If \( f(x)=\left(x^{2}+3 x+3\right)^{3} \), then \( f^{\prime}(x)= \) \( f^{\prime}(5)= \)

To find \( f^{\prime}(x) \) for the function \( f(x) = (x^2 + 3x + 3)^3 \), we will use the chain rule. Let \( u = x^2 + 3x + 3 \), then \( f(x) = u^3 \). Using the chain rule: \[ f^{\prime}(x) = 3u^2 \cdot u^{\prime} \] First, we need to find \( u^{\prime} \): \[ u^{\prime} = \frac{d}{dx}(x^2 + 3x + 3) = 2x + 3 \] Substituting back: \[ f^{\prime}(x) = 3(x^2 + 3x + 3)^2 \cdot (2x + 3) \] Next, we evaluate \( f^{\prime}(5) \): First, calculate \( u \) at \( x = 5 \): \[ u(5) = 5^2 + 3(5) + 3 = 25 + 15 + 3 = 43 \] Now calculate \( u^{\prime} \) at \( x = 5 \): \[ u^{\prime}(5) = 2(5) + 3 = 10 + 3 = 13 \] Finally, substituting these into \( f^{\prime}(5) \): \[ f^{\prime}(5) = 3(43)^2 \cdot 13 \] Calculating \( (43)^2 \): \[ 43^2 = 1849 \] Thus, \[ f^{\prime}(5) = 3 \cdot 1849 \cdot 13 \] Calculating this step-by-step: \[ 3 \cdot 1849 = 5547 \] Now, \[ 5547 \cdot 13 = 72011 \] Therefore, \[ f^{\prime}(5) = 72011 \]