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