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: 90

Answer

$f''(x)=-2{\pi}^2\sec^4({\pi}x)(\cos(2{\pi}x)-2)$

Work Step by Step

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