Call Now to Set Up Tutoring:

(888) 888-0446

High School Math : Finding Second Derivative of a Function

Study concepts, example questions & explanations for High School Math

Create An Account

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Derivatives

What is the second derivative of $2x^4+\frac{1}{2}x$ ?

Possible Answers:

$24x^{2}+\frac{1}{2x}$

$24x^{2}$

$24x^{-2}$

Correct answer:

$24x^{2}$

Explanation:

To take the second derivative, we need to start with the first derivative.

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{4-1})+(1*\frac{1}{2}x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

Now we repeat the process, but using $8x^3+\frac{1}{2}$ as our expression.

We're going to treat $\frac{1}{2}$ as being $\frac{1}{2}x^0$ since anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(3*8x^{3-1})+(0*\frac{1}{2}x^{0-1})$

Notice that $(0*\frac{1}{2}x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(3*8x^{2})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}$

Report an Error

Example Question #32 : Derivatives

What is the second derivative of 3x+12 ?

Possible Answers:

Undefined

$\frac{3}{2}x^2$

x+8

Correct answer:

Explanation:

To take the second derivative, we need to start with the first derivative.

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat as 12x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(0*12x^{0-1})$

Notice that $(0*12x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3$

Now we repeat the process but use , or 3x^0 , as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^0)=0*3x^{0-1}$

As stated before, anything times zero is zero.

Therefore, $\frac{\mathrm{d} }{\mathrm{d} x}(3x^0)=0$ .

Report an Error

Example Question #33 : Derivatives

What is the second derivative of 5x^2+12x ?

Possible Answers:

10x+12

Correct answer:

Explanation:

To find the second derivative, first we need to start with the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

Now we repeat the process using 10x+12 as our expression.

We're going to treat as 12x^0 .

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})+(0*12x^{0-1})$

Notice that $(0*12x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$

Report an Error

Example Question #34 : Derivatives

What is the second derivative of -8x^2+15 ?

Possible Answers:

16x

Correct answer:

Explanation:

To find the second derivative, we start by taking the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as 15x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Now we repeat the process, but using -16x as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

Report an Error

Example Question #35 : Derivatives

What is the second derivative of x^3+2x^2+5 ?

Possible Answers:

6x+4

$\frac{2}{3}x+\frac{1}{4}$

3x^2+4x

Correct answer:

6x+4

Explanation:

To find the second derivative, we need to start by finding the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as 5x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{2})+(2*2x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=3x^2+4x$

Now we repeat the process but using 3x^2+4x as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(2*3x^{2-1})+(1*4x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(2*3x^{1})+(1*4x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(6x)+(4(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=6x+4$

Report an Error

Example Question #36 : Derivatives

What is the second derivative of $4x^2+\frac{2}{3}x$ ?

Possible Answers:

$8x+\frac{2}{3}$

Correct answer:

Explanation:

To find the second derivative, we need to start by finding the first derivative.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

Now we repeat the process, but we use $8x+\frac{2}{3}$ as our expression.

For this problem, we're going to say that $\frac{2}{3}=\frac{2}{3}x^0$ since, as stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(1*8x^{1-1})+(0*\frac{2}{3}x^{0-1})$

Notice that $(0*\frac{2}{3}x^{0-1})=0$ as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(1*8x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(8x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=8$

Report an Error

Example Question #1 : Finding Second Derivative Of A Function

What is the second derivative of (x+6)^2 ?

Possible Answers:

$\frac{1}{2}$

6x+12

2x+12

Correct answer:

Explanation:

To find the second derivative, we need to start with the first derivative.

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))*g'(x)$

Plug in our equations.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6)^{1})*(1+0)$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6))$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=2x+12$

From here, we can use our normal power rule to find the second derivative.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(1*2x^{1-1})+(0*12x^{0-1})$

Anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(1*2x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(2x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=2$

Report an Error

Example Question #2121 : High School Math

What is the second derivative of 4x^2+5x-3 ?

Possible Answers:

y=32x+5x^2

y=8

y=3

y=16x

y=8x

Correct answer:

y=8

Explanation:

To find the second derivative, we need to find the first derivative first. To find the first derivative, we can use the power rule.

For each variable, multiply by the exponent and reduce the exponent by one:

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(2*4x^{2-1})+(1*5x^{1-1})-(0*3x^0-1)$

Treat as 3x^0 since anything to the zero power is one.

Remember, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(8x^{1})+(5x^{0})-0$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=8x+5$

Now follow the same process but for 8x+5 .

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(1*8x^{1-1})+(0*5x^{0-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(1*8x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=8$

Therefore the second derivative will be the line y=8 .

Report an Error

Example Question #1 : Finding Second Derivative Of A Function

Let f(x)=sin(x)-cos(x) .

Find the second derivative of f(x) .

Possible Answers:

f''(x)=sin^2(x)-cos^2(x)

f''(x)=cos(x)-sin(x)

f''(x)=sin^2(x)+cos^2(x)

f''(x)=sin(x)+cos(x)

f''(x)=sin(x)-cos(x)

Correct answer:

f''(x)=cos(x)-sin(x)

Explanation:

The second derivative is just the derivative of the first derivative. So first we find the first derivative of f(x) . Remember the derivative of $\sin x$ is $\cos x$ , and the derivative for $\cos x$ is $-\sin x$ .

f'(x)= cos(x)+sin(x)

Then to get the second derivative, we just derive this function again. So

f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)

Report an Error

Example Question #2 : Finding Second Derivative Of A Function

Define $f (x) = 6x^{3} - 12x^{2} + 4x -8$ .

What is f ' ' (x) ?

Possible Answers:

f ' '(x) = 36x - 24

f ' '(x) = 6x - 1 2

f ' '(x) = 36x

f ' '(x) = 18

f ''(x) = 18x - 24

Correct answer:

f ' '(x) = 36x - 24

Explanation:

Take the derivative of , then take the derivative of .

$f (x) = 6x^{3} - 12x^{2} + 4x -8$

$f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0$

$f '(x) = 18x^{2} - 24x+ 4$

$f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0$

f ' '(x) = 36x - 24

Report an Error

View Tutors

Edmund
Certified Tutor

University of Ghana, Bachelor of Science, Statistics. university of Ghana, Master of Philosophy, Statistics.

View Tutors

Muddasir
Certified Tutor

National University of Science and Technology (NUST), Bachelor, Electrical Engineering.

View Tutors

Alvin
Certified Tutor

CUNY Brooklyn College, Bachelor of Science, Computer Science.

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in Atlanta, SAT Tutors in Philadelphia, SAT Tutors in Atlanta, Statistics Tutors in San Diego, English Tutors in Miami, Statistics Tutors in Los Angeles, English Tutors in Dallas Fort Worth, ACT Tutors in Denver, French Tutors in Dallas Fort Worth, ACT Tutors in Phoenix

Popular Courses & Classes

GMAT Courses & Classes in Atlanta, Spanish Courses & Classes in San Diego, ACT Courses & Classes in Philadelphia, ACT Courses & Classes in San Diego, SSAT Courses & Classes in Chicago, Spanish Courses & Classes in Boston, MCAT Courses & Classes in New York City, GRE Courses & Classes in Chicago, MCAT Courses & Classes in Seattle, ISEE Courses & Classes in Denver

Popular Test Prep

SAT Test Prep in Phoenix, SSAT Test Prep in Chicago, GMAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Seattle, GMAT Test Prep in Washington DC, SAT Test Prep in San Diego, LSAT Test Prep in Philadelphia, LSAT Test Prep in Los Angeles, ACT Test Prep in Dallas Fort Worth, SSAT Test Prep in Seattle

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

High School Math : Finding Second Derivative of a Function

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #31 : Derivatives

Example Question #32 : Derivatives

Example Question #33 : Derivatives

Example Question #34 : Derivatives

Example Question #35 : Derivatives

Example Question #36 : Derivatives

Example Question #1 : Finding Second Derivative Of A Function

Example Question #2121 : High School Math

Example Question #1 : Finding Second Derivative Of A Function

Example Question #2 : Finding Second Derivative Of A Function

All High School Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: