Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises - Page 267: 5

Answer

$y′=x(2sin(πx)+πxcos(πx))$

Work Step by Step

Start with the function: $y=x^2sin(πx)$. Use the product rule to differentiate: $y′=2xsin(πx)+πx^2cos(πx)$. Factor out the common $x$ term to arrive at the answer: $y′=x(2sin(πx)+πxcos(πx))$.
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