We first observe that \(h\) is the product of two functions: \(h(t) = a(t) \cdot b(t)\text{,}\) where \(a(t) = 3^{t^2 + 2t}\) and \(b(t) = \sec^4(t)\text{.}\) We will need to use the product rule to differentiate \(h\text{.}\) And because \(a\) and \(b\) are composite functions, we will need the chain rule. We therefore begin by computing \(a'(t)\) and \(b'(t)\text{.}\)
Writing \(a(t) = f(g(t)) = 3^{t^2 + 2t}\text{,}\) and finding the derivatives of \(f\) and \(g\text{,}\) we have
\(f(t) = 3^t\) 

\(g(t) = t^2 + 2t\) 
\(f'(t) = 3^t \ln(3)\) 

\(g'(t) = 2t+2\) 
\(f'(g(t)) = 3^{t^2 + 2t}\ln(3)\) 


Thus, by the chain rule, it follows that \(a'(t) = f'(g(t))g'(t) = 3^{t^2 + 2t}\ln(3) (2t+2)\text{.}\)
Turning next to \(b\text{,}\) we write \(b(t) = r(s(t)) = \sec^4(t)\) and find the derivatives of \(r\) and \(s\text{.}\)
\(r(t) = t^4\) 

\(s(t) = \sec(t)\) 
\(r'(t) = 4t^3\) 

\(s'(t) = \sec(t)\tan(t)\) 
\(r'(s(t)) = 4\sec^3(t)\) 


By the chain rule,
\begin{equation*}
b'(t) = r'(s(t))s'(t) = 4\sec^3(t)\sec(t)\tan(t) = 4 \sec^4(t) \tan(t)\text{.}
\end{equation*}
Now we are finally ready to compute the derivative of the function \(h\text{.}\) Recalling that \(h(t) = 3^{t^2 + 2t}\sec^4(t)\text{,}\) by the product rule we have
\begin{equation*}
h'(t) = 3^{t^2 + 2t} \frac{d}{dt}[\sec^4(t)] + \sec^4(t) \frac{d}{dt}[3^{t^2 + 2t}]\text{.}
\end{equation*}
From our work above with \(a\) and \(b\text{,}\) we know the derivatives of \(3^{t^2 + 2t}\) and \(\sec^4(t)\text{,}\) and therefore
\begin{equation*}
h'(t) = 3^{t^2 + 2t} 4\sec^4(t) \tan(t) + \sec^4(t) 3^{t^2 + 2t}\ln(3) (2t+2)\text{.}
\end{equation*}