To find the differential of \((1 + 2x^2)(x - x^2)\), we will use the **product rule** of differentiation. The product rule states that:

\[
\frac{d}{dx}(uv) = u'v + uv'
\]

Where:
- \( u = 1 + 2x^2 \)
- \( v = x - x^2 \)

Now, let's differentiate both parts.

### Step 1: Differentiate \( u = 1 + 2x^2 \)
\[
u' = \frac{d}{dx}(1 + 2x^2) = 0 + 4x = 4x
\]

### Step 2: Differentiate \( v = x - x^2 \)
\[
v' = \frac{d}{dx}(x - x^2) = 1 - 2x
\]

### Step 3: Apply the product rule
Now, we apply the product rule:
\[
\frac{d}{dx}[(1 + 2x^2)(x - x^2)] = u'v + uv'
\]

Substitute the values of \( u, u', v, v' \):
\[
= (4x)(x - x^2) + (1 + 2x^2)(1 - 2x)
\]

### Step 4: Simplify both terms

- First term: \( 4x(x - x^2) = 4x^2 - 4x^3 \)
- Second term: \( (1 + 2x^2)(1 - 2x) = 1 - 2x + 2x^2 - 4x^3 = 1 - 2x + 2x^2 - 4x^3 \)

### Step 5: Combine like terms
Now add both expressions together:
\[
4x^2 - 4x^3 + 1 - 2x + 2x^2 - 4x^3
\]
\[
= 4x^2 + 2x^2 - 4x^3 - 4x^3 + 1 - 2x
\]
\[
= 6x^2 - 8x^3 + 1 - 2x
\]

Thus, the differential is:
\[
\frac{d}{dx}[(1 + 2x^2)(x - x^2)] = 6x^2 - 8x^3 + 1 - 2x
\]

Would you like more details on this or any specific step?

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Here are 5 related questions you might find helpful:
1. What is the product rule in differentiation?
2. How do you differentiate polynomials?
3. Can the chain rule be applied in this problem, and why or why not?
4. How can we use second derivatives in solving problems like this?
5. How would the result change if we had a more complex expression for \( u \) or \( v \)?

**Tip**: Always simplify expressions after differentiation to make sure you identify and combine like terms.

Learn how to find the differential of (1 + 2x^2)(x - x^2) using the product rule of differentiation. This step-by-step guide covers key calculus concepts like polynomial differentiation and the product rule, suitable for Grades 10-12.

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