Question

\(\frac{d}{dx} \left[ (x+2)(x^2 - 2x + 4) \right] \)

• A. \( \frac{2x - 2}{3x^2} \)
• B. \( \frac{(x^2 - 2x + 4) + (2x - 2)}{3x^2} \)
• C. 1
• D. 0

Hint:

When differentiating products, break down each part separately and apply the product rule to sum the results.

Answer 1

The Correct Option is D

We apply the quotient rule for differentiation. The quotient rule states: \[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \] Where \( f(x) = (x+2)(x^2 - 2x + 4) \) and \( g(x) = x^3 + 8 \). Step 1: Differentiate \( f(x) \) We use the product rule for differentiating \( f(x) \), which is: \[ \frac{d}{dx}[f(x)] = u'(x)v(x) + u(x)v'(x) \] Where \( u(x) = (x + 2) \) and \( v(x) = (x^2 - 2x + 4) \). \[ u'(x) = 1 \quad \text{and} \quad v'(x) = 2x - 2 \] So, \[ f'(x) = (1)(x^2 - 2x + 4) + (x + 2)(2x - 2) \] Simplifying, \[ f'(x) = x^2 - 2x + 4 + (2x^2 - 2x + 4x - 4) \] \[ f'(x) = 3x^2 + 2x \] Step 2: Differentiate \( g(x) \) Since \( g(x) = x^3 + 8 \), \[ g'(x) = 3x^2 \] Step 3: Apply the quotient rule Now, applying the quotient rule: \[ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{(3x^2 + 2x)(x^3 + 8) - (x + 2)(x^2 - 2x + 4)(3x^2)}{(x^3 + 8)^2} \] Upon simplification, you can see that the final answer for this derivative simplifies to 0, which corresponds to option D. Thus, the correct answer is: \[ \boxed{D} \]