Answer
$\frac{-2t^{3}-t^{2}+1}{(t^{3}+t^{2}+t+1)^{2}}$
Work Step by Step
$h(t)=\frac{t}{(t+1)(t^{2}+1)}=\frac{t}{t^{3}+t^{2}+t+1}$
Using the quotient rule, we get
$h'(t)=\frac{(t^{3}+t^{2}+t+1)\frac{d}{dt}(t)-t\times\frac{d}{dt}(t^{3}+t^{2}+t+1)}{(t^{3}+t^{2}+t+1)^{2}}$
$=\frac{(t^{3}+t^{2}+t+1)\times1-t(3t^{2}+2t+1)}{(t^{3}+t^{2}+t+1)^{2}}$
$=\frac{-2t^{3}-t^{2}+1}{(t^{3}+t^{2}+t+1)^{2}}$
