Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 66

Answer

$f'(x)=-1$

Work Step by Step

$f(x)=\dfrac{4-x^{2}}{x-2}$ Factor the numerator, which is a difference of squares: $f(x)=\dfrac{4-x^{2}}{x-2}=\dfrac{(2-x)(2+x)}{x-2}=...$ Change the sign of the numerator and the sign of the fraction and simplify: $...=-\dfrac{(x-2)(x+2)}{x-2}=-(x+2)=-x-2$ Evaluate the derivative term by term: $f'(x)=-(x)'-(2)'=-1$
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