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: 46

Answer

$h'(x)=2x\sec{x^2}\tan{x^2}.$

Work Step by Step

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