Answer
$g'(t)=-5\pi\sin{2\pi t}.$
Work Step by Step
Using the Chain Rule:
$u=\cos{\pi t}$; $\dfrac{du}{dt}=-\pi\sin{\pi t}.$
$\dfrac{d}{du}g(u)=10u$
$\dfrac{d}{dt}g(t)=\dfrac{d}{du}g(u)\times\dfrac{du}{dt}=(-\pi\sin{\pi t})(10\cos{\pi t})$
$=-10\pi\cos{\pi t}\sin{\pi t}=-5\pi\sin{2\pi t}$
