Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Implicit differentiation

Study concepts, example questions & explanations for AP Calculus AB

Create An Account

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 Next →

Example Question #1 : Implicit Differentiation

Use implicit differentiation to find $\frac{\mathrm{d}y }{\mathrm{d} x}$ given

sin(x^2)+cos(y^2)=1

Possible Answers:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{ysin(y^2)}{xcos(x^2)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{cos(2x)}{sin(2y)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2xcos(x^2)}{1+2ycos(y^2)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{xcos(x^2)}{ysin(y^2)}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{xsin(x^2)}{ycos(y^2)}$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{xcos(x^2)}{ysin(y^2)}$

Explanation:

We simply differentiate both sides of the equation

sin(x^2)+cos(y^2)=1

2xcos(x^2)dx-2ysin(y^2)dy=0 (Don't forget the chain rule)

Now we solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$

xcos(x^2)dx=ysin(y^2)dy

$xcos(x^2)=\frac{ysin(y^2)dy}{dx}$

$\frac{xcos(x^2)}{ysin(y^2)}=\frac{\mathrm{d}y }{\mathrm{d} x}$

Report an Error

Example Question #1 : Implicit Differentiation

Find $\frac{dy}{dx}.$

x^2+3y^2=9

Possible Answers:

$\frac{dy}{dx}=\frac{-x}{3y}$

$\frac{dy}{dx}=\frac{x}{3y}$

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{dy}{dx}=\frac{-x^2}{3y^2}$

$\frac{dy}{dx}=\frac{x}{y}$

Correct answer:

$\frac{dy}{dx}=\frac{-x}{3y}$

Explanation:

Implicit differentiation is very similar to normal differentiation, but every time we take a derivative with respect t , we need to multiply the result by $\frac{dy}{dx}.$ We also differentiate the entire equation from left to right, including any numbers. Then, we solve for that $\frac{dy}{dx}$ for our final answer.

x^2+3y^2=9

$2x+6y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-2x}{6y}=\frac{-x}{3y}.$

Report an Error

Example Question #3 : Implicit Differentiation

Find $\frac{dy}{dx}.$

y^2=4sinx+3x^2

Possible Answers:

$\frac{dy}{dx}=\frac{-2cosx+3x}{y^2}$

$\frac{dy}{dx}=\frac{2sinx+3x}{y}$

$\frac{dy}{dx}=\frac{2cosx+3x}{y^2}$

$\frac{dy}{dx}=-2cosx+3x$

$\frac{dy}{dx}=\frac{2cosx+3x}{y}$

Correct answer:

$\frac{dy}{dx}=\frac{2cosx+3x}{y}$

Explanation:

Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to , we need to implicitly write $\frac{dy}{dx}$ after it. Hence, the name of this method. Then, we solve for $\frac{dy}{dx}.$

$(y^2=4sinx+3x^2)'=2y\frac{dy}{dx}=4cosx+6x.$

$\frac{dy}{dx}=\frac{4cosx+6x}{2y}=\frac{2cosx+3x}{y}.$

Report an Error

Example Question #4 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function with respect to x

$y=xy^2+2\tan(y)$

Possible Answers:

$y'=\frac{y}{1-2xy-2\sec^2{y}}$

$y'=\frac{y^2}{1-2xy-2\sec^2{x}}$

$y'=\frac{y^2}{1+2xy-2\sec^2{y}}$

$y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

Correct answer:

$y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

Explanation:

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as .

$y'=\frac{\mathrm{d} }{\mathrm{d} x}(xy^2)+\frac{\mathrm{d} }{\mathrm{d} x}(2\tan{y})$

$y'=y^2+y'(2xy)+2y'\sec^2(y)$

Bringing the terms with to one side and factoring it out, we get

$y'(1-2xy-2\sec^2y)=y^2$

$y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

Report an Error

Example Question #5 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function

y=x^3y+y^2

Possible Answers:

$y'=\frac{4x^2y}{1-x^2-2y}$

$y'=\frac{3x^2y}{1-x^3-2y}$

$y'=\frac{3x^2y}{1-x^2-y}$

$y'=\frac{3x^2y}{1+x^3+2y}$

Correct answer:

$y'=\frac{3x^2y}{1-x^3-2y}$

Explanation:

$y'=\frac{\mathrm{d} }{\mathrm{d} x}(x^3y)+\frac{\mathrm{d} }{\mathrm{d} x}(y^2)$

y'=y'x^3+3x^2y+2yy'

Bringing the terms with to one side and factoring it out, we get

y'(1-x^3-2y)=3x^2y

$y'=\frac{3x^2y}{1-x^3-2y}$

Report an Error

Example Question #1 : Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

y+x^2+e^xy^2=2y^2

Possible Answers:

$\frac{-2x+e^xy^2}{1+2ye^x-4y}$

$\frac{-2x-e^xy^2}{2ye^x-4y}$

$\frac{-2x-e^xy^2}{1+2ye^x-4y}$

$\frac{-2x-e^xy^2}{1-2ye^x+4y}$

$\frac{2x-e^xy^2}{1+2ye^x-4y}$

Correct answer:

$\frac{-2x-e^xy^2}{1+2ye^x-4y}$

Explanation:

To find $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we must take the derivative of both sides of the equation with respect to x. When we do this, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}+2x+e^xy^2+2ye^x\frac{\mathrm{d}y }{\mathrm{d} x}=4y\frac{\mathrm{d} y}{\mathrm{d} x}$

The derivatives were found using the following rules:

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

Note that for every derivative of a function of y with respect to x, the chain rule was used, which accounts for $\frac{\mathrm{d} y}{\mathrm{d} x}$ appearing.

Algebraic simplification gets us our final answer,

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2x-e^xy^2}{1+2ye^x-4y}$

Report an Error

Example Question #1 : Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

$\cos(xy)=2x$

Possible Answers:

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

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

$\frac{2+y\sin(xy)}{x\sin(xy)}$

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

$\frac{2+x\sin(xy)}{y\sin(xy)}$

Correct answer:

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

Explanation:

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

$-\sin(xy)(y+x\frac{\mathrm{d} y}{\mathrm{d} x}) =2$

The following derivative rules were used:

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

Note that the chain rule was used for both the cosine function (which contains an inner product of two functions), and for the derivative of y, whose derivative with respect to x we want to solve for.

Solving, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2+y\sin(xy)}{x\sin(xy)}$

Report an Error

Example Question #8 : Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

$\frac{y^3}{3}+x^3=ye^{2x}$

Possible Answers:

$\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

$\frac{2ye^{2x}+3x^2}{y^2-e^{2x}}$

$\frac{2ye^{2x}-y^2}{3x^2-e^{2x}}$

$\frac{y^2-e^{2x}}{2ye^{2x}-3x^2}$

$\frac{2ye^{2x}-3x^2}{y^2+e^{2x}}$

Correct answer:

$\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

Explanation:

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

$y^2 \frac{\mathrm{d} y}{\mathrm{d} x}+3x^2=2ye^{2x}+e^{2x}\frac{\mathrm{d} y}{\mathrm{d} x}$

The following rules were used:

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

Note that the chain rule was used everywhere we took the derivative of a function containing y, as well as in the exponential function. The product rule was used because both x and y are both functions of x.

Using algebra to solve, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

Report an Error

Example Question #1 : Implicit Differentiation

Find $\frac{\mathrm{d} \alpha}{\mathrm{d} x}$ , where $\alpha$ is a function of x:

$\alpha^2x^2+2\alpha x=\alpha$

Possible Answers:

$\frac{2\alpha^2 x+2 \alpha}{1-2\alpha x^2 -2x}$

None of the other answers

$\frac{2\alpha^2 x+2 \alpha}{1+2\alpha x^2+2x}$

$\frac{2\alpha x^2+2 \alpha}{1-2\alpha^2 x -2x}$

Correct answer:

$\frac{2\alpha^2 x+2 \alpha}{1-2\alpha x^2 -2x}$

Explanation:

To find the derivative of $\alpha$ with respect to x, we must take the differentiate both sides of the equation with respect to x:

$2\alpha x^2 \frac{\mathrm{d} \alpha}{\mathrm{d} x}+2\alpha^2x+2x \frac{\mathrm{d} \alpha}{\mathrm{d} x}+2x=\frac{\mathrm{d} \alpha}{\mathrm{d} x}$

The following derivative rules were used:

Note that the chain rule was used for the derivative of any function containing $\alpha$ , whose derivative with respect to x we want to solve for.

Solving, we get

$\frac{\mathrm{d} \alpha}{\mathrm{d} x}=\frac{2\alpha^2 x+2 \alpha}{1-2\alpha x^2 -2x}$

Report an Error

Example Question #10 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function

$y=\sin(xy)+\tan(y^2)$

Possible Answers:

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-2y\sec^2(y^2)}$

$y'=\frac{y\cos(x)}{1-x\cos(xy)-2y\sec^2(y^2)}$

$y'=\frac{x\cos(xy)}{1-x\cos(xy)-\sec^2(y^2)}$

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-y\sec^2(y^2)}$

Correct answer:

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-2y\sec^2(y^2)}$

Explanation:

To solve this using implicit differentiation, we must always treat y as a function of x, and therefore when we differentiate y with respect to x, we denote it as

Step by step, we get the following:

$y'=\cos(xy)(y+xy')+\sec^2(y^2)(2yy')$

This resulted from the product rule and chain rule

The next steps are:

$y'=y\cos(xy)+xy'\cos(xy)+2yy'\sec^2(y^2)$

$y'-xy'\cos(xy)-2yy'\sec^2(y^2)=y\cos(xy)$

$y'(1-x\cos(xy)-2y\sec^2(y^2))=y\cos(xy)$

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-2y\sec^2(y^2)}$

Report an Error

← Previous 1 2 3 4 Next →

View AP Calculus AB Tutors

Joseph
Certified Tutor

University of Waterloo, Doctor of Philosophy, Applied Mathematics.

View AP Calculus AB Tutors

Raha
Certified Tutor

Amirkabir University of Technology, Bachelor of Science, Chemical Engineering. Wayne State University, Master of Science, Che...

View AP Calculus AB Tutors

Erin Diadem
Certified Tutor

St. Mary's University, Bachelor of Science, Industrial Engineering.

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in Seattle, French Tutors in New York City, Reading Tutors in Atlanta, Statistics Tutors in Atlanta, Math Tutors in San Francisco-Bay Area, ISEE Tutors in Houston, Calculus Tutors in Denver, Physics Tutors in Phoenix, GRE Tutors in Los Angeles, Spanish Tutors in Washington DC

Popular Courses & Classes

ISEE Courses & Classes in Washington DC, LSAT Courses & Classes in Seattle, SSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Chicago, ISEE Courses & Classes in Boston, GRE Courses & Classes in Seattle, ACT Courses & Classes in Houston, SAT Courses & Classes in Miami, Spanish Courses & Classes in Phoenix, SAT Courses & Classes in Denver

Popular Test Prep

SAT Test Prep in Seattle, GRE Test Prep in Boston, GRE Test Prep in Chicago, ACT Test Prep in Boston, ISEE Test Prep in Denver, ACT Test Prep in Houston, LSAT Test Prep in Boston, GRE Test Prep in Dallas Fort Worth, SAT Test Prep in Boston, ACT Test Prep in Philadelphia

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

AP Calculus AB : Implicit differentiation

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #3 : Implicit Differentiation

Example Question #4 : Implicit Differentiation

Example Question #5 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #8 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #10 : Implicit Differentiation

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: