Answer
$$F'(t)=\frac{t(t^3+4)}{2(t^3+1)^{3/2}}$$
Work Step by Step
$$F'(t)=\left(\frac{t^2}{\sqrt{t^3+1}}\right)'=\frac{(t^2)'\sqrt{t^3+1}-t^2(\sqrt{t^3+1)'}}{(\sqrt{t^3+1})^2}=\frac{2t\sqrt{t^3+1}-t^2\frac{1}{2\sqrt{t^3+1}}(t^3+1)'}{t^3+1}=\frac{4t(t^3+1)-t^2\cdot3t^3}{2(t^3+1)^{3/2}}=\frac{4t^4+4t-3t^4}{2(t^3+1)^{3/2}}=
\frac{t(t^3+4)}{2(t^3+1)^{3/2}}$$
