To find the derivative $h'(x)$ of the function $h(x) = (x - x^2)(2x^2 - x^3)$, we can apply the product rule. The product rule states that if $h(x) = f(x) \cdot g(x)$, then:

$h'(x) = f'(x)g(x) + f(x)g'(x).$

In this case, let's define: $f(x) = x - x^2, \quad g(x) = 2x^2 - x^3.$

Step 1: Find $f'(x)$ and $g'(x)$

1. Compute $f'(x)$: $f(x) = x - x^2 \implies f'(x) = 1 - 2x.$

2. Compute $g'(x)$: $g(x) = 2x^2 - x^3 \implies g'(x) = 4x - 3x^2.$

Step 2: Apply the product rule

Using the product rule, we have: $h'(x) = f'(x)g(x) + f(x)g'(x).$

Substitute $f(x), f'(x), g(x),$ and $g'(x)$ into the equation:

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

Step 3: Simplify the expression

Let's expand each term:

1. Expand $f'(x)g(x)$: $(1 - 2x)(2x^2 - x^3) = (1)(2x^2 - x^3) + (-2x)(2x^2 - x^3).$ $= 2x^2 - x^3 - 4x^3 + 2x^4.$ $= 2x^2 - 5x^3 + 2x^4.$

2. Expand $f(x)g'(x)$: $(x - x^2)(4x - 3x^2) = (x)(4x - 3x^2) + (-x^2)(4x - 3x^2).$ $= 4x^2 - 3x^3 - 4x^3 + 3x^4.$ $= 4x^2 - 7x^3 + 3x^4.$

Step 4: Combine like terms

Now combine both expressions:

$h'(x) = (2x^2 - 5x^3 + 2x^4) + (4x^2 - 7x^3 + 3x^4).$

Combine like terms:

$h'(x) = (2x^2 + 4x^2) + (-5x^3 - 7x^3) + (2x^4 + 3x^4).$ $h'(x) = 6x^2 - 12x^3 + 5x^4.$

Final Answer

$\boxed{h'(x) = 5x^4 - 12x^3 + 6x^2.}$

Would you like more details on this problem or have any questions?

Here are 5 related questions for further exploration:

1. How would you find the second derivative $h''(x)$ of the function $h(x)$?
2. What are the critical points of the function $h(x)$?
3. Can you determine the intervals where $h(x)$ is increasing or decreasing?
4. How can you find the inflection points of $h(x)$?
5. What would the graph of $h(x)$ look like, and where would it have local maxima or minima?

Tip: When dealing with derivatives of products of functions, always consider using the product rule for a structured approach.