Answer
$$ y''=-\frac{8}{x^3}, \quad y'''=\frac{24}{x^4}.$$
Work Step by Step
Let $ y=\frac{x-4}{x}$; first we simplify $ y $ as follows
$$ y= 1-\frac{4}{x}.$$
Now, we get
$$ y'=\frac{4}{x^2}$$
$$ y''=-\frac{8}{x^3}, \quad y'''=\frac{24}{x^4}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 135: 13
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