Answer
$y' = 3sec(3x+1)tan(3x+1)$
Work Step by Step
$y = sec (3x+1)$
$y' = (sec(3x+1))'$
The inner function is $g(x) = 3x+1$
and the outer function is $f(u) = sec (u)$
$f'(u) = sec(u)tan(u)$
$y' = (sec(3x+1))' = sec(g(x))tan(g(x)) g'(x)= sec(3x+1)tan(3x+1) \times 3= 3sec(3x+1)tan(3x+1)$
