Answer
$s'=5\sec t(\sec t+\tan t)^5$
Work Step by Step
Given that $s=(\sec t+\tan t)^5$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[(\sec t+\tan t)^5]$
$=5(\sec t+\tan t)^4\dfrac{d}{dx}(\sec t+\tan t)$
$=5(\sec t+\tan t)^4(\sec t\tan t+\sec^2 t)$
Hence, $s'=5\sec t(\sec t+\tan t)^5$
