Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$. Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$.
So we have the (Macluarin) series
\begin{equation*}
\sinh{(x)} = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}.
\end{equation*}
at $x = 0$ so
\begin{equation*}
\sinh{(x^4)} = \sum_{n=0}^{\infty} \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{8n+4}}{(2n+1)!}.
\end{equation*}
The Maclaurin series for $\sinh{(x^4)}$ is
\begin{equation*}
\sum_{n=0}^{\infty} \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{8n+4}}{(2n+1)!} = \sum_{m=0}^{\infty} \frac{x^m}{\left(\frac{m}{4}\right)!}.
\end{equation*}
We can see that the coefficient for $x^m$ is $\frac{1}{\left(\frac{m}{4}\right)!}$. On the other hand, this is the Maclaurin series for $g(x)$, and so the coefficient on $x^m$ is equal to $\frac{g^{(m)}(0)}{m}$. Equating these two, we have
\begin{equation*}
\frac{g^{(m)}(0)}{m} = \frac{1}{\left(\frac{m}{4}\right)!} \Longleftrightarrow g^{(m)}(0) = \frac{m}{\left(\frac{m}{4}\right)!}.
\end{equation*}
Substituting $m = 30$ gives $g^{(30)}(0) = \frac{30}{7.5!}$. Am I on the right track here?

Definitely on the right track. Last thing you need to know is that the coefficient of $x^n$ in the Maclaurin series is given by $a_n=\frac{f^{n}(0)}{n!}$
