Answer
$h'(\pi)=\dfrac{1}{\pi^2}$
Work Step by Step
Using the quotient rule: $h'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$.
$u(t)=\sec(t) ;u'(t)=\sec(t)\tan(t)$.
$v(t)=t ;v'(t)=1$
$h'(t)=\frac{(\sec(t)\tan(t))(t)-\sec(t)}{t^2}$.
To evaluate, substitute $x=\pi$ :
$h'(\pi)=\frac{\sec(\pi)((\pi)(\tan(\pi)-1)}{\pi^2}=\frac{1}{\pi^2}$.
