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 $f^{\prime}(3)$.
\[
f^{\prime}(3)=
\]

Answer

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Answer

So, the final answer is $\boxed{636}$

Steps

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 $\boxed{636}$