\( \begin{array}{l} \text { Find the value of the given } \\ \text { function's derivative at } x=2 \text {. } \\ f(x)=k(g(x)) \\ g(2)=16 \\ g^{\prime}(2)=1 \\ k(6)=2 \\ k^{\prime}(16)=7 \\ f^{\prime}(2)=[?]\end{array} \)

To find the value of \( f'(2) \), we can use the chain rule. The function \( f(x) = k(g(x)) \) implies that \( f'(x) = k'(g(x)) \cdot g'(x) \). At \( x = 2 \): 1. We know \( g(2) = 16 \), so \( g(x) \) at \( x = 2 \) gives us \( g(2) = 16 \). 2. We also know \( g'(2) = 1 \), which tells us the derivative of \( g(x) \) at that point. 3. Now we need \( k'(g(2)) = k'(16) \). Since \( k'(16) = 7 \), we can substitute this value into our equation. Using the chain rule: \[ f'(2) = k'(g(2)) \cdot g'(2) = k'(16) \cdot g'(2) = 7 \cdot 1 = 7. \] Therefore, the value of the derivative at \( x = 2 \) is \( f'(2) = 7 \).