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: 93

Answer

$f''(0)=0.$

Work Step by Step

$u=x^2$; $\dfrac{du}{dx}=2x$ $\dfrac{d}{du}f(u)=-\sin{u}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=-2x\sin{x^2}.$ Product Rule $(f’'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=-2x ;u’(x)=-2$ $v(x)=\sin{x^2} ;v’(x)=2x\cos{x^2} $ $f''(x)=(-2)(\sin{x^2})(-2x)(2x\cos{x^2})$ $=-2(\sin{x^2}+2x^2\cos{x^2}).$ $f''(0)=-2(\sin{0^2}+2(0)^2\cos{0^2})=0$.
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