Answer
$f'(t)=\frac{-t\sin(t)-\cos(t)}{t^2}$.
Work Step by Step
Using the quotient rule: $g'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$.
$u(t)=\cos(t) ;u'(t)=-\sin(t)$.
$v(t)=t ;v'(t)=1$. $f'(t)=\frac{(-\sin(t))(t)-(1)(\cos(t))}{t^2}$.
$=\frac{-t\sin(t)-\cos(t)}{t^2}$.
