Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : First and Second Derivatives of Functions

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

← Previous 1 2 3 4 5 6 7 8 9 … 25 26 Next →

Example Question #21 : First And Second Derivatives Of Functions

Use the chain rule to find the derivative of the function $f(x) = e^{-6x^2}$

Possible Answers:

$-6x^2e^{-6x^2}$

$e^{-6x^2}$

None of these answers

$-12xe^{-6x^2}$

-6x^2e

Correct answer:

$-12xe^{-6x^2}$

Explanation:

The formula for the chain rule works as follows;

$[f(g(x))]' = f'(g(x))\times g'(x)$

Setting f(x) = e^x, g(x)=-6x^2 , we have

$f(g(x)) = e^{-6x^2}$

Hence

$[f(g(x))]' = f'(g(x))\times g'(x) = e^{-6x^2} \times -12x = -12xe^{-6x^2}$ .

(Keep in mind that the derivative of e^x is e^x )

Report an Error

Example Question #22 : First And Second Derivatives Of Functions

Find the second derivative of: y= sec(2x)

Possible Answers:

8sec(2x)tan^2(2x)+ 8sec^3(2x)

2sec^3(2x)+2sec(2x)tan^2(2x)

2sec^2(2x)+2sec(2x)tan(2x)

4sec^3(2x)+4sec(2x)tan^2(2x)

8sec(4x)tan(4x) +4sec^3(2x)

Correct answer:

4sec^3(2x)+4sec(2x)tan^2(2x)

Explanation:

The derivative of secant is:

$\frac{d}{dx}sec(x) = sec(x)tan(x)$

Since the inner function of secant is not , we will need to use chain rule, which is to multiply the derivative of the inner function .

Solve for the first derivative.

$\frac{d}{dx}sec(2x) = sec(2x)tan(2x) \cdot 2 = 2sec(2x)tan(2x)$

To take the second derivative, we will need to use the product rule and chain rule.

The product rule is: f(x) g'(x) + g(x)f'(x)

To avoid confusion, let f(x) =sec(2x) and g(x)= tan(2x) . Their derivatives are then:

f'(x) = 2sec(2x)tan(2x)

g'(x) = 2sec^2(2x)

The derivative of 2sec(2x)tan(2x) in terms of f(x) and g(x) by product rule is:

$\frac{d^2}{dx^2}:\:2[f(x) g'(x) + g(x)f'(x)]$

Resubstitute the functions and derivative functions into the equation.

$2[sec(2x)\cdot 2sec^2(2x)+ tan(2x)\cdot 2sec(2x)tan(2x)]$

Simplify. The answer is:

4sec^3(2x)+4sec(2x)tan^2(2x)

Report an Error

Example Question #23 : First And Second Derivatives Of Functions

Find the derivative of: $y=sec(\frac{1}{x})$

Possible Answers:

$ln(x)sec(\frac{1}{x})tan(\frac{1}{x})$

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

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

$-ln(x)sec(\frac{1}{x})tan(\frac{1}{x})$

$sec(\frac{1}{x})tan(\frac{1}{x})$

Correct answer:

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

Explanation:

Write the derivative of secant.

$\frac{d}{dx} sec(x) = sec(x)tan(x)$

Since the inner function is $\frac{1}{x}$ , we will need to apply the chain rule to solve this derivative.

Take the derivative of $\frac{1}{x}$ . Do not mix this with the integration, which is ln(x) .

$\frac{d}{dx}\:\frac{1}{x} = \frac{d}{dx}\:x^{-1} = -1\cdot x^{-2} = -\frac{1}{x^2}$

The derivative of sec(x) is then:

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

The answer is: $-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

Report an Error

Example Question #24 : First And Second Derivatives Of Functions

What is the second derivative of $\sin(x^2)$ ?

Possible Answers:

$-\sin(x^2)$

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

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

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

Correct answer:

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

Explanation:

We use the chain rule to get the first derivative:

$(f \circ g)' = (f'\circ g)\cdot g'$

This gives us $2x\cdot \cos(x^2)$ . Now we must take the derivative again, which means we have to use the product rule *and* the chain rule. As a reminder, the product rule says that

$(f\cdot g)' = f'\cdot g + f\cdot g'$

Which means the second derivative of the original function is

$2\cos(x^2)+2x\cdot 2x (-\sin(x^2))$ . With some simplifying we can see that this is equal to

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

Report an Error

Example Question #25 : First And Second Derivatives Of Functions

Evaluate the first and second derivative of the function

$f(x)=\pi x$

when x=4

Possible Answers:

$f'(4)=\pi$

f''(4)=0

$f'(4)=4\pi$

f''(4)=4

$f'(4)=4x\pi$

$f''(4)=4\pi$

f'(4)=4x

f''(4)=x

Correct answer:

$f'(4)=\pi$

f''(4)=0

Explanation:

We must find the first and second derivatives.

To do so we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[x^n]=nx^{n-1}$

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]=1*\pi x^{1-1}=\pi x^0 = \pi$

To find the second derivative we apply the power rule again

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x^0]=0*\pi x^{0-1}=0$

We find that

$f'(x)=\pi$ and f''(x)=0

We then evaluate each for x=4 and get

$f'(4)=\pi$

f''(4)=0

Report an Error

Example Question #1 : Velocity, Speed, Acceleration

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=cot(x)

f''(x)=csc^2(x)

f'(x)=sec^2(x)

f''(x)=sec^3(x)

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

f'(x)=csc(x)

f''(x)=-sec(x)

Correct answer:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Report an Error

Example Question #26 : First And Second Derivatives Of Functions

Find the derivative of the following function:

$f(x)=\frac{x^3}{\sin(x)}+\ln(x^2)$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=-\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Report an Error

Example Question #27 : First And Second Derivatives Of Functions

Find the derivative of the function:

$f(x)=\cos(e^{2x})+x^5+3x$

Possible Answers:

$-2e^{2x}\sin(e^{2x})+5x^4+3$

$2e^{2x}\cos(e^{2x})+5x^4+3$

$-2e^{2x}\sin(e^{2x})+5x^4+3x$

$2e^{2x}\sin(e^{2x})+5x^4+3$

Correct answer:

$-2e^{2x}\sin(e^{2x})+5x^4+3$

Explanation:

The derivative of the function is equal to

$f'(x)=-2e^{2x}\sin(e^{2x})+5x^4+3$

and was found using the following rules:

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

Report an Error

Example Question #28 : First And Second Derivatives Of Functions

Find the second derivative of the function:

$\ln(5x)+\cos(x^2)$

Possible Answers:

$-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

$\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

$-\frac{5}{x^2}-2\sin(x^2)+4x^2\cos(x^2)$

$-\frac{5}{x^2}-\sin(x^2)-4x^2\cos(x^2)$

Correct answer:

$-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{5}{x}-2x\sin(x^2)$

and was found using the following rules:

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

The second derivative of the function is equal to

$f''(x)=-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

and was found using the same rules as above, along with

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

Report an Error

Example Question #29 : First And Second Derivatives Of Functions

Find the derivative of the function

$f(x)= \frac{x^2}{x\ln(x)}$

Possible Answers:

$\frac{(x^2)(1-\ln(x)^2)}{[\ln(x)]^4}$

$\frac{1-\ln(x)}{[\ln(x)]^2}$

None of the other answers.

$\frac{\ln(x)-1}{[\ln(x)]^2}$

$\frac{1-\ln(x)^2}{[\ln(x)]^4}$

Correct answer:

$\frac{\ln(x)-1}{[\ln(x)]^2}$

Explanation:

We find the answer using the quotient rule

$(\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

(f(x)g(x))' = f(x)g'(x)+g(x)f'(x)

and then simplifying.

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

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

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

$=\frac{\ln(x)-1}{\ln(x)^2}$ or $\frac{\ln(x)-1}{[\ln(x)]^2}$ . The extra brackets in the denominator are optional.

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 … 25 26 Next →

View Calculus 2 Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View Calculus 2 Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, Electrical Engineering.

View Calculus 2 Tutors

Filiberto
Certified Tutor

University of Arizona, Bachelor of Science, Electrical Engineering.

All Calculus 2 Resources

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

Popular Subjects

English Tutors in Philadelphia, ISEE Tutors in New York City, Statistics Tutors in Denver, ACT Tutors in Chicago, Reading Tutors in Los Angeles, GMAT Tutors in Atlanta, Math Tutors in Denver, LSAT Tutors in Miami, Physics Tutors in Atlanta, GRE Tutors in Miami

Popular Courses & Classes

MCAT Courses & Classes in Atlanta, LSAT Courses & Classes in Boston, LSAT Courses & Classes in Los Angeles, SAT Courses & Classes in Denver, ACT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Seattle, LSAT Courses & Classes in San Diego, ISEE Courses & Classes in Los Angeles, ISEE Courses & Classes in San Diego, GMAT Courses & Classes in Boston

Popular Test Prep

ISEE Test Prep in Houston, SSAT Test Prep in Philadelphia, GRE Test Prep in Philadelphia, ACT Test Prep in Washington DC, ISEE Test Prep in San Diego, MCAT Test Prep in Houston, ACT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Los Angeles, LSAT Test Prep in Los Angeles, GRE Test Prep in San Diego

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 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : First And Second Derivatives Of Functions

Example Question #22 : First And Second Derivatives Of Functions

Example Question #23 : First And Second Derivatives Of Functions

Example Question #24 : First And Second Derivatives Of Functions

Example Question #25 : First And Second Derivatives Of Functions

Example Question #1 : Velocity, Speed, Acceleration

Example Question #26 : First And Second Derivatives Of Functions

Example Question #27 : First And Second Derivatives Of Functions

Example Question #28 : First And Second Derivatives Of Functions

Example Question #29 : First And Second Derivatives Of Functions

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: