Find Derivative \( h(t)=\left(\frac{t^{2}}{t^{3}+2}\right)^{2} \)

To find the derivative of \( h(t)=\left(\frac{t^{2}}{t^{3}+2}\right)^{2} \), we can use the chain rule and the quotient rule. First, let's denote \( u(t) = \frac{t^{2}}{t^{3}+2} \). Then, we have \( h(t) = u(t)^2 \). Using the chain rule: \[ h'(t) = 2u(t) \cdot u'(t) \] Now, to find \( u'(t) \), we apply the quotient rule: \[ u'(t) = \frac{(t^{3}+2)(2t) - t^{2}(3t^{2})}{(t^{3}+2)^{2}} \] This simplifies to: \[ u'(t) = \frac{2t(t^{3}+2) - 3t^{4}}{(t^{3}+2)^{2}} = \frac{2t^{4} + 4t - 3t^{4}}{(t^{3}+2)^{2}} = \frac{-t^{4} + 4t}{(t^{3}+2)^{2}} \] Substituting \( u(t) \) and \( u'(t) \) back into our expression for \( h'(t) \): \[ h'(t) = 2 \left(\frac{t^{2}}{t^{3}+2}\right) \cdot \frac{-t^{4} + 4t}{(t^{3}+2)^{2}} \] Thus, the derivative simplifies to: \[ h'(t) = \frac{2t^{2}(-t^{4} + 4t)}{(t^{3}+2)^{3}} \] This gives you the derivative of the function \( h(t) \).