1. **State the problem:** We need to find the derivative of the composite function $ (f \circ g)(x) = f(g(x)) $ at $ x=2 $. Given functions are $ f(x) = 3x^{2} - 2x + 1 $ and $ g(x) = x^{2} - 5 $. 2. **Recall the chain rule:** The derivative of a composite function $ (f \circ g)(x) $ is given by $$ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) $$ This means we first find the derivative of $ f $ evaluated at $ g(x) $, then multiply by the derivative of $ g $ at $ x $. 3. **Find derivatives of $ f $ and $ g $:** - Derivative of $ f(x) $: $$ f'(x) = \frac{d}{dx}(3x^{2} - 2x + 1) = 6x - 2 $$ - Derivative of $ g(x) $: $$ g'(x) = \frac{d}{dx}(x^{2} - 5) = 2x $$ 4. **Evaluate $ g(2) $:** $$ g(2) = 2^{2} - 5 = 4 - 5 = -1 $$ 5. **Evaluate $ f'(g(2)) = f'(-1) $:** $$ f'(-1) = 6(-1) - 2 = -6 - 2 = -8 $$ 6. **Evaluate $ g'(2) $:** $$ g'(2) = 2 \times 2 = 4 $$ 7. **Apply the chain rule:** $$ (f \circ g)'(2) = f'(g(2)) \cdot g'(2) = (-8) \times 4 = -32 $$ **Final answer:** The derivative of the composite function at $ x=2 $ is $$ \boxed{-32} $$