Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find differential functions

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 #681 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=xe^{x^2}$ at the point x=2

Possible Answers:

3e^4

8e^4

2e^4

9e^4

e^4

Correct answer:

9e^4

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

d[e^u]=e^udu

Product rule: d[uv]=udv+vdu

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=xe^{x^2}$ at the point x=2

The slope of the tangent is

$f'(x)=e^{x^2}+2x^2e^{x^2}$

$f'(2)=e^{2^2}+2(2)^2e^{2^2}=9e^4$

Report an Error

Example Question #682 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=sin(xe^x) at the point x=0

Possible Answers:

e+1

Correct answer:

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

d[e^u]=e^udu

Trigonometric derivative:

d[sin(u)]=cos(u)du

Product rule: d[uv]=udv+vdu

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=sin(xe^x) at the point x=0

The slope of the tangent is

$f'(x)=(e^x+xe^x)cos(xe^{x})$

f'(0)=(e^0+(0)e^0)cos((0)e^0})=1

Report an Error

Example Question #683 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=tan(\frac{\pi x^2}{4})$ at the point x=6

Possible Answers:

$3\pi$

$-6\pi$

$-3\pi$

$6\pi$

Correct answer:

$3\pi$

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=tan(\frac{\pi x^2}{4})$ at the point x=6

The slope of the tangent is

$f'(x)=\frac{\frac{x}{2}\pi}{cos^2(\frac{\pi x^2}{4})}$

$f'(6)=\frac{\frac{6}{2}\pi}{cos^2(\frac{\pi (6)^2}{4})}=3\pi$

Report an Error

Example Question #684 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=cos(ln(cos(x))) at the point $x=2\pi$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

d[cos(u)]=-sin(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function

The slope of the tangent is f(x)=cos(ln(cos(x))) at the point $x=2\pi$

$f'(x)=-\frac{-sin(x)}{cos(x)}sin(ln(cos(x)))=tan(x)sin(ln(cos(x)))$

$f'(2\pi)=tan(2\pi)sin(ln(cos(2\pi)))=0$

Report an Error

Example Question #685 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=sin(7x+2\pi)$ at the point $x=\pi$

Possible Answers:

$9\pi$

$-9\pi$

Correct answer:

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=sin(7x+2\pi)$ at the point $x=\pi$

The slope of the tangent is

$f'(x)=7cos(7x+2\pi)$

$f'(\pi)=7cos(7\pi+2\pi)=-7$

Report an Error

Example Question #1902 : Calculus

Find the derivative of the following: $f(x) = (6x+3)^3 + \cos(4x)$

Possible Answers:

$f'(x) = 18(6x+3)^2 + 4\sin(4x)$

$f'(x) = 6x(6x+3)^2 + \sin(4x)$

$f'(x) = 18(6x+3)^2 -4\sin(4x)$

$f'(x) = 6x(6x+3)^2 -4\sin(4x)$

$f'(x) = 6x(6x+3)^2 -\sin(4x)$

Correct answer:

$f'(x) = 18(6x+3)^2 -4\sin(4x)$

Explanation:

If f(x)=C , then the derivative is f'(x)= 0 .

If f(x)= x^C , the the derivative is $f'(x)= (C)x^{(C-1)}$ .

If $f(x)= \sin x$ , then the derivative is $f'(x)= \cos x$ .

If f(x)= e^x , then the derivative is f'(x)= e^x .

If f(x)= ln (x) , then the derivative is $f'(x)= \frac{1}{x}$ .

There are many other rules for the derivatives for trig functions.

If F(x)= (f*g)(x) , then the derivative is F'(x)= f'(g(x)) g'(x) . This is known as the chain rule.

In this case, we must find the derivative of the following: $f(x) = (6x+3)^3 + \cos(4x)$

That is done by doing the following: $f'(x) = {\color{Blue} (3)}(6x+3)^{\color{Blue} {(3-1)}}({\color{Blue} (1)}6x^{\color{Blue} {(1-1)}}+0) + ({\color{Blue} -\sin}(4x))({\color{Blue} (1)}4x^{\color{Blue} {(1-1)}})$

Therefore, the answer is: $f'(x) = 18(6x+3)^2 -4\sin(4x)$

Report an Error

Example Question #686 : How To Find Differential Functions

Let f(x)=x^3cos^3(x) on the interval (0,3) . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

7.8

2.1

6.4

3.4

1.7

Correct answer:

2.1

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint (a,b);a<b , there exists at a minimum one point, , within the interval (a,b) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of f(x)=x^3cos^3(x) on the interval (0,3)

f(0)=0^3cos^3(0)=0

f(3)=3^3cos^3(3)=-26.2

Then take the difference of the two and divide by the interval.

$\frac{-26.2-0}{3-0}=-8.7$

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of an exponential:

Trigonometric derivative:

d[cos(u)]=-sin(u)du

Product rule: d[uv]=udv+vdu

f'(x)=3x^2cos^3(x)-3x^3sin(x)cos^2(x)

3x^2cos^3(x)-3x^3sin(x)cos^2(x)=-8.7

Using a calculator, we find the solution x=2.1 , which fits within the interval (0,3) , satisfying the mean value theorem.

Report an Error

Example Question #1904 : Calculus

Let f(x)=xsin^3(x) on the interval $(0,\frac{\pi}{2})$ . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

0.54

0.89

0.72

1.31

0.12

Correct answer:

0.72

Explanation:

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of f(x)=xsin^3(x) on the interval $(0,\frac{\pi}{2})$

f(0)=(0)sin^3(0)=0

$f(\frac{\pi}{2})=(\frac{\pi}{2})sin^3(\frac{\pi}{2})=\frac{\pi}{2}$

Then take the difference of the two and divide by the interval.

$\frac{\frac{\pi}{2}-0}{\frac{\pi}{2}-0}=1$

Now find the derivative of the function; this will be solved for the value(s) found above.

Trigonometric derivative:

d[sin(u)]=cos(u)du

Product rule: d[uv]=udv+vdu

f'(x)=sin^3(x)+3xcos(x)sin^2(x)

sin^3(x)+3xcos(x)sin^2(x)=1

Using a calculator, we find the solutions x=0.72 , which fits within the open interval $(0,\frac{\pi}{2})$ , satisfying the mean value theorem.

Report an Error

Example Question #1905 : Calculus

Let f(x)=xcos^2(x) on the interval $(\frac{\pi}{4},\frac{\pi}{2})$ . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

0.91

1.25

1.04

0.83

1.49

Correct answer:

0.91

Explanation:

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of f(x)=xcos^2(x) on the interval $(\frac{\pi}{4},\frac{\pi}{2})$

$f(\frac{\pi}{4})=(\frac{\pi}{4})cos^2(\frac{\pi}{4})=\frac{\pi}{8}$

$f(\frac{\pi}{4})=(\frac{\pi}{2})cos^2(\frac{\pi}{2})=0$

Then take the difference of the two and divide by the interval.

$\frac{0-\frac{\pi}{8}}{\frac{\pi}{2}-\frac{\pi}{4}}=-\frac{1}{2}$

Now find the derivative of the function; this will be solved for the value(s) found above.

Trigonometric derivative:

d[cos(u)]=-sin(u)du

Product rule: d[uv]=udv+vdu

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

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

Using a calculator, we find the solutions x=0.91,1.37 , which fit within the interval $(\frac{\pi}{4},\frac{\pi}{2})$ , satisfying the mean value theorem.

Report an Error

Example Question #1906 : Calculus

Let f(x)=-ln(cos(x)) on the interval (0,1) . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

0.56

0.67

0.23

0.34

0.11

Correct answer:

0.56

Explanation:

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of f(x)=-ln(cos(x)) on the interval (0,1)

f(0)=-ln(cos(0))=0

f(1)=-ln(cos(1))=0.62

Then take the difference of the two and divide by the interval.

$\frac{0.62-0}{1-0}=0.62$

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

d[cos(u)]=-sin(u)du

$f'(x)=-\frac{-sin(x)}{cos(x)}=tan(x)$

tan(x)=0.62

Using a calculator, we find the solution x=0.56 , which fits within the interval (0,1) , satisfying the mean value theorem.

Report an Error

View Calculus Tutors

Sharon
Certified Tutor

Tabor College, Bachelor of Science, Mathematics Teacher Education. Tabor College, Masters in Education, Education.

View Calculus Tutors

Usman
Certified Tutor

New York University, Bachelor in Arts, Political Science and Government.

View Calculus Tutors

Rainier
Certified Tutor

Norfolk State University, Bachelors, Physics.

All Calculus 1 Resources

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

Popular Subjects

Reading Tutors in Los Angeles, ISEE Tutors in Philadelphia, ACT Tutors in Miami, Computer Science Tutors in Philadelphia, Chemistry Tutors in Phoenix, GMAT Tutors in Phoenix, LSAT Tutors in Los Angeles, MCAT Tutors in San Francisco-Bay Area, MCAT Tutors in Philadelphia, GMAT Tutors in San Francisco-Bay Area

Popular Courses & Classes

GMAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in New York City, MCAT Courses & Classes in Denver, MCAT Courses & Classes in Seattle, ISEE Courses & Classes in Phoenix, GMAT Courses & Classes in Chicago, LSAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in Atlanta, GMAT Courses & Classes in Seattle

Popular Test Prep

SSAT Test Prep in Boston, GRE Test Prep in San Diego, ISEE Test Prep in Houston, ACT Test Prep in Philadelphia, LSAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Chicago, MCAT Test Prep in Chicago, ISEE Test Prep in Miami, MCAT Test Prep in Denver, SSAT Test Prep in Dallas Fort Worth

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 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #681 : How To Find Differential Functions

Example Question #682 : How To Find Differential Functions

Example Question #683 : How To Find Differential Functions

Example Question #684 : How To Find Differential Functions

Example Question #685 : How To Find Differential Functions

Example Question #1902 : Calculus

Example Question #686 : How To Find Differential Functions

Example Question #1904 : Calculus

Example Question #1905 : Calculus

Example Question #1906 : Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: