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: 48

Answer

$y'=\dfrac{(\sec(x))(x\tan(x)-1)}{x^2}$.

Work Step by Step

Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=\sec(x); u'(x)=\sec(x)\tan(x)$ $v(x)=x; v'(x)=1$ $y'=\dfrac{(\sec(x)\tan(x))(x)-(1)(\sec(x))}{x^2}$ $=\dfrac{(\sec(x))(x\tan(x)-1)}{x^2}$
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