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: 25

Answer

$y' = 3sec(3x+1)tan(3x+1)$

Work Step by Step

$y = sec (3x+1)$ $y' = (sec(3x+1))'$ The inner function is $g(x) = 3x+1$ and the outer function is $f(u) = sec (u)$ $f'(u) = sec(u)tan(u)$ $y' = (sec(3x+1))' = sec(g(x))tan(g(x)) g'(x)= sec(3x+1)tan(3x+1) \times 3= 3sec(3x+1)tan(3x+1)$
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