Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 191: 20


$y '= 8(x^2+2x+7)^7\times (x+2)$

Work Step by Step

$y =(x^2+2x+7)^{8}$ $y ' =((x^2+2x+7)^{8})' $ The inner function sis $g(x) = x^2+2x+7$ and the outer function is $f(u) = u^{8}$ $f'(u) = 8u^7$ $((x^2+2x+7)^{8})' = 8(g(x))^7\times g'(x)$ $= 8(x^2+2x+7)^7\times (x+2)$
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