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: 24

Answer

$y' = -5sin(5t)$

Work Step by Step

$y = cos(5t)$ $y' = (cos(5t))'$ The inner function is $g(t) = 5t$ and the outer function is $f(u) = cos (u)$ $f'(u) = -sin (u)$ $y' = (cos(5t))' = - sin (g(t))\times g'(t) = -sin (5t)\times5 = -5sin(5t)$
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