Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.3 Exercises: 98

Answer

$f''(x)=\sec^3{x}+\sec{x}\tan^2{x}$

Work Step by Step

First Derivative By Theorem $2.9$: $f'(x)=\sec{x}\tan{x}$ Second Derivative: Product Rule: $f’'(x)=\frac{d}{dx}f'(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x)$ $u(x)=\sec{x} ;u’(x)=\sec{x}\tan{x} $ $v(x)=tan{x} ;v’(x)=\sec^2{x} $ $f''(x)=\sec^3{x}+\sec{x}\tan^2{x}$
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