Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1: 22

Answer

$f'(0)=2$

Work Step by Step

To find the slope of a function at any given point, all one has to do would be to look for the value of the derivative function at the same x value. Given that $f(x)=\frac{x-1}{x+1}$, then the derivative function by the quotient rule will be $f′(x)=\frac{(-1)(x-1)}{(x+1)^2}+\frac{1}{x+1}=\frac{2}{(x+1)^2}$ As such, the slope at the point $x=0$ would be $f′(0)=\frac{0+2}{(0+1)^2}=2$
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