If \( f: f(x) \) is differentiable at \( x=2, f(2)=3, \dot{f}(2)=\frac{1}{2}, g(x)=x^{2} f(x) \) ,then \( \dot{g}(2)=\cdots \) \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } \frac{3}{2} & \text { (c) } 12 & \text { (d) } 14\end{array} \)

We have \( g(x)=x^2 f(x) \). To find \(\dot{g}(2)\), we first differentiate \( g(x) \) using the product rule: \[ \dot{g}(x)=\frac{d}{dx}\left(x^2\right)f(x)+x^2\frac{d}{dx}\left(f(x)\right) \] Since \(\frac{d}{dx}\left(x^2\right)=2x\) and \(\frac{d}{dx}\left(f(x)\right)=\dot{f}(x)\), we obtain: \[ \dot{g}(x)=2x f(x)+x^2 \dot{f}(x) \] Now, substitute \( x=2 \), \( f(2)=3 \), and \(\dot{f}(2)=\frac{1}{2}\): \[ \dot{g}(2)=2(2) \cdot 3 + 2^2 \cdot \frac{1}{2} \] Perform the calculations step by step: 1. Compute the first term: \[ 2(2) \cdot 3 = 4 \cdot 3 = 12 \] 2. Compute the second term: \[ 2^2 \cdot \frac{1}{2} = 4 \cdot \frac{1}{2} = 2 \] 3. Sum the results: \[ 12 + 2 = 14 \] Thus, the correct answer is: \[ \boxed{14} \]