To solve this problem, we need to find the derivative of the function $f(x) = (1 + 2x^2)(x - x^2)$ using two methods: the Product Rule and expanding the expression first.

1. Using the Product Rule

The Product Rule states that if $f(x) = g(x) \cdot h(x)$, then $f'(x) = g'(x)h(x) + g(x)h'(x)$.

Here, let:

• $g(x) = 1 + 2x^2$
• $h(x) = x - x^2$

First, find $g'(x)$ and $h'(x)$:

• $g(x) = 1 + 2x^2$
  $g'(x) = 0 + 4x = 4x$

• $h(x) = x - x^2$
  $h'(x) = 1 - 2x$

Now, apply the Product Rule: $f'(x) = g'(x)h(x) + g(x)h'(x)$ Substitute the expressions: $f'(x) = (4x)(x - x^2) + (1 + 2x^2)(1 - 2x)$ Expand each term:

1. $(4x)(x - x^2) = 4x^2 - 4x^3$
2. $(1 + 2x^2)(1 - 2x) = 1 - 2x + 2x^2 - 4x^3$

Combine the terms: $f'(x) = (4x^2 - 4x^3) + (1 - 2x + 2x^2 - 4x^3)$ Simplify: $f'(x) = 4x^2 - 4x^3 + 1 - 2x + 2x^2 - 4x^3$ $f'(x) = 6x^2 - 8x^3 - 2x + 1$

So, the derivative using the Product Rule is: $f'(x) = 6x^2 - 8x^3 - 2x + 1$

2. By Expanding First

Now let's expand $f(x)$ before differentiating.

1. Expand $f(x) = (1 + 2x^2)(x - x^2)$: $f(x) = x - x^2 + 2x^3 - 2x^4$

2. Differentiate term by term: $f'(x) = 1 - 2x + 6x^2 - 8x^3$

Thus, the derivative by expanding first is: $f'(x) = -2x + 6x^2 - 8x^3 + 1$

Both methods give the same result: $f'(x) = 6x^2 - 8x^3 - 2x + 1$

Would you like more details on any step?

Here are some related questions to explore:

1. How does the Product Rule apply to more than two functions?
2. Why does expanding first sometimes make finding the derivative easier?
3. What are some common mistakes to avoid when using the Product Rule?
4. How can the Chain Rule be used in combination with the Product Rule?
5. What are some other ways to simplify derivatives involving polynomials?

Tip: When using the Product Rule, always ensure you correctly identify and differentiate both functions involved before combining them.