Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : First And Second Derivatives Of Functions

Find the derivative of the following function at t=0 :

$f(t)=\frac{4e^{\cos(t)}}{50t^2-7t+1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

To find the derivative, we must use the following rule:

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

Now, using the above rule, write out the derivative:

$f'(t)=\frac{(50t^2-7t+1)(-4\sin(t)e^{\cos(t)})-4e^{\cos(t)}(100t-7)}{(50t^2-7t+1)^2}$

The internal derivatives were found using the following rules:

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

Evaluated at t=0 , we get

$f'(0)=\frac{-4(-7)}{1}=28$

Report an Error

Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function $f(x) = 3xe^{\sin(x)}$ .

Possible Answers:

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

None of the other answers

$3\sin(x)e^{sin(x)}(x\sin(x)+1)$

$3e^{sin(x)}(x\cos(x)+1)$

$3xe^{sin(x)}(x^2\cos(x)+1)$

Correct answer:

$3e^{sin(x)}(x\cos(x)+1)$

Explanation:

We will need to use the product rule and the chain rule to find the derivative.

$y = 3xe^{\sin(x)}$ . Start

$y' = (3x)(e^{\sin(x)})' + (3x)'(e^{\sin(x)})$ . Product Rule

$y' = (3x)(\cos(x)e^{\sin(x)}) + (3)(e^{\sin(x)})$ . Use the Chain Rule for $(e^{\sin(x)})'$ .

$y' = 3x\cos(x)e^{\sin(x)} + 3e^{\sin(x)}$ . Multiply

$y' = 3e^{\sin(x)}(x\cos(x)+1)$ . Factor out a , and an $e^{\sin(x)}$ .

Report an Error

Example Question #231 : Derivative Review

Find the derivative of the function $y = 50\cos(x)\tan(x)$

Possible Answers:

None of the other answers.

$50\cos^2(x)-\sin^2(x)$

$50\cos(x)sec^{2}(x)$

$50\cos(x)$

$25\sin(x)$

Correct answer:

$50\cos(x)$

Explanation:

Although we could use the Product Rule to compute the derivative, it becomes much easier to find if we rewrite $\tan(x) = \frac{\sin(x)}{\cos(x)}$ .

$y=50\cos(x)\tan(x)$ . Start

$y=50\cos(x)\frac{\sin(x)}{\cos(x)}$

$y = 50\sin(x)$

$y' = 50\cos(x)$ .

Report an Error

Example Question #31 : First And Second Derivatives Of Functions

If x^y = xy , find in terms of and .

Possible Answers:

$\frac{xy\ln(y)^2}{\ln(x)-xy\ln(x)^2}$

$\frac{xy^2\ln(y)^2}{x\ln(x)-x^2\ln(x)^2}$

None of the other answers

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

$\frac{xy\ln(y)^2}{\ln(x)-x^2y\ln(x)^2}$

Correct answer:

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

Explanation:

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

x^y=yx

$\ln(x^y)=\ln(xy)$

$y\ln(x)=\ln(x)+\ln(y)$

$y = \frac{\ln(y)}{\ln(x)}+1$

$y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}$ . Quotient Rule + implicit differentiation.

$y' = \frac{y'}{y\ln(x)}-\frac{\ln(y)}{x\ln(x)^2}$

$y' -\frac{1}{y\ln(x)}y' = -\frac{\ln(y)}{x\ln(x)^2}$

$(\frac{1}{y\ln(x)}-1)y' = \frac{\ln(y)}{x\ln(x)^2}$

$(\frac{1-y\ln(x)}{y\ln(x)})y' = \frac{\ln(y)}{x\ln(x)^2}$

$y' = \frac{\ln(y)}{x\ln(x)^2}(\frac{y\ln(x)}{1-y\ln(x)})$

$y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

Report an Error

Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function $y= \frac{\tan^{-1}(x)}{x}$

Possible Answers:

$\frac{x^2}{\tan(x)\sqrt{1-x^2}}$

$\frac{x^2}{\tan^{-1}(x)\sqrt{1-x^2}}$

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

$\frac{\tan^{-1}(x)}{x^2}$

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}$ .

Using the Quotient Rule and the fact $\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$ , we have

$y = \frac{\tan^{-1}(x)}{x}$

$y' = \frac{x\frac{1}{1+x^2}-\tan^{-1}(x)}{x^2}$

$y' = \frac{\frac{x}{1+x^2}-\tan^{-1}(x)}{x^2}$

$y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)}{x^2}$

$y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}$

$y' = \frac{x-\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}$ .

Report an Error

Example Question #1361 : Calculus Ii

Find the derivative of the function $y=e^{x^{\frac{1}{2}}+x^2-\ln(x)}$

Possible Answers:

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

$(\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x})e^{x^\frac{1}{2}+x^2-\ln(x)}$

None of the other answers

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

$e^{x^\frac{1}{2}+x^2-\ln(x)}$

Correct answer:

$(\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x})e^{x^\frac{1}{2}+x^2-\ln(x)}$

Explanation:

The Chain Rule is required here.

$y=e^{x^{\frac{1}{2}}+x^2-\ln(x)}$ . Start

The Chain Rule Proceeds as follows: $f(g(x)) = f'(g(x))\times g'(x)$ . In this case

f(x)= e^x

and

$g(x) = \frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x}$ .

Putting these into the Chain Rule, we get

$y' = e^{x^\frac{1}{2}+x^2-\ln(x)}(\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x})$

Or the same to say

$y' = (\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x})e^{x^\frac{1}{2}+x^2-\ln(x)}$ .

Report an Error

Example Question #231 : Derivatives

Evaluate the derivative of y = 6x^2 +C , where is any constant.

Possible Answers:

None of the other answers

12x+1

6x^2 +C

12x

Correct answer:

12x

Explanation:

For the 6x^2 term, we simply use the power rule to abtain 12x . Since is a constant (not a variable), we treat it as such. The derivative of any constant (or "stand-alone number") is .

Report an Error

Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function y=7x^2e^x .

Possible Answers:

14x^2(x-1)

$\frac{7}{3}x^3e^x$

14xe^x

7xe^x(x+2)

None of the other answers

Correct answer:

7xe^x(x+2)

Explanation:

We use the Product Rule to find our answer here. The Product Rule formula is (fg)'(x) = f(x)g'(x)+f'(x)g(x) .

Let f(x) = 7x^2 , g(x)= e^x , then we have f'(x) = 14x , g'(x)=e^x .

Putting these into our formula, we have

y' = (7x^2)(e^x)+(e^x)(14x)

=e^x(7x^2+14x)

=7xe^x(x+2) .

Report an Error

Example Question #1364 : Calculus Ii

What is the derivative of

$f(x)=5\sqrt{x}$ ?

Possible Answers:

$f'(x)=\frac{5}{2\sqrt{x}}$

$f'(x)=\frac{5}{2}\sqrt{x}$

$f'(x)=\frac{5}{\sqrt{x}}$

$f'(x)=\frac{2}{\sqrt{x}}$

Correct answer:

$f'(x)=\frac{5}{2\sqrt{x}}$

Explanation:

We can find the derivative of

$f(x)=5\sqrt{x}$

using the power rule

$\frac{d}{dx}x^a=ax^{a-1}$

with $a=\frac{1}{2}$

so we have

$f'(x)=5\frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{5}{2x^{1/2}}=\frac{5}{2\sqrt{x}}$

Report an Error

Example Question #1365 : Calculus Ii

Find the velocity function given the displacement function:

$f(t)=\pi t^{2/5}$

Possible Answers:

$f'(t)=\frac{\pi}{ t^{3/5}}$

$f'(t)=\frac{2\pi}{ 5t^{2/5}}$

$f'(t)=\frac{2\pi}{ 5t^{3/5}}$

$f'(t)=\frac{5\pi}{ 2t^{3/5}}$

Correct answer:

$f'(t)=\frac{2\pi}{ 5t^{3/5}}$

Explanation:

The derivative of the displacement function is the velocity, so we need to find f'(t) . We can use the power rule

$\frac{d}{dx} x^a=ax^{a-1}$

with $a=\frac{2}{5}$

to get

$f'(t)=\pi\frac{2}{ 5}t^{-3/5}=\frac{2\pi}{ 5t^{3/5}}$

Report an Error

View Calculus 2 Tutors

Jozsef
Certified Tutor

University of Waterloo, Doctor of Philosophy, Applied Mathematics.

View Calculus 2 Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus 2 Tutors

Jack
Certified Tutor

University of Washington-Seattle Campus, Bachelor in Arts, Mathematics. Western Washington University, Master of Science, Mat...

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Houston, English Tutors in San Francisco-Bay Area, Math Tutors in Boston, Physics Tutors in Boston, GRE Tutors in Los Angeles, Physics Tutors in New York City, Chemistry Tutors in Miami, Chemistry Tutors in Philadelphia, French Tutors in Los Angeles, LSAT Tutors in Philadelphia

Popular Courses & Classes

Spanish Courses & Classes in Chicago, ACT Courses & Classes in Boston, GMAT Courses & Classes in Washington DC, Spanish Courses & Classes in Seattle, MCAT Courses & Classes in Los Angeles, GMAT Courses & Classes in San Diego, SSAT Courses & Classes in Houston, ISEE Courses & Classes in Seattle, Spanish Courses & Classes in Houston, GRE Courses & Classes in San Francisco-Bay Area

Popular Test Prep

GRE Test Prep in Chicago, LSAT Test Prep in Seattle, ISEE Test Prep in Washington DC, SAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Philadelphia, GMAT Test Prep in Dallas Fort Worth, ACT Test Prep in Miami, MCAT Test Prep in New York City, GMAT Test Prep in Seattle, LSAT Test Prep in Houston

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 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #31 : First And Second Derivatives Of Functions

Example Question #32 : First And Second Derivatives Of Functions

Example Question #231 : Derivative Review

Example Question #31 : First And Second Derivatives Of Functions

Example Question #32 : First And Second Derivatives Of Functions

Example Question #1361 : Calculus Ii

Example Question #231 : Derivatives

Example Question #32 : First And Second Derivatives Of Functions

Example Question #1364 : Calculus Ii

Example Question #1365 : Calculus Ii

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: