## Calculus (3rd Edition)

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