Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 32

$$ p^{\prime}=-3 \sin q+\frac{q \cos q-\sin q}{q^{2}}$$

Given $$ p=\frac{3 q+\tan q}{q \sec q}$$ So, we have \begin{aligned} p&=\frac{3 q+\tan q}{q \sec q}\\ &=\frac{3 q}{q \sec q}+\frac{\tan q}{q \sec q}\\ &=\frac{3}{\sec q}+\frac{\frac{\sin q}{\cos q}}{\frac{ q}{\cos q}}\\ &=3 \cos q+\frac{\sin q }{ q}\\ \text{so, we get}\\ p^{\prime}&=-3 \sin q+\frac{q \cos q-\sin q}{q^{2}}\end{aligned}