Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 19


$f'(\theta) = 9(-2\sin \theta + 5\cos \theta)(2\cos \theta + 5 \sin \theta)^8$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make an «u» substitution to make it easier $u = 2\cos \theta + 5 \sin \theta $ $f(u) = u^9 $ Derivate the function: $f'(u) = 9u^8u'$ Now let's find u' $u' = -2\sin \theta + 5\cos \theta$ Then undo the substitution, simplify and get the answer: $f'(\theta) = 9(-2\sin \theta + 5\cos \theta)(2\cos \theta + 5 \sin \theta)^8$
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