Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 8

Answer

$(198x + 99)( x^2 + x + 1)^{98}$

Work Step by Step

Original Expression: $F(u) = (1 + x + x^2)^{99}$ $ u = (1 + x + x^2)$ $ F(u) = x^{99}$ Apply the chain rule: $f′(u)\times u′$ $F’(u) = 99(u)^{98} (2x+1)$ $F'(x) = 99(x^2 + x + 1)^{98} (2x+1)$ Simplify: $(198x + 99)( x^2 + x + 1)^{98}$
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