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: 89

Answer

$f''(x)=2\cos{x^2}-4x^2\sin{x^2}.$

Work Step by Step

Using the Chain Rule: $u=x^2$; $\dfrac{du}{dx}=2x$ $\dfrac{d}{du}f(u)=\cos{u}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=2x\cos{x^2}$ Product Rule $(f’'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=2x ;u’(x)=2 $ $v(x)=\cos{x^2} ;v’(x)=-2x\sin{x^2} $ $f''(x)=(2)(\cos{x^2})+(-2x\sin{x^2})(2x)$ $=2\cos{x^2}-4x^2\sin{x^2}.$
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