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