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 - Page 151: 45

Answer

$f'(x)=20x^{3}+30x^{2}+3$ $f"(x)=60x^{2}+60x$ $f^{(3)}(x)=120x+60$

Work Step by Step

$f(x)=5x^{4}+10x^{3}+3x+6$ $f'(x)=5(4)x^{4-1}+10(3)x^{3-1}+3x^{1-1}+0$ $f'(x)=20x^{3}+30x^{2}+3x^{0}$ $f'(x)=20x^{3}+30x^{2}+3$ $f"(x)=20(3)x^{3-1}+30(2)x^{2-1}+0$ $f"(x)=60x^{2}+60x^1$ $f"(x)=60x^{2}+60x$ $f^{(3)}(x)=60(2)x^{2-1}+60x^{1-1}$ $f^{(3)}(x)=120x^{1}+60x^{0}$ $f^{(3)}(x)=120x+60$
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