## Thomas' Calculus 13th Edition

$p^{\prime}=-sinq-csc^2q$
Given that $p=(1+csc q)cosq$ $\implies p=cosq+cscq\space cosq$ $p=cosq+cosq.\frac{1}{sinq}$ (because cscq and sinq are reciprocals) $p=cosq+cotq$ Now differentiating both sides, we obtain: $p^{\prime}=-sinq-csc^2q$