Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 15


$$ y'=\frac{\sec \theta (\theta \tan \theta -1) }{\theta ^2}$$

Work Step by Step

Recall that $(\sec x)'=\sec x\tan x$. Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $ y= \frac{\sec \theta }{\theta }$, then, using the quotient rule, we have \begin{align*} y'&=\frac{\theta (\sec \theta)'- \sec \theta (\theta )'}{\theta ^2}\\ &=\frac{\theta \sec \theta \tan \theta -\sec \theta }{\theta ^2}\\ &=\frac{\sec \theta (\theta \tan \theta -1) }{\theta ^2}. \end{align*}
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