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: 19

Answer

$f'(x) = 12x^3 + 7$

Work Step by Step

$f(x) = 3x^4 + 7x$ $f'(x) = (3x^4 + 7x)' = (3x^4)' + ( 7x)'$ $(3x^4)' = 3 \times (x^4)' $ $(x^4)' = 4x^3 $ therefore $(3x^4)' = 3 \times (x^4)' = 12x^3$ $(7x)' = 7 \times (x)' $ $(x)' = 1x^0 = 1$ therefore $(7x)' = 7 \times (x)' = 7$ Therefore $f'(x) = (3x^4 + 7x)' = (3x^4)' + ( 7x)' = 12x^3 + 7$
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