## Thomas' Calculus 13th Edition

$sec^{2}q$
Using quotient rule, we have $\frac{dp}{dq}= \frac{[(sin q+cos q)'\times cos q]-[(cosq)'\times(sin q+cos q)]}{cos^{2}q}$ $= \frac{[(cosq-sinq)cosq]-[-sin q(sinq+cosq)]}{cos^{2}q}$ $=\frac{[cos^{2}q-sin q cos q]-[-sin^{2}q-sinqcosq]}{cos^{2}q}$ $=\frac{cos^{2}q+sin^{2}q}{cos^{2}q}$= $1+tan^{2}q=sec^{2}q$