Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 75


$y'=(2x+1)^3(10x+1)$ $y''=16(2x+1)^2(5x+1)$

Work Step by Step

y=$ x(2x+1)^4$ y $'=x.4(2x+1)^3(2)+1.((2x+1)^3)(8x+(2x+1))$ =$(2x+1)^3(10x+1)$ $ y''=(2x+1)^3(10)+3(2x+1)^2(2)(10x+1)$ =2$(2x+1)^2(5(2x+1)+3(10x+1))$ =2$(2x+1)^2(40x+8)$ =$16(2x+1)^2(5x+1)$
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