Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Chain Rule and Implicit Differentiation

Study concepts, example questions & explanations for AP Calculus BC

Create An Account

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 Next →

Example Question #1 : Chain Rule And Implicit Differentiation

Find dy/dx by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{e^y+3x^2}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

$\frac{3x^2-\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{e^y+\tan^{-1}(y)-x^3}{x}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Correct answer:

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Explanation:

To find dy/dx we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dx}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dx}))+(1)(\frac{dx}{dx})(\tan^{-1}(y))]=e^y\frac{dy}{dx}$

Notice that anytime we take the derivative of a term with x involved we place a "dx/dx" next to it, but this is equal to "1".

So this now becomes:
$3x^2+[\frac{x}{1+y^2}\frac{dy}{dx}+\tan^{-1}(y)}]=e^y\frac{dy}{dx}$

Now if we place all the terms with a "dy/dx" onto one side and factor out we can solved for it:

$3x^2+\tan^{-1}(y)=e^y\frac{dy}{dx}-\frac{x}{1+y^2}\frac{dy}{dx}$

$3x^2+\tan^{-1}(y)=\frac{dy}{dx}(e^y-\frac{x}{1+y^2})$

$\frac{dy}{dx}=\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

This is one of the answer choices.

Report an Error

Example Question #2 : Chain Rule And Implicit Differentiation

Find dx/dy by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{\frac{x}{1+y^2}+e^y}{5x^2-\tan^{-1}(y)}$

$\frac{3x^2+e^y}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

$\frac{\tan^{-1}(y)-\frac{x}{1+y^2}}{3x^2+e^y}$

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Correct answer:

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

Explanation:

To find dx/dy we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dy}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dy}))+(1)(\frac{dx}{dy})(\tan^{-1}(y))]=e^y\frac{dy}{dy}$

Notice that anytime we take the derivative of a term with y involved we place a "dy/dy" next to it, but this is equal to "1".

So this now becomes:
$3x^2\frac{dx}{dy}+[\frac{x}{1+y^2}+\tan^{-1}(y)\frac{dx}{dy}]=e^y$

Now if we place all the terms with a "dx/dy" onto one side and factor out we can solved for it:

$3x^2\frac{dx}{dy}+\tan^{-1}(y)\frac{dx}{dy}=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}[3x^2+\tan^{-1}(y)]=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}=\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

This is one of the answer choices.

Report an Error

Example Question #8 : Derivative At A Point

Use implicit differentiation to find the slope of the tangent line to $\small y^3-lny=x^2$ at the point $\small (1,1)$ .

Possible Answers:

$\small 0$

$\small 2$

$\small -1$

$\small 1$

$\small -2$

Correct answer:

$\small 1$

Explanation:

We must take the derivative $\small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get

$\small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$ , and on the right side we'll get $\small 2x$ .

We include the $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\small y$ is a function of $\small x$ , so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\small 3y^2-1/y$ in order to solve for $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ .

Doing this gives you

$\small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$ .

We want to find the slope at $\small (1,1)$ , so we can sub in $\small 1$ for $\small x$ and $\small y$ .

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$ .

Report an Error

Example Question #1 : Chain Rule And Implicit Differentiation

$g(x) = e ^{x^{2}+x}$

Evaluate g '(3 ) .

Possible Answers:

$7 e^{9}$

$7 e^{12}$

$12 e^{12}$

$e^{7}$

$12 e^{11}$

Correct answer:

$7 e^{12}$

Explanation:

To find g'(x) , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

Report an Error

Example Question #1 : Chain Rule And Implicit Differentiation

$g (x) = \frac{1}{4x - 7}$

Evaluate g ' (1 ) .

Possible Answers:

$\frac{4}{3}$

$\frac{4}{9}$

$- \frac{4}{9}$

Undefined

$-\frac{4}{3}$

Correct answer:

$- \frac{4}{9}$

Explanation:

To find g'(x) , substitute u =4x - 7 and use the chain rule:

$g (x) = \frac{1}{4x - 7}$

$g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$= \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$=4 \cdot (-1 ) u ^{-1-1}$

$=-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

$g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$= - \frac{4}{(4 -7)^{2}}$

$= - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

Report an Error

Example Question #1 : Chain Rule And Implicit Differentiation

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

Evaluate $g' \left ( \pi )$ .

Possible Answers:

$-\sqrt{2}$

Undefined

$\sqrt{2}$

Correct answer:

Explanation:

To find g'(x) , substitute $u = x +\frac{ \pi }{4}$ and use the chain rule:

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right )\cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right ) \cdot \sec ^{2} u$

$=1 \cdot \sec ^{2}\left ( x +\frac{ \pi }{4}\right ) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(x) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\left ( \pi +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\frac{ 5 \pi }{4}$

$= \left ( - \sqrt{2 }\right )^{2} = 2$

Report an Error

Example Question #1 : Chain Rule And Implicit Differentiation

$g (x) = 3 ^{x^{2}}$

Evaluate g ' (-1 ) .

Possible Answers:

$-\frac{3 \ln 3}{2}$

$-6 \ln 3$

Correct answer:

$-6 \ln 3$

Explanation:

To find g'(x) , substitute $u = x ^{2}$ and use the chain rule:

$g (x) = 3 ^{x^{2}}$

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

$= \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= \frac{\mathrm{d}}{\mathrm{d} x} x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= 2x \cdot \ln 3 \cdot 3 ^{u}$

$= 2x \ln 3 \cdot 3 ^{x^2}$

So $g'(x)= 2x \ln 3 \cdot 3 ^{x^2}$

and

$g'(-1)= 2(-1) \ln 3 \cdot 3 ^{(-1)^2}$

$= -2 \ln 3 \cdot 3 ^{1} = -2 \ln 3 \cdot 3 = -6 \ln 3$

Report an Error

Example Question #1 : Chain Rule And Implicit Differentiation

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

Evaluate $g ' \left ( \frac{1}{2} \right )$ .

Possible Answers:

$- \pi \sqrt{2}$

$\frac{ \pi \sqrt{2}}{2}$

$\pi \sqrt{2}$

$- \frac{ \sqrt{2}}{2 \pi}$

$- \frac{ \pi \sqrt{2}}{2}$

Correct answer:

$- \frac{ \pi \sqrt{2}}{2}$

Explanation:

To find g'(x) , substitute $u = \pi x^{2}$ and use the chain rule:

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

$g'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( \pi x^{2}\right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} \cos u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \pi x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$= \pi \cdot 2 x\cdot \left ( - \sin u \right )$

$= \pi \cdot 2 x\cdot \left [- \sin \left ( \pi x^{2} ) \right ]$

$=-2 \pi x \sin \left ( \pi x^{2} )$

So $g'(x)=-2 \pi x \sin \left ( \pi x^{2} )$

and

$g ' \left ( \frac{1}{2} \right )=-2 \pi\cdot \frac{1}{2} \sin \left [\pi \cdot \left ( \frac{1}{2} \right )^{2} \right ]$

$=- \pi \sin \left ( \frac{\pi}{4} \right )$

$=- \pi \cdot \frac{\sqrt{2}}{2} = - \frac{ \pi \sqrt{2}}{2}$

Report an Error

Example Question #2 : Chain Rule And Implicit Differentiation

$\frac{d}{dx} \ln(3x-7)$

Possible Answers:

$\ln(3)$

$\frac{1}{3x-7}$

$3 \ln (3x-7)$

$\frac{3}{3x-7}$

Correct answer:

$\frac{3}{3x-7}$

Explanation:

Consider this function a composition of two functions, f(g(x)). In this case, f(x) is ln(x) and g(x) is 3x - 7. The derivative of ln(x) is 1/x, and the derivative of 3x - 7 is 3. The derivative is then $\frac{1}{3x-7} \cdot 3 = \frac{3}{3x-7}$ .

Report an Error

Example Question #1 : Chain Rule And Implicit Differentiation

$\frac{d}{dx } 2 \sin (x^2 )$

Possible Answers:

$2\cdot \cos(x^2)$

$2x \cdot \cos(x^2)$

$4x \cdot \cos(x^2)$

$-4x \cdot \cos(x^2)$

Correct answer:

$4x \cdot \cos(x^2)$

Explanation:

Consider this function a composition of two functions, f(g(x)). In this case, $f(x) =2 \sin (x)$ and g(x) = x^2 . According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . Here, $f'(g(x)) = 2 \cos (x^2 )$ and g'(x) = 2x , so the derivative is $2 \cos (x^2 ) \cdot 2x = 4x \cdot \cos(x^2)$

Report an Error

← Previous 1 2 3 Next →

View AP Calculus BC Tutors

Shakeel
Certified Tutor

View AP Calculus BC Tutors

Nene
Certified Tutor

University of Science and Arts of Oklahoma, Bachelor of Science, Mathematics. Florida Institute of Technology, Master of Scie...

View AP Calculus BC Tutors

Emma
Certified Tutor

University of South Carolina, Bachelor of Science, Physics. University of South Carolina-Columbia, Bachelor of Science, Psych...

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Seattle, Computer Science Tutors in San Diego, SSAT Tutors in Los Angeles, GMAT Tutors in Philadelphia, Computer Science Tutors in Atlanta, Chemistry Tutors in San Diego, Physics Tutors in New York City, Biology Tutors in Boston, SAT Tutors in Miami, ACT Tutors in Seattle

Popular Courses & Classes

LSAT Courses & Classes in San Diego, MCAT Courses & Classes in Chicago, GRE Courses & Classes in Boston, SAT Courses & Classes in Chicago, GRE Courses & Classes in Chicago, GMAT Courses & Classes in Atlanta, Spanish Courses & Classes in Boston, Spanish Courses & Classes in Los Angeles, ISEE Courses & Classes in Washington DC, GRE Courses & Classes in Washington DC

Popular Test Prep

LSAT Test Prep in Phoenix, ACT Test Prep in Houston, GMAT Test Prep in Los Angeles, LSAT Test Prep in New York City, LSAT Test Prep in Atlanta, GMAT Test Prep in New York City, SAT Test Prep in Atlanta, SAT Test Prep in Chicago, ISEE Test Prep in Atlanta, SSAT Test Prep in Phoenix

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 BC : Chain Rule and Implicit Differentiation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Chain Rule And Implicit Differentiation

Find dy/dx by implicit differentiation:

Example Question #2 : Chain Rule And Implicit Differentiation

Find dx/dy by implicit differentiation:

Example Question #8 : Derivative At A Point

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #2 : Chain Rule And Implicit Differentiation

Example Question #1 : Chain Rule And Implicit Differentiation

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: