Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 45

Answer

$g'(x)=15\sec^2{3x}.$

Work Step by Step

$u=3x$; $\dfrac{du}{dx}=3$ $g(u)=5\tan{u};\dfrac{d}{du}g(u)=5\sec^2{u}=5\sec^2{3x}$ $\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=15\sec^2{3x}.$
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