$ , and finally $\frac{\partial }{\partial y}$ , in that order (the notation in the problem statement dictates that).

The first derivative you obtain will be

9x^2y+yz+z^3 .

The subsequent derivative is 18xy .

The final derivative with respect to y is 18x .

The rule used for all derivatives is

$\frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}$ ,

and we treat all other variables as constants.

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Example Question #941 : Partial Derivatives

Find the determinant of the 2x2 matrix:

$\begin{bmatrix} 1&4 \\ 3 &11 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

The formula for the determinant for a matrix

$\begin{bmatrix} a&b \\ c &d \end{bmatrix}$

ad-cb .

Using our values, this becomes

(1*11)-(4*3)=-1

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #941 : Partial Derivatives

Example Question #942 : Partial Derivatives

Example Question #943 : Partial Derivatives

Example Question #944 : Partial Derivatives

Example Question #945 : Partial Derivatives

Example Question #946 : Partial Derivatives

Example Question #947 : Partial Derivatives

Example Question #948 : Partial Derivatives

Example Question #949 : Partial Derivatives

Example Question #941 : Partial Derivatives

All Calculus 3 Resources

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