To compute the first derivative of the function $f(x) = ((2-x)^2) \cdot (x^3 + 2x)$, we will apply the product rule along with the chain rule.

Step 1: Apply the product rule

The product rule states: $\frac{d}{dx} \left[ u(x) \cdot v(x) \right] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ Here, we identify:

• $u(x) = (2 - x)^2$
• $v(x) = x^3 + 2x$

Now, we'll differentiate $u(x)$ and $v(x)$ separately.

Step 2: Differentiate $u(x) = (2 - x)^2$

Using the chain rule, we differentiate $u(x)$: $u'(x) = 2 \cdot (2 - x) \cdot (-1) = -2(2 - x)$ So, $u'(x) = -2(2 - x)$.

Step 3: Differentiate $v(x) = x^3 + 2x$

The derivative of $v(x)$ is: $v'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(2x) = 3x^2 + 2$

Step 4: Apply the product rule

Now that we have $u'(x)$ and $v'(x)$, we apply the product rule: $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ Substituting the values: $f'(x) = [-2(2 - x)] \cdot (x^3 + 2x) + (2 - x)^2 \cdot (3x^2 + 2)$

Step 5: Simplify the expression

Now let's expand each term.

1. For $[-2(2 - x)] \cdot (x^3 + 2x)$: $-2(2 - x)(x^3 + 2x) = -2(2x^3 + 4x - x^4 - 2x^2)$ Simplifying: $= -2x^4 - 4x^2 + 4x^3 + 8x$

2. For $(2 - x)^2 \cdot (3x^2 + 2)$: $(2 - x)^2 = 4 - 4x + x^2$ Now multiply: $(4 - 4x + x^2)(3x^2 + 2) = 12x^2 - 12x^3 + 3x^4 + 8 - 8x + 2x^2$ Simplifying: $= 3x^4 - 12x^3 + 14x^2 - 8x + 8$

Step 6: Combine the results

Now, we combine both parts: $f'(x) = (-2x^4 - 4x^2 + 4x^3 + 8x) + (3x^4 - 12x^3 + 14x^2 - 8x + 8)$ Simplifying the final expression: $f'(x) = x^4 - 8x^3 + 10x^2 + 0x + 8$ So, the derivative is: $f'(x) = x^4 - 8x^3 + 10x^2 + 8$

This is the first derivative of the given function.

Would you like any further details or clarifications? Here are some related questions:

1. How is the product rule applied in other scenarios?
2. What happens if you change the powers in $f(x)$?
3. Can we analyze the critical points of $f(x)$?
4. How does the chain rule influence this solution?
5. What happens to the behavior of the graph for higher powers of $x$?

Tip: Always simplify carefully after applying rules to avoid errors when combining terms!