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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #12 : Differentials

Find the differential of the function:

$k=z^3\tan(y^2)$

Possible Answers:

$2yz^3\sec^2(y^2)+3z^2\tan(y^2)$

$2yz^3\sec^2(y^2)dy+3z^2\tan(y^2)dz$

$2z^3\sec^2(y^2)dy+3z^2\tan(y^2)dz$

$2y\sec^2(y^2)dy+3z^2\tan(y^2)dz$

Correct answer:

$2yz^3\sec^2(y^2)dy+3z^2\tan(y^2)dz$

Explanation:

The differential of a function k=f(x,y, z) is given by

dk=f_xdx+f_ydy+f_zdz

The partial derivatives of the function are

f_x=0

$f_y=2yz^3\sec^2(y^2)$

$f_z=3z^2\tan(y^2)$

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Example Question #11 : Differentials

Find the differential of the function:

g=2x^3yz

Possible Answers:

$\left \langle 6x^2yzdx, 2x^3zdy, 2x^3ydz \right \rangle$

6x^2yzdx+2x^3zdy+2x^3ydz

6x^2yz+2x^3z+2x^3y

Correct answer:

6x^2yzdx+2x^3zdy+2x^3ydz

Explanation:

The differential of the function g=f(x, y, z) is given by

dg=f_xdx+f_ydy+f_zdz

The partial derivatives are

f_x=6x^2yz

f_y=2x^3z

f_z=2x^3y

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Example Question #1431 : Partial Derivatives

Find the total differential, , of the following function

z=xy+2x+2y

Possible Answers:

dz=(x+2)dx+(y+2)dy

dz=(y+2)dx+(x+2)dy

dz=(2xy)dx+(2xy)dy

Correct answer:

dz=(y+2)dx+(x+2)dy

Explanation:

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

For the function

z=xy+2x+2y

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=y+2$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=x+2$

We then substitute these partial derivatives into the first equation to get the total differential

dz=(y+2)dx+(x+2)dy

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Example Question #11 : Differentials

Find the total differential, , of the following function

$z=x^2y^2+e^x+cos\,y$

Possible Answers:

$dz=(x^2y^2+e^x)dx+(x^2y^2+cos\,y)dy$

$dz=(2xy^2+e^x)dx+(2x^2y-sin\,y)dy$

dz=(2xy^2)dx+(2x^2y)dy

Correct answer:

$dz=(2xy^2+e^x)dx+(2x^2y-sin\,y)dy$

Explanation:

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

For the function $z=x^2y^2+e^x+cos\,y$

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

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

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=2x^2y-sin\,y$

We then substitute these partial derivatives into the first equation to get the total differential

$dz=(2xy^2+e^x)dx+(2x^2y-sin\,y)dy$

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Example Question #16 : Differentials

Find the total differential, , of the following function

$z=x+sin\,y+cos\,x+y$

Possible Answers:

dz=dx+dy

$dz=(1+sin\,x+y)dx+(x-cos\,y+1)dy$

$dz=(1-sin\,x)dx+(cos\,y+1)dy$

Correct answer:

$dz=(1-sin\,x)dx+(cos\,y+1)dy$

Explanation:

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

For the function $z=x+sin\,y+cos\,x+y$

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=1-sin\,x$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=cos\,y+1$

We then substitute these partial derivatives into the first equation to get the total differential

$dz=(1-sin\,x)dx+(cos\,y+1)dy$

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Example Question #17 : Differentials

If f(x,y) = xe^y , calculate the differential when moving from (x_0,y_0)=(1,0) to (0,1) .

Possible Answers:

e-1

e+1

Correct answer:

Explanation:

The equation for is

dz = f_x(x_0,y_0)dx +f_y(x_0,y_0)dy .

Evaluating partial derivatives and substituting, we get

$f_x(x_0,y_0) = f_x(1,0) = e^{(0)} =1$

$f_y(x_0,y_0) = f_y(1,0) = (1)e^{(0)}=1$

$dx = \Delta x = -1, dy=\Delta y = 1$

Plugging in, we get

dz = (1)(-1)+(1)(1)=0 .

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Example Question #18 : Differentials

If f(x,y) = x^2-y^2 , calculate the differential when moving from the point (x_0,y_0)=(1,1) to the point (2,3) .

Possible Answers:

-1.5

-.5

Correct answer:

Explanation:

The equation for is

dz = f_x(x_0,y_0)dx +f_y(x_0,y_0)dy .

Evaluating partial derivatives and substituting, we get

f_x(x_0,y_0) = f_x(1,1) = 2(1) =2

f_y(x_0,y_0) = f_y(1,1) = -2(1) =-2

$dx = \Delta x = 1, dy=\Delta y = 2$

Plugging in, we get

dz = (2)(1)+(-2)(2) =-2 .

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Example Question #19 : Differentials

If $f(x,y) = \ln(xy)$ , calculate the differential when moving from the point (x_0,y_0)=(1,1) to the point (e,e) .

Possible Answers:

e-1

-e-1

2(e-1)

3e-2

Correct answer:

2(e-1)

Explanation:

The equation for is

dz = f_x(x_0,y_0)dx +f_y(x_0,y_0)dy .

Evaluating partial derivatives and substituting, we get

$f_x(x_0,y_0) = f_x(1,1) = \frac{1}{(1)} =1$

$f_y(x_0,y_0) = f_y(1,1) = \frac{1}{(1)}=1$

$dx = \Delta x = e-1, dy=\Delta y = e-1$

Plugging in, we get

dz = (1)(e-1)+(1)(e-1) =2e-2=2(e-1).

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Example Question #1431 : Partial Derivatives

Find the differential of the following function:

g=y^4+13xz^2+3e^x

Possible Answers:

(13z^2+3e^x)dx+4y^3dy+13xzdz

13z^2+3e^x+4y^3+26xz

(z^2+3e^x)dx+4y^3dy+26xzdz

(13z^2+3e^x)dx+4y^3dy+26xzdz

Correct answer:

(13z^2+3e^x)dx+4y^3dy+26xzdz

Explanation:

The differential of the function is given by

f_xdx+f_ydy+f_zdz

The partial derivatives are

f_x=13z^2+3e^x , f_y=4y^3dy , f_z=26xzdz

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Example Question #21 : Differentials

Find the differential of the following function:

$h=z\cos(xyz)+y^2\ln(xz^2)$

Possible Answers:

$-yz^2\sin(xyz)+\frac{y^2}{x}-xz^2\sin(xyz)+2y\ln(xz^2)+\cos(xyz)-xyz\sin(xyz)+\frac{2y^2}{z}$

$(-yz^2\sin(xyz)+\frac{y^2}{x})dx+(-xz^2\sin(xyz)+2y\ln(xz^2))dy+(\cos(xyz)-xyz\sin(xyz)+\frac{2y^2}{z})dz$

$(-yz^2\sin(xyz)+\frac{y^2}{x})dx+(xz^2\sin(xyz)+2y\ln(xz^2))dy+(\cos(xyz)+xyz\sin(xyz)+\frac{2y^2}{z})dz$

$(yz^2\sin(xyz)+\frac{y^2}{x})dx+(-xz^2\sin(xyz)+2y\ln(xz^2))dy+(\cos(xyz)-xyz\sin(xyz)+\frac{2y^2}{z})dz$

Correct answer:

$(-yz^2\sin(xyz)+\frac{y^2}{x})dx+(-xz^2\sin(xyz)+2y\ln(xz^2))dy+(\cos(xyz)-xyz\sin(xyz)+\frac{2y^2}{z})dz$

Explanation:

The differential of the function is given by

f_xdx+f_ydy+f_zdz

The partial derivatives are

$f_x=-yz^2\sin(xyz)+\frac{y^2}{x}$ , $f_y=-xz^2\sin(xyz)+2y\ln(xz^2)$ , $f_z=\cos(xyz)-xyz\sin(xyz)+\frac{2y^2}{z}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #12 : Differentials

Example Question #11 : Differentials

Example Question #1431 : Partial Derivatives

Example Question #11 : Differentials

Example Question #16 : Differentials

Example Question #17 : Differentials

Example Question #18 : Differentials

Example Question #19 : Differentials

Example Question #1431 : Partial Derivatives

Example Question #21 : Differentials

All Calculus 3 Resources

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