AllExperts > Calculus 
Search      
Calculus
Volunteer
Answers to thousands of questions
 Home · More Calculus Questions · Answer Library  · Encyclopedia ·
More Calculus Answers
Question Library

Ask a question about Calculus
Volunteer
Experts of the Month
Expert Login

Awards

About Us
Tell friends
Link to Us
Disclaimer

 
 
 
 
About Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience
Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook


 
   

You are here:  Experts > Teens > Homework/Study Tips > Calculus > finding the derivative

Calculus - finding the derivative


Expert: Abe Mantell - 7/24/2008

Question
hi, i have summer calculas home work to do but i forgot how to do the the derivative part because we spent so little time on it. my question is:

let f(x)= (4x-8)/(x^2+5x-14)

a-find the derivative of f(x)at x=1
b- for what values is the derivative of f, f'(x), not continous
c- determine the limit of the derivative of f, f'(x), at each point of discontinuity found in part (b)

thank you very much for your help

Answer
Hello Kiko,

The usual way to take the derivative of this would be via the
quotient rule.  Thus,
a. f'(x)=[(4x-8)'(x^2+5x-14)-(4x-8)(x^2+5x-14)']/(x^2+5x-14)^2
--- = [4(x^2+5x-14)-(4x-8)(2x+5)]/(x^2+5x-14)^2
--- which simplifies to: f'(x)=-4/(x+7)^2, provided x is not equal to 2.
b. at x=2 and x=-7
c. limit (x->2) f'(x)=-4/(2+7)^2=-4/81, limit (x->-7) f'(x) is undefined

I will soon be taking a vacation from AllExperts.com, so e-mail me
directly at mantell@alum.rpi.edu if you have any more questions about
this!

Abe


Add to this Answer   Ask a Question


 
User Agreement | Privacy Policy | Kids' Privacy Policy | Help
Copyright  © 2008 About, Inc. AllExperts, AllExperts.com, and About.com are registered trademarks of About, Inc. All rights reserved.