Answer
The derivative is $\frac{-tsin(t)-3cos(t)}{t^4}$
Work Step by Step
Using the quotient rule: $f'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$
$u(t)=cos(t) ;u'(t)=-sin(t)$
$v(t)=t^3 ;v'(t)=3t^2$
$f'(t)=\frac{(-sin(t))(t^3)-(cos(t))(3t^2)}{(t^3)^2}$
$=\frac{(t^2)(-tsin(t)-3cos(t))}{t^6}=\frac{-tsin(t)-3cos(t)}{t^4}$.
