The differentiable functions $f$ and $g$ are defined for all real numbers $x$. Values of $f, f^{\prime}, g$, and $g^{\prime}$ for various values of $x$ are given in the table.
\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & $f(x)$ & $f^{\prime}(x)$ & $g(x)$ & $g^{\prime}(x)$ \\
\hline 1 & 3 & 4 & 2 & 6 \\
\hline 2 & 1 & 5 & 8 & 7 \\
\hline 3 & 7 & 7 & 2 & 9 \\
\hline
\end{tabular}
(a) If $h(x)=f(g(x))$, find $h^{\prime}(1)$.
\[
h^{\prime}(1)=
\]
(b) If $H(x)=g(f(x))$, find $H^{\prime}(2)$.
\[
H^{\prime}(2)=
\]

Step 1 ：Use the chain rule to find \(h^\prime(1)\), which states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function. In this case, \(h(x)=f(g(x))\), so \(h^\prime(x)=f^\prime(g(x)) \cdot g^\prime(x)\).

Step 2 ：Substitute \(x=1\) into the equation to get \(h^\prime(1)=f^\prime(g(1)) \cdot g^\prime(1)\).

Step 3 ：From the table, we know that \(g(1)=2\) and \(g^\prime(1)=6\).

Step 4 ：So, \(h^\prime(1)=f^\prime(2) \cdot 6\).

Step 5 ：Again from the table, we know that \(f^\prime(2)=5\).

Step 6 ：So, \(h^\prime(1)=5 \cdot 6\).

Step 7 ：\(\boxed{h^\prime(1)=30}\)

Step 8 ：Similarly, use the chain rule to find \(H^\prime(2)\). In this case, \(H(x)=g(f(x))\), so \(H^\prime(x)=g^\prime(f(x)) \cdot f^\prime(x)\).

Step 9 ：Substitute \(x=2\) into the equation to get \(H^\prime(2)=g^\prime(f(2)) \cdot f^\prime(2)\).

Step 10 ：From the table, we know that \(f(2)=1\) and \(f^\prime(2)=5\).

Step 11 ：So, \(H^\prime(2)=g^\prime(1) \cdot 5\).

Step 12 ：Again from the table, we know that \(g^\prime(1)=6\).

Step 13 ：So, \(H^\prime(2)=6 \cdot 5\).

Step 14 ：\(\boxed{H^\prime(2)=30}\)