Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - Review Exercises: 6



Work Step by Step

$f(x)=\frac{1}{x+4}$ $c=3$ $f'(c)=\lim\limits_{x \to c}\frac{f(x)-f(c)}{x-c}$ $f'(3)=\lim\limits_{x \to 3}\frac{\frac{1}{x+4}-\frac{1}{3+4}}{x-3}=\lim\limits_{x \to 3}\frac{\frac{1}{x+4}-\frac{1}{7}}{x-3}=\lim\limits_{x \to 3}\frac{(\frac{7-(x+4)}{7(x+4)})}{x-3}=\lim\limits_{x \to 3}{\frac{-(x-3)}{7(x+4)(x-3)}}=\lim\limits_{x \to 3}{\frac{-(x-3)}{7(x+4)(x-3)}}=\lim\limits_{x \to 3}\frac{-1}{7x+28}=\frac{-1}{7(3)+28}=\frac{-1}{21+28}=\frac{-1}{49}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.