Answer
$h'(x)=2x\sec{x^2}\tan{x^2}.$
Work Step by Step
$u=x^2$; $\dfrac{du}{dx}=2x$
$\dfrac{d}{du}h(u)=\sec{u}\tan{u}.$
$\dfrac{d}{dx}h(x)=\dfrac{d}{du}h(u)\times\dfrac{du}{dx}=2x\sec{x^2}\tan{x^2}.$
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: 46
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