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Calculus 3 : Differentials

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Example Questions

Example Question #1 : Differentials

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=2xe^{x^2+4y^2}\sin(y)dx$

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

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Example Question #1 : Differentials

Find the total differential , , of the function

z=e^x + cos (y)

Possible Answers:

dz=xe^x dx+sin(y)dy

dz=e^x dx-sin(y)dy

dz=cos(y) dx-e^xdy

dz=e^x dx+sin(y)dy

Correct answer:

dz=e^x dx-sin(y)dy

Explanation:

The total differential is defined as

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

We first find

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

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

$\frac{\partial z}{\partial x}=e^x$

We then find

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

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

$\frac{\partial z}{\partial y}=-sin(y)$

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

dz=e^x dx-sin(y)dy

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Example Question #3 : Differentials

Find the total differential, , of the function

$z=e^{x^2+y^2}+sec(2x)$

Possible Answers:

$dz=[2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

$dz=[2xe^{x^2+y^2}]dx+2ye^{x^2+y^2}dy$

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

Correct answer:

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

Explanation:

The total differential is defined as

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

We first find $\frac{\partial z}{\partial x}$ by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}[e^{x^2+y^2}+sec(2x)]=2xe^{x^2+y^2}+2sec(2x)tan(2x)$

We then find $\frac{\partial z}{\partial y}$ by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}[e^{x^2+y^2}+sec(2x)]=2ye^{x^2+y^2}$

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

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

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Example Question #4 : Differentials

Find the total differential, , of the function

$w=e^{x+y}tan(4z)+sin(z)$

Possible Answers:

$dw=e^{x+y}tan(4z)dx+e^{x+y}tan(4z)}dy +[4e^{x+y}sec^2(4z)+cos(z)]dz$

$dw=e^{x+y}tan(4z)dx+e^{x+y}tan(4z)}dy +[sec^2(4z)]dz$

$dw=e^{x+y}dx+e^{x+y}}dy +[4e^{x+y}sec^2(4z)+cos(z)]dz$

Correct answer:

$dw=e^{x+y}tan(4z)dx+e^{x+y}tan(4z)}dy +[4e^{x+y}sec^2(4z)+cos(z)]dz$

Explanation:

The total differential is defined as

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

We first find $\frac{\partial w}{\partial x}$ by taking the derivative with respect to and treating the other variables as constants.

$\frac{\partial w}{\partial x}=\frac{\partial }{\partial x}[e^{x+y}tan(4z)+sin(z)]=e^{x+y}tan(4z)$

We then find $\frac{\partial w}{\partial y}$ by taking the derivative with respect to and treating the other variables as constants.

$\frac{\partial w}{\partial y}=\frac{\partial }{\partial y}[e^{x+y}tan(4z)+sin(z)]=e^{x+y}tan(4z)$

We then find $\frac{\partial w}{\partial z}$ by taking the derivative with respect to and treating the other variables as constants.

$\frac{\partial w}{\partial z}=\frac{\partial }{\partial y}[e^{x+y}tan(4z)+sin(z)]=4e^{x+y}sec^2(4z)+cos(z)$

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

$dw=e^{x+y}tan(4z)dx+e^{x+y}tan(4z)}dy +[4e^{x+y}sec^2(4z)+cos(z)]dz$

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Example Question #1 : Differentials

Find the total derivative of the function:

$f(x, y)=x\sec(y)+6y^4$

Possible Answers:

$(\sec(y)+x\sec(y)\tan(y)+24y^3)dy$

$\sec(y)dx+(x\sec(y)\tan(y)+24y^3)dy$

$\sec(y)dy+(x\sec(y)\tan(y)+24y^3)dx$

$-\sec(y)dx+(x\sec(y)\tan(y)+24y^3)dy$

Correct answer:

$\sec(y)dx+(x\sec(y)\tan(y)+24y^3)dy$

Explanation:

The total derivative of a function of two variables is given by the following:

f_x dx+f_ydy

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

The partial derivatives of the function are

$f_x=\sec(y)$

$f_y=x\sec(y)\tan(y)+24y^3$

The derivatives were found using the following rules:

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

So, our final answer is

$\sec(y)dx+(x\sec(y)\tan(y)+24y^3)dy$

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Example Question #1 : Differentials

Find the total derivative of the function:

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

Possible Answers:

$(\frac{1}{2\sqrt{x+y^2}}+\frac{1}{x})dx+(\frac{y}{\sqrt{x+y^2}})dy$

$(\frac{1}{2\sqrt{x+y^2}}+\frac{2}{x})dy+(\frac{y}{\sqrt{x+y^2}})dx$

$(\frac{1}{2\sqrt{x+y^2}}+\frac{2}{x})dx+(\frac{y}{\sqrt{x+y^2}})dy$

$\frac{1}{2\sqrt{x+y^2}}+\frac{2}{x}+\frac{y}{\sqrt{x+y^2}}$

Correct answer:

$(\frac{1}{2\sqrt{x+y^2}}+\frac{2}{x})dx+(\frac{y}{\sqrt{x+y^2}})dy$

Explanation:

The total derivative of a function of two variables is given by

f_xdx+f_ydy

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

The partial derivatives are

$f_x=\frac{1}{2\sqrt{x+y^2}}+\frac{2}{x}$

$f_y=\frac{y}{\sqrt{x+y^2}}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$

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Example Question #7 : Differentials

Find the total derivative of the function:

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

Possible Answers:

$-y^2\sin(xy^2)dx-2xy\sin(xy^2)dy+2ze^{z^2}dz$

$-y^2\sin(xy^2)-2xy\sin(xy^2)+2ze^{z^2}$

$-y^2\sin(xy^2)dx-2xy\sin(xy^2)dy$

$-y^2\sin(xy^2)dx+2xy\sin(xy^2)dy+2ze^{z^2}dz$

Correct answer:

$-y^2\sin(xy^2)dx-2xy\sin(xy^2)dy+2ze^{z^2}dz$

Explanation:

The total derivative of a function f(x, y, z) is given by

f_xdx+f_ydy+f_zdz

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

The partials are

$f_x=-y^2\sin(xy^2)$

$f_y=-2xy\sin(xy^2)$

$f_z=2ze^{z^2}$

The derivatives were found using the following rules:

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

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Example Question #1 : Differentials

Find the total derivative of the following function:

$f(x, y,z)=x^2y+\csc(z)$

Possible Answers:

$2xydx+x^2dy+\csc(z)\cot(z)dz$

$2xy+x^2-\csc(z)\cot(z)$

$2xydx+x^2dy-\csc(z)\cot(z)dz$

$2xdx-\csc(z)\cot(z)dz$

Correct answer:

$2xydx+x^2dy-\csc(z)\cot(z)dz$

Explanation:

The total derivative of a function is given by

f_xdx+f_ydy+f_zdz

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

The partial derivatives are

f_x=2xy

f_y=x^2

$f_z=-\csc(z)\cot(z)$

The derivatives were found using the following rules:

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

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Example Question #9 : Differentials

Find the differential of the following function:

$f(x, y, z)=x^5+\tan(y^2z)$

Possible Answers:

$5x^4dx+2yz\sec^2(y^2z)dy+y^2\sec^2(y^2z)dz$

$5x^4dx+y\sec^2(y^2z)dy+z\sec^2(y^2z)dz$

$5x^4dx+2yz\sec^2(y^2z)dy+zy^2\sec^2(y^2z)dz$

$5x^4+\sec^2(y^2z)(2yz+y^2)$

Correct answer:

$5x^4dx+2yz\sec^2(y^2z)dy+y^2\sec^2(y^2z)dz$

Explanation:

The differential of a function is given by

f_xdx+f_ydy+f_zdz

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

The partial derivatives are

f_x=5x^4

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

$f_z=y^2\sec^2(y^2z)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #10 : Differentials

Find the differential of the function

$f(x, y, z)=ze^y+\ln(x^2)$

Possible Answers:

$\frac{2}{x}+e^y(z+1)$

$\frac{2}{x}dx+ze^ydy+e^ydz$

$\frac{2}{x}dx+e^ydy+ze^ydz$

$\frac{2}{x^2}dx+ze^ydy+e^ydz$

Correct answer:

$\frac{2}{x}dx+ze^ydy+e^ydz$

Explanation:

The differential of a function is given by

f_xdx+f_ydy+f_zdz

The partial derivatives of the function are

$f_x=\frac{2}{x}$

f_y=ze^y

f_z=e^y

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Calculus 3 : Differentials

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Differentials

Example Question #1 : Differentials

Example Question #3 : Differentials

Example Question #4 : Differentials

Example Question #1 : Differentials

Example Question #1 : Differentials

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