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To find the first derivative of the given function \( f(x) = |x|(x + 1)^2 \), we need to take into account the absolute value and apply the product rule since this is a product of two functions. Here's the process step-by-step:

### 1. Rewrite the function:
We can break the function into two cases based on the absolute value of \(x\):

\[
f(x) = 
\begin{cases} 
x(x + 1)^2 & \text{for } x \geq 0 \\
-x(x + 1)^2 & \text{for } x < 0 
\end{cases}
\]

Now, we compute the derivative for each case separately.

### 2. Case 1: \( x \geq 0 \)

Here, \( f(x) = x(x + 1)^2 \). Apply the product rule for derivatives:
\[
f'(x) = \frac{d}{dx}[x] \cdot (x + 1)^2 + x \cdot \frac{d}{dx}[(x + 1)^2]
\]

First term:
\[
\frac{d}{dx}[x] = 1, \quad \text{so the first part is } (x + 1)^2
\]

Second term:
\[
\frac{d}{dx}[(x + 1)^2] = 2(x + 1)
\]

Thus, the derivative is:
\[
f'(x) = (x + 1)^2 + 2x(x + 1)
\]

Simplify:
\[
f'(x) = (x + 1)[(x + 1) + 2x] = (x + 1)(3x + 1)
\]

### 3. Case 2: \( x < 0 \)

In this case, \( f(x) = -x(x + 1)^2 \). Again, apply the product rule:
\[
f'(x) = \frac{d}{dx}[-x] \cdot (x + 1)^2 + (-x) \cdot \frac{d}{dx}[(x + 1)^2]
\]

First term:
\[
\frac{d}{dx}[-x] = -1, \quad \text{so the first part is } -(x + 1)^2
\]

Second term:
\[
\frac{d}{dx}[(x + 1)^2] = 2(x + 1)
\]

Thus, the derivative is:
\[
f'(x) = -(x + 1)^2 + 2x(x + 1)
\]

Simplify:
\[
f'(x) = -(x + 1)^2 + 2x(x + 1)
\]
\[
f'(x) = -(x + 1)(x + 1) + 2x(x + 1)
\]
\[
f'(x) = (x + 1)(2x - (x + 1)) = (x + 1)(x - 1)
\]

### 4. Final Answer:

\[
f'(x) = 
\begin{cases} 
(x + 1)(3x + 1) & \text{for } x \geq 0 \\
(x + 1)(x - 1) & \text{for } x < 0 
\end{cases}
\]

Would you like more details on any part of this solution or further clarification?

Here are five related questions for further exploration:
1. How would you apply the second derivative to this function?
2. What is the interpretation of the derivative at \(x = 0\), where the absolute value changes?
3. How can we sketch the graph of \(f'(x)\) based on this derivative?
4. How would the derivative change if \(f(x)\) involved \( |x + 1| \) instead of \( |x| \)?
5. Can we analyze the critical points and concavity of \(f(x)\) from the derivative?

**Tip**: Always be cautious with absolute values when differentiating; they often lead to piecewise functions depending on the sign of the variable inside the absolute value.

This page provides a step-by-step solution for finding the first derivative of the function f(x) = |x|(x + 1)^2. The solution applies the product rule and handles the absolute value function with a piecewise approach. Suitable for advanced high school or early university calculus students.

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