Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises: 20

Answer

$g'(x)=30x^4 - 1$

Work Step by Step

$g(x) = 6x^5 - x$ Lets use THEOREM 3.5 $g'(x) = (6x^5 -x)' = (6x^5)' + ( -x)'$ $(6x^5)' = 6 \times (x^5)' $ THEOREM 3.4 $(x^4)' = 5x^4 $ THEOREM 3.3 therefore $(6x^5)' = 6 \times (x^5)' = 30x^4$ $(-x)' = -1 \times (x)' $ THEOREM 3.4 $(x)' = 1x^0 = 1$ THEOREM 3.3 therefore $(-x)' = -1 \times (x)' = -1$ Therefore $g'(x) = (6x^5 -x)' = (6x^5)' + ( -x)' = 30x^4 - 1$
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