Let $f$ and $g$ be the functions in the table below. \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 & 3 & 7 \\ \hline 3 & 2 & 7 & 1 & 9 \\ \hline \end{tabular} (a) If $F(x)=f(f(x))$, find $F^{\prime}(2)$. \[ F^{\prime}(2)= \] (b) If $G(x)=g(g(x))$, find $G^{\prime}(3)$. \[ G^{\prime}(3)= \]
Solution
Step 1 ：For part (a), we need to find the derivative of \(F(x)=f(f(x))\) at \(x=2\). We can use the chain rule, which states that the derivative of a composition of functions is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. In this case, the outer function is \(f(x)\) and the inner function is also \(f(x)\). So, \(F'(x) = f'(f(x)) * f'(x)\). We need to evaluate this at \(x=2\).
Step 2 ：For part (b), we need to find the derivative of \(G(x)=g(g(x))\) at \(x=3\). We can use the chain rule again. In this case, the outer function is \(g(x)\) and the inner function is also \(g(x)\). So, \(G'(x) = g'(g(x)) * g'(x)\). We need to evaluate this at \(x=3\).
Step 3 ：Final Answer: (a) \(F^{\prime}(2)=\boxed{20}\) (b) \(G^{\prime}(3)=\boxed{54}\)
