Let $f(t)=\left(t^{2}+6 t+4\right)\left(2 t^{2}+4\right)$. Find $f^{\prime}(t)$. \[ f^{\prime}(t)=8 t^{\wedge} 3+36 t^{\wedge} 2+24 t+24 \] Find $f^{\prime}(3)$. \[ f^{\prime}(3)= \]

Step 1 ：Substitute \(t=3\) into the derivative of the function \(f(t)\) to find \(f^\prime(3)\)

Step 2 ：So, \(f^\prime(3)=8(3)^3+36(3)^2+24(3)+24\)

Step 3 ：Calculate it step by step:

Step 4 ：\(f^\prime(3)=8(27)+36(9)+72+24\)

Step 5 ：\(f^\prime(3)=216+324+72+24\)

Step 6 ：\(f^\prime(3)=636\)

Step 7 ：So, the final answer is \(\boxed{636}\)