Answer
$f'(t)=6\pi\sec^2{(\pi t-1)}\tan{(\pi t-1)}.$
Work Step by Step
$u=\sec{(\pi t-1)}$; $\dfrac{du}{dt}=\pi\sec{(\pi t-1)}\tan{(\pi t-1)}$
$\dfrac{d}{du}f(u)=6u.$
$\dfrac{d}{dt}f(t)=\dfrac{d}{du}f(u)\times\dfrac{du}{dt}$
$=(6\sec{(\pi t-1)})(\pi\sec{(\pi t-1)}\tan{(\pi t-1)})$
$=6\pi\sec^2{(\pi t-1)}\tan{(\pi t-1)}.$
