Differentiate the function. $$ f(x)=x^{3}(x+4) $$ $$ f^{\prime}(x)=\square $$

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**Step 1: Rewrite the function**  
We have  
$$
f(x) = x^3 (x+4)
$$

**Step 2: Apply the Product Rule**  
The product rule states that if  
$$
f(x) = u(x)v(x),
$$  
then  
$$
f'(x) = u'(x)v(x) + u(x)v'(x).
$$  
Let  
$$
u(x) = x^3 \quad \text{and} \quad v(x) = x+4.
$$

**Step 3: Differentiate $$ u(x) $$ and $$ v(x) $$**  
Differentiate $$ u(x) $$:  
$$
u'(x) = 3x^2.
$$  
Differentiate $$ v(x) $$:  
$$
v'(x) = 1.
$$

**Step 4: Substitute into the Product Rule**  
$$
f'(x) = u'(x)v(x) + u(x)v'(x) = 3x^2(x+4) + x^3(1).
$$

**Step 5: Simplify the Expression**  
Expand the term $$ 3x^2(x+4) $$:  
$$
3x^2(x+4) = 3x^3 + 12x^2.
$$  
Now, add the term $$ x^3 $$:  
$$
f'(x) = 3x^3 + 12x^2 + x^3 = 4x^3 + 12x^2.
$$

**Step 6: Factor the Derivative**  
Factor out $$ 4x^2 $$:  
$$
f'(x) = 4x^2 (x + 3).
$$

$$
\boxed{f'(x) = 4x^2(x+3)}
$$
$$ f^{\prime}(x) = 4x^2(x + 3) $$

Differentiate the function. \[ f(x)=x^{3}(x+4) \] \( f^{\prime}(x)=\square \)

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