We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1231 : Calculus

What is the first derivative of f(x) = x²sin(4x³)?

Possible Answers:

2x * cos(4x³)

2x(sin(4x³) + 6x³cos(4x³))

2x * sin(4x³) + 12x⁵cos(4x³)

24x⁴ * cos(4x³)

sin(4x³) + 12x²cos(4x³)

Correct answer:

2x(sin(4x³) + 6x³cos(4x³))

Explanation:

This is a product rule combined with a chain rule. Let's do the chain rule for sin(4x³) first:

cos(4x³) * 12x² = 12x²cos(4x³)

With this in mind, let's solve our whole problem:

2x * sin(4x³) + x² * 12x²cos(4x³) = 2x * sin(4x³) + 12x⁴cos(4x³) = 2x(sin(4x³) + 6x³cos(4x³))

Report an Error

Example Question #1232 : Calculus

What is the first derivative of f(x) = x⁴ – x * sin(x^–5)?

Possible Answers:

None of the other answers

4x³ + sin(x^–5) – (5 * cos(x^–5)/x⁵)

4x³ + cos(x^–5)

4x³ – cos(x^–5)

4x³ – sin(x^–5) + (5 * cos(x^–5)/x⁵)

Correct answer:

4x³ – sin(x^–5) + (5 * cos(x^–5)/x⁵)

Explanation:

The first element is merely differentiated as 4x³

The second element is a relatively simple product rule:

sin(x^–5) + x * cos(x^–5) * –5 * x^–6 = sin(x^–5) + cos(x^–5) * –5 * x^–5 = sin(x^–5) – (5 * cos(x^–5)/x⁵)

Put everything back together:

4x³ – ( sin(x^–5) – (5 * cos(x^–5)/x⁵) ) = 4x³ – sin(x^–5) + (5 * cos(x^–5)/x⁵)

Report an Error

Example Question #17 : Other Differential Functions

What is the first derivative of f(x) = 5x * ln(2x)?

Possible Answers:

5(ln(2) + 1)

5(ln(2) + (1/2))

5ln(2)/2x

5/x

5/2x

Correct answer:

5(ln(2) + 1)

Explanation:

This is just a normal product rule problem:

5 * ln(2x) + 5x * 2 * (1/2x)

Simplify: 5ln(2) + 5 = 5(ln(2) + 1)

Report an Error

Example Question #1233 : Calculus

What is the slope of the tangent line at x = 2 for f(x) = 6/x²?

Possible Answers:

–12

–1.5

1.5

–3

Correct answer:

–1.5

Explanation:

First rewrite your function to make this easier:

f(x) = 6/x²= 6x^–2

Now, we must find the first derivative:

f'(x) = –2 * 6 * x^–3 = –12/x³

The slope of the tangent line of f(x) at x = 2 is: f'(2) = –12/2³ = –12/8 = –3/2 = –1.5

Report an Error

Example Question #1234 : Calculus

What is the first derivative of f(x) = 2ln(cos(x)sin(x))?

Possible Answers:

cot(x)cos²(x)

2cos(2x)sec(x)csc(x)

2cot(x)cos²(x)

2cos²(x)sec(x)csc(x)

None of the other answers

Correct answer:

2cos(2x)sec(x)csc(x)

Explanation:

This requires both the use of the chain rule and the product rule. Start with the natural logarithm: 2/(cos(x)sin(x))

Now, multiply by d/dx cos(x)sin(x), which is: –sin(x)sin(x) + cos(x)cos(x) = cos²(x) – sin²(x)

Therefore, f'(x) = (cos²(x) – sin²(x)) * 2/(cos(x)sin(x)) = 2(cos²(x) – sin²(x))sec(x)csc(x)

From our trigonometric identities, we know cos²(x) – sin²(x) = cos(2x)

Therefore, we can finilize our simplification to f'(x) = 2cos(2x)sec(x)csc(x)

Report an Error

Example Question #1235 : Calculus

Take the derivative of

$y=(7+x)^{(7+x)^2}$

Possible Answers:

${y}'=(7+x)^{(7+x)^2}[21+\ln(7+x)+x]$

${y}'=(7+x)^2(7+x)^{(7+x)^2-1}$

${y}'=(7+x)^{(7+x)^2}(1+\ln(7+x)^{(7+x)^2)})$

Derivative does not exist

${y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]$

Correct answer:

${y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]$

Explanation:

We need to use logarithm differentiation to do this problem. Take the natural log of both sides.

$\ln y=\ln(7+x)^{(7+x)^2}$

Apply the power rule of natural log

$\ln y=(7+x)^2\ln(7+x)$

Perform implicit differentiation to both sides

$\frac{{y}'}{y}=2(7+x)\ln(7+x)+7+x$

Solve for $y^{'}$

${y}'=y[2(7+x)\ln(7+x)+7+x]$

${y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]$

Report an Error

Example Question #1236 : Calculus

Find $\small d(\tan x)/dx$

Possible Answers:

$\small \sec^2x$

$\small -\sec^2x$

$\small \csc^2x$

$\small \cot^2x$

$\small -\csc^2x$

Correct answer:

$\small \sec^2x$

Explanation:

$\small \frac{d}{dx}(\tan x) = \frac{d}{dx}(\frac{\sin x}{\cos x}) = \frac {\cos x \frac{d}{dx}(\sin x) - \sin x \frac{d}{dx}(\cos x)}{\cos^2x}$

$\small = \frac{ \cos x \cos x- \sin x (- \sin x)} {\cos^2x}$

$\small = \frac{\cos^2x + \sin^2x}{\cos^2x}$
$\small = \frac{1}{\cos^2x} = \sec^2x$

Report an Error

Example Question #1237 : Calculus

Differentiate:

$\small \sin(x^2+x)$ with respect to $\small x$ .

Possible Answers:

$\small (x^2+x)(\cos(2x+1))$

$\small -(2x+1)(cos(x^2+x))$

$\small (2x+1)(\sin(x^2+x))$

$\small (2x+1)(\cos(x^2+x))$

$\small (x^2+x)(\sin(2x+1))$

Correct answer:

$\small (2x+1)(\cos(x^2+x))$

Explanation:

Apply the chain rule: differentiate the "outside" function first. Let $u=x^{2}+x$ .
$\small \small \small \small \frac{d}{du} \sin(u) = \cos(u) = \cos(x^2+x)$
Differentiate the "inside" function next.
$\small \frac{d}{dx}(x^2+x) = 2x+1$
Multiply these two functions to find the derivative of the original function.
$\small \cos(x^2+x) * (2x+1)$

Report an Error

Example Question #1238 : Calculus

Evaluate $\small \int (x^2 - 3x + 6) dx$

Possible Answers:

$\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

$\small 2x-3+C$

$\small 2x - 3$

$\small \small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x$

$\small \small \small x^3 - 3x^2 + 6x + C$

Correct answer:

$\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

Explanation:

To integrate the function, integrate each term of the function. e.g., integrate $\small x^2$ by increasing the exponent by 1 integer and dividing the term by this new integer: $\small \frac {x^3}{3}$ .

Do this for the rest to get $\small \small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x$ .

But remember that every integration requires an arbitrary constant, $\small C$ . Thus, the integral of the function is $\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

Report an Error

Example Question #1239 : Calculus

Integrate

$\small \int\sin^2x\hspace2dx$

Possible Answers:

$\small \small \small \frac{x}{2}+\frac{\cos2x}{4}+C$

$\small \frac{x}{2}-\frac{\sin2x}{4}+C$

$\small \small \frac{x}{2}+\frac{\sin2x}{4}+C$

$\small \small \small \frac{x}{2}+\frac{\sin2x}{4}$

$\small \small \small \frac{x}{2}-\frac{\cos2x}{4}+C$

Correct answer:

$\small \frac{x}{2}-\frac{\sin2x}{4}+C$

Explanation:

We can use trigonometric identities to transform integrals that we typically don't know how to integrate.

$\small \small \sin^2x = \frac{1-\cos2x}{2}$
Thus,
$\small \small \small \int\sin^2x\hspace2dx = \int \frac{1-\cos2x}{2}\hspace2dx$
$\small = \frac{1}{2} \int (1-\cos2x) \hspace2 dx = \frac{1}{2}x-\frac{1}{2}\frac{sin2x}{2}+C = \frac{x}{2} - \frac{sin2x}{4}+C$

Report an Error

View Calculus Tutors

Michael
Certified Tutor

Carthage College, Bachelor of Science, Finance.

View Calculus Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

View Calculus Tutors

Asad
Certified Tutor

NED University of Engineering and Technology, Bachelor of Science, Electrical Engineering. University of Cincinnati-Main Camp...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in Houston, ISEE Tutors in San Diego, LSAT Tutors in New York City, Spanish Tutors in Washington DC, SAT Tutors in Atlanta, ACT Tutors in New York City, Reading Tutors in Seattle, ACT Tutors in Dallas Fort Worth, GMAT Tutors in Atlanta, GRE Tutors in Philadelphia

Popular Courses & Classes

GRE Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Boston, MCAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in San Diego, GRE Courses & Classes in Chicago, MCAT Courses & Classes in Houston, GRE Courses & Classes in Los Angeles, ISEE Courses & Classes in Seattle, Spanish Courses & Classes in Houston

Popular Test Prep

SSAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in San Diego, ACT Test Prep in Seattle, GRE Test Prep in Chicago, MCAT Test Prep in San Diego, GRE Test Prep in Seattle, MCAT Test Prep in Chicago, GMAT Test Prep in Dallas Fort Worth, GRE Test Prep in Dallas Fort Worth, GRE Test Prep in New York City

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 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1231 : Calculus

Example Question #1232 : Calculus

Example Question #17 : Other Differential Functions

Example Question #1233 : Calculus

Example Question #1234 : Calculus

Example Question #1235 : Calculus

Example Question #1236 : Calculus

Example Question #1237 : Calculus

Example Question #1238 : Calculus

Example Question #1239 : Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: