University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 137: 68



Work Step by Step

$$A=\lim_{x\to-1}\frac{x^{2/9}-1}{x+1}=\lim_{x\to-1}\frac{x^{2/9}-(-1)^{2/9}}{x-(-1)}$$ According to the definition of derivative, we know that $$\lim_{x\to z}\frac{f(x)-f(z)}{x-z}=f'(z)$$ So here if we take $f(x)=x^{2/9}$, we have $$f'(z)=\lim_{x\to z}\frac{x^{2/9}-z^{2/9}}{x-z}=\frac{d(z^{2/9})}{dz}=\frac{2}{9}z^{2/9-1}=\frac{2}{9}z^{-7/9}$$ Take $z=-1$, then $$f'(-1)=\lim_{x\to1}\frac{x^{2/9}-(-1)^{2/9}}{x-(-1)}=\frac{2}{9}\times(-1)^{-7/9}=\frac{2}{9}\times(-1)=-\frac{2}{9}$$ $$A=-\frac{2}{9}$$
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.