Let's solve the problem step by step as requested. Given: \[ y = \frac{(4x - 5x^2 + 8)^5 \cdot (3e^x + 1)^9}{3x^2(x^2 + 2x + 3)} \] --- **Step 1: Simplify the expression and identify which rules to use** Let's denote: - \( u = (4x - 5x^2 + 8)^5 \) - \( v = (3e^x + 1)^9 \) - \( w = 3x^2(x^2 + 2x + 3) \) So, \[ y = \frac{u \cdot v}{w} \] To find \( y' \), we will use the quotient rule: \[ \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \] where \( f = u \cdot v \) and \( g = w \). **Explanation (Step 1):** We have rewritten the function in terms of simpler components (u, v, w) and identified that we need to use the **quotient rule** and the **product rule** (since the numerator is a product). --- **Step 2: Apply the quotient and product rules** First, find the derivatives: - \( f = u \cdot v \), so \( f' = u'v + uv' \) (product rule) - \( g = w \), so \( g' = w' \) Now, compute \( u', v', w' \): - \( u' = 5(4x-5x^2+8)^4 \cdot (4 - 10x) \) (chain rule) - \( v' = 9(3e^x+1)^8 \cdot (3e^x) \) (chain rule) - \( w = 3x^2(x^2 + 2x + 3) \) Use the product rule for \( w \): \( w' = 3x^2 \cdot \frac{d}{dx}(x^2 + 2x + 3) + (x^2 + 2x + 3) \cdot \frac{d}{dx}(3x^2) \) \( = 3x^2 \cdot (2x + 2) + (x^2 + 2x + 3) \cdot 6x \) \( = 3x^2(2x + 2) + 6x(x^2 + 2x + 3) \) \( = 6x^3 + 6x^2 + 6x^3 + 12x^2 + 18x \) Combine like terms: \( = (6x^3 + 6x^3) + (6x^2 + 12x^2) + 18x \) \( = 12x^3 + 18x^2 + 18x \) Now, plug into the quotient rule: \[ y' = \frac{(u'v + uv')w - (uv)w'}{w^2} \] **Final Answer:** \[ \boxed{ y' = \frac{[(u'v + uv')w - (uv)w']}{w^2} } \] where: - \( u = (4x - 5x^2 + 8)^5 \) - \( u' = 5(4x-5x^2+8)^4 (4-10x) \) - \( v = (3e^x + 1)^9 \) - \( v' = 9(3e^x + 1)^8 \cdot 3e^x \) - \( w = 3x^2(x^2 + 2x + 3) \) - \( w' = 12x^3 + 18x^2 + 18x \) --- **Explanation (Step 2):** We've applied the quotient rule and the product/chain rules to find the derivative. The final answer is written in terms of the original components for clarity and brevity. If needed, you can expand further, but this form is sufficient for most purposes.