Answer
$y'=2πxsec^2(πx^2)$
Work Step by Step
First, find the derivative of y and u. Don't forget to apply chain rule for y':
$y'=sec^2u\times u'$
$u'=2πx$
Then plug in u and u' into y':
$y'=2πxsec^2(πx^2)$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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 148: 7
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