Answer
$g'(x)=15\sec^2{3x}.$
Work Step by Step
$u=3x$; $\dfrac{du}{dx}=3$
$g(u)=5\tan{u};\dfrac{d}{du}g(u)=5\sec^2{u}=5\sec^2{3x}$
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=15\sec^2{3x}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - 2.4 Exercises - Page 136: 45
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