Answer
$h'(t)=\dfrac{2t^3(4-t^3)}{(t^3+2)^3}.$
Work Step by Step
Using the quotient rule: $h'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$.
$u(t)=t^4 ;u'(t)=4t^3$.
$v(t)=(t^3+2)^2 $
$v'(t)$ is found using the chain rule:
$u=t^3+2$; $\dfrac{du}{dt}=3t^2$
$\dfrac{d}{du}v(u)=2u$
$\dfrac{d}{dt}v(t)=\dfrac{d}{du}v(u)\times\dfrac{du}{dt}=6t^2(t^3+2).$
$h'(t)=\dfrac{(4t^3)(t^3+2)^2-(6t^2(t^3+2))(t^4)}{(t^3+2)^4}$
$=\dfrac{2t^3(4-t^3)}{(t^3+2)^3}.$
