We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #91 : How To Find Solutions To Differential Equations

Find $\frac{dy}{dx}$ for the equation:

ln(xy)=x^2-2y

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

For this problem, note that:

$d[ln(u)]=\frac{du}{u}$

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

ln(xy)=x^2-2y

The derivative of each side is

$\frac{y}{xy}dx+\frac{x}{xy}dy=2xdx+0dy-0dx-2dy$

$\frac{1}{x}dx+\frac{1}{y}dy=2xdx-2dy$

Now move and terms to opposite sides of the equation:

$(\frac{1}{x}-2x)dx=-(2+\frac{1}{y})dy$

Finally rearrange variables to get $\frac{dy}{dx}$ :

$\frac{2x-\frac{1}{x}}{2+\frac{1}{y}}=\frac{dy}{dx}$

Report an Error

Example Question #92 : How To Find Solutions To Differential Equations

Find $\frac{dx}{dy}$ for the equation:

$ln(x^2y)=e^{xy}$

Possible Answers:

$\frac{\frac{2}{x}-ye^{xy}}{xe^{xy}-\frac{1}{y}}$

$\frac{\frac{2y}{x}-ye^{xy}}{xe^{xy}-\frac{x}{y}}$

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

$\frac{xe^{xy}-\frac{x}{y}}{\frac{2y}{x}-ye^{xy}}$

$\frac{xe^{xy}-\frac{1}{y}}{\frac{2}{x}-ye^{xy}}$

Correct answer:

$\frac{xe^{xy}-\frac{1}{y}}{\frac{2}{x}-ye^{xy}}$

Explanation:

For this problem, note that:

$d[ln(u)]=\frac{du}{u}$

d[e^u]=du(e^u)

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

$ln(x^2y)=e^{xy}$

The derivative is then

$\frac{2xy}{x^2y}dx+\frac{x^2}{x^2y}dy=ye^{xy}dx+xe^{xy}dy$

$\frac{2}{x}dx+\frac{1}{y}dy=ye^{xy}dx+xe^{xy}dy$

Now bring and terms to opposite sides of the equation:

$(\frac{2}{x}-ye^{xy})dx=(xe^{xy}-\frac{1}{y})dy$

Now rearraging variables gives $\frac{dx}{dy}$ :

$\frac{dx}{dy}=\frac{xe^{xy}-\frac{1}{y}}{\frac{2}{x}-ye^{xy}}$

Report an Error

Example Question #93 : How To Find Solutions To Differential Equations

Find $\frac{dy}{dx}$ for the equation:

sin(x^2y)cos(3xy^3)=1

Possible Answers:

$\frac{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}$

$\frac{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}$

$\frac{xycos(x^2y)cos(3xy^3)-y^3sin(x^2y)sin(3xy^3)}{xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}$

$\frac{9xy^2sin(x^2y)sin(3xy^3)+x^2cos(x^2y)cos(3xy^3)}{2xycos(x^2y)cos(3xy^3)+3y^3sin(x^2y)sin(3xy^3)}$

$\frac{2xycos(x^2y)cos(3xy^3)+3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)+x^2cos(x^2y)cos(3xy^3)}$

Correct answer:

$\frac{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}$

Explanation:

For this problem, note that:

d[sin(u)]=du(cos(u))

d[cos(u)]=-du(sin(u))

Product rule: d[uv]=udv+vdu

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

sin(x^2y)cos(3xy^3)=1

The derivative is then

${2xycos(x^2y)cos(3xy^3)dx-3y^3sin(x^2y)sin(3xy^3)dx+x^2cos(x^2y)cos(3xy^3)dy-9xy^2sin(x^2y)sin(3xy^3)dy=0}$

Remember to utilize the chain rule!

Now bring and terms to opposite sides of the equation:

${(2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3))dx=(9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3))dy}$

Now rearraging variables gives $\frac{dy}{dx}$ :

${\frac{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}=\frac{dy}{dx}}$

Report an Error

Example Question #94 : How To Find Solutions To Differential Equations

Find $\frac{dy}{dx}$ for the equation:

x^2sin(xy^2)=y

Possible Answers:

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1+2x^3ycos(xy^2)}$

$\frac{1+2x^3ycos(xy^2)}{x(2sin(xy^2)+xy^2cos(xy^2))}$

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1-2x^3ycos(xy^2)}$

$\frac{x(sin(xy^2)+xy^2cos(xy^2))}{1+x^3ycos(xy^2)}$

$\frac{1-2x^3ycos(xy^2)}{x(2sin(xy^2)+xy^2cos(xy^2))}$

Correct answer:

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1-2x^3ycos(xy^2)}$

Explanation:

Note that:

d[sin(u)]=du(cos(u))

Product Rule: d[uv]=udv+vdu

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

x^2sin(xy^2)=y

The derivative is then found using the product rule to be:

2xsin(xy^2)dx+x^2y^2cos(xy^2)dx+2x^3ycos(xy^2)dy=dy

Notice how the chain rule needs to be utilized an additional time when taking the derivative of the x^2sin(xy^2) term with respect to .

Now bring and terms to opposite sides of the equation:

(2xsin(xy^2)+x^2y^2cos(xy^2))dx=(1-2x^3ycos(xy^2))dy

Now rearraging variables gives $\frac{dy}{dx}$ :

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1-2x^3ycos(xy^2)}=\frac{dy}{dx}$

Report an Error

Example Question #2431 : Calculus

Find $f_{xyz}$ if $f(x,y,z)=e^{yz}cos(x^2z)$

Possible Answers:

$-2xz^2e^{yz}sin(x^2z)$

$-2xze^{yz}sin(x^2z)$

$-2x(2z^2e^{yz}sin(x^2z)+y^3ze^{yz}sin(x^2z)))$

$-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

$-2yz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

Correct answer:

$-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

Explanation:

For this problem, note that:

d[e^u]=du(e^u)

d[sin(u)]=du(cos(u))

d[cos(u)]=-du(sin(u))

Product Rule d[uv]=udv+vdu

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

If we're looking for $f_{xyz}$ for the function $f(x,y,z)=e^{yz}cos(x^2z)$ then we'll begin by differentiating with respect to first:

$f_x=-2xze^{yz}sin(x^2z)$

Next, differentiate with respect to :

$f_{xy}=-2xz^2e^{yz}sin(x^2z)$

Now finally we'll differentiate with respect to ; remember to use the product rule:

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xz^2(ye^{yz}sin(x^2z)+x^2e^{yz}cos(x^2z))$

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xyz^2e^{yz}sin(x^2z)-2x^3z^2e^{yz}cos(x^2z))$

$f_{xyz}=-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

Report an Error

Example Question #351 : Equations

Find $f_{xyy}$ for the function $f(x,y)=x^2e^{y^2}$

Possible Answers:

$12xy^2e^{y^2}$

$4xe^{y^2}+4xy^2e^{y^2}$

$4xe^{y^2}+8xy^2e^{y^2}$

$4ye^{y^2}$

$12xe^{y^2}$

Correct answer:

$4xe^{y^2}+8xy^2e^{y^2}$

Explanation:

For this problem, note that:

d[e^u]=du(e^u)

Product Rule: d[uv]=udv+vdu

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

To find $f_{xyy}$ for the function $f(x,y)=x^2e^{y^2}$ , begin by differentiating with respect to :

$f_{x}=2xe^{y^2}$

Next, differentiate with respect to :

$f_{xy}=4xye^{y^2}$

Finally, differentiate with respect to once more, remembering to utilize the product rule:

$f_{xyy}=4xe^{y^2}+8xy^2e^{y^2}$

Report an Error

Example Question #2433 : Calculus

Find $\frac{dy}{dx}$ for the equation

sin(x^2y)=cos(xy^2)

Possible Answers:

$\frac{2xycos(x^2y)+y^2sin(xy^2)}{2xysin(xy^2)+x^2cos(x^2y)}$

$\frac{2xysin(xy^2)+x^2cos(x^2y)}{2xycos(x^2y)+y^2sin(xy^2)}$

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

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

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

Correct answer:

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

Explanation:

For this problem, note that:

d[sin(u)]=du(cos(u))

d[cos(u)]=-du(sin(u))

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

sin(x^2y)=cos(xy^2)

The derivative is then

2xycos(x^2y)dx+x^2cos(x^2y)dy=-y^2sin(xy^2)dx-2xysin(xy^2)dy

Now bring and terms to opposite sides of the equation:

(2xycos(x^2y)+y^2sin(xy^2))dx=(-2xysin(xy^2)-x^2cos(x^2y))dy

Finally, rearrange terms to find $\frac{dy}{dx}$ :

$\frac{2xycos(x^2y)+y^2sin(xy^2)}{-2xysin(xy^2)-x^2cos(x^2y)}=\frac{dy}{dx}$

Report an Error

Example Question #2434 : Calculus

Find the derivative of the following function.

$f(x)=\sin (3x)e^{4x}$

Possible Answers:

$12e^{x}\cos(x)$

$3e^{4x}\cos(3x)$

$12e^{4x}\cos(3x)$

None of these

$4e^{4x}\cos(3x)$

Correct answer:

$12e^{4x}\cos(3x)$

Explanation:

To solve this derivative, we must realize that there are two parts to the function and we must use the product rule. The rule states that [f(x)g(x)]'=f'(x)g(x)+f(x)g'(x) .

We must asle recognize that the derivative of $\sin$ is $\cos$ and the derivative of e^x is e^x . By these rules, the derivative is

$f'(x)=3\cos (3x)\cdot4e^{4x}$

$=12e^{4x}\cos(3x)$

Report an Error

Example Question #101 : Solutions To Differential Equations

Find the derivative of the function.

$f(x)=\ln(e^{15x})$

Possible Answers:

None of these

$\frac{1}{15x}$

$e^{15x}$

Correct answer:

Explanation:

This function is just a function inside of a function. This means we have to use the chain rule. The chain rule states that the derivative of [f(g(x))]'=f'(g(x))g'(x) .

The derivative of $\ln(x)$ is $\frac{1}{x}$ and the derivative of e^x is e^x . This makes the derivative

$f'(x)=\frac{1}{e^{15x}}*15e^{15x}$

$=15\frac{e^{15x}}{e^{15x}}$

=15

This makes sense because ln(e^x)=x .

Report an Error

Example Question #101 : Solutions To Differential Equations

Find the derivative of the function.

$f(x)=\frac{\cos (x)}{x^3}$

Possible Answers:

$-\frac{x\sin (x)+\cos (x)}{x^3}$

None of these

$-\frac{x\sin (x)+2\cos (x)}{x^4}$

$-\frac{x\sin (x)+2\cos (x)}{x^3}$

$\frac{x\sin (x)+2\cos (x)}{x^3}$

Correct answer:

$-\frac{x\sin (x)+2\cos (x)}{x^3}$

Explanation:

To find the derivative of this function, we must use the division rule. This rule states that the derivative of $\frac{f(x)}{g(x)}$ is $\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$ . The derivative of $\cos$ is $-\sin$ and the derivative of x^n is $nx^{n-1}$ .

Thus the derivative is

$f'(x)=\frac{-x^2\sin (x)-2x\cos (x)}{x^4}$

$=-\frac{x\sin (x)+2\cos (x)}{x^3}$

Report an Error

View Calculus Tutors

MEER
Certified Tutor

IIT , Master's/Graduate, Microelectronics .

View Calculus Tutors

Muhammad
Certified Tutor

Quaid-E-Azam University, Master of Science, Applied Mathematics. Quaid-E-Azam University, Master of Philosophy, Applied Lingu...

View Calculus Tutors

Fatimah
Certified Tutor

Yarmuke University, Bachelor of Science, Mathematics.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Los Angeles, Calculus Tutors in Miami, Spanish Tutors in San Diego, LSAT Tutors in Dallas Fort Worth, Physics Tutors in Washington DC, MCAT Tutors in Phoenix, SAT Tutors in San Francisco-Bay Area, SSAT Tutors in Washington DC, GRE Tutors in Washington DC, ISEE Tutors in Phoenix

Popular Courses & Classes

GRE Courses & Classes in Boston, SSAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Phoenix, MCAT Courses & Classes in Seattle, LSAT Courses & Classes in Miami, GRE Courses & Classes in New York City, MCAT Courses & Classes in Miami, MCAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Boston, GRE Courses & Classes in San Francisco-Bay Area

Popular Test Prep

SSAT Test Prep in New York City, GMAT Test Prep in Seattle, ACT Test Prep in Miami, LSAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Philadelphia, LSAT Test Prep in San Diego, MCAT Test Prep in Miami, MCAT Test Prep in Chicago, ISEE Test Prep in Philadelphia, SSAT Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #91 : How To Find Solutions To Differential Equations

Example Question #92 : How To Find Solutions To Differential Equations

Example Question #93 : How To Find Solutions To Differential Equations

Example Question #94 : How To Find Solutions To Differential Equations

Example Question #2431 : Calculus

Example Question #351 : Equations

Example Question #2433 : Calculus

Example Question #2434 : Calculus

Example Question #101 : Solutions To Differential Equations

Example Question #101 : Solutions To Differential Equations

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: