1. **State the problem:** Find the derivative of the function $$y = \frac{(x+2)^2}{(x+1)^3 (x+3)^4}$$. 2. **Formula and rules:** We will use the quotient rule for derivatives: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $$u = (x+2)^2$$ and $$v = (x+1)^3 (x+3)^4$$. 3. **Find derivatives of numerator and denominator:** - $$u' = 2(x+2)$$ by the power rule. - To find $$v'$$, use the product rule: $$v = f(x)g(x)$$ with $$f(x) = (x+1)^3$$ and $$g(x) = (x+3)^4$$. 4. **Derivatives of $$f$$ and $$g$$:** - $$f' = 3(x+1)^2$$ - $$g' = 4(x+3)^3$$ 5. **Apply product rule for $$v'$$:** $$v' = f'g + fg' = 3(x+1)^2 (x+3)^4 + (x+1)^3 4(x+3)^3$$ 6. **Write the derivative using quotient rule:** $$y' = \frac{2(x+2)(x+1)^3 (x+3)^4 - (x+2)^2 \left[3(x+1)^2 (x+3)^4 + 4(x+1)^3 (x+3)^3\right]}{\left[(x+1)^3 (x+3)^4\right]^2}$$ 7. **Simplify numerator:** Factor common terms: - Common factor in numerator: $$(x+1)^2 (x+3)^3 (x+2)$$ Rewrite numerator as: $$ (x+1)^2 (x+3)^3 (x+2) \left[ 2(x+1)(x+3) - (x+2) \left(3(x+3) + 4(x+1)\right) \right] $$ 8. **Simplify inside the bracket:** - $$2(x+1)(x+3) = 2(x^2 + 4x + 3) = 2x^2 + 8x + 6$$ - $$3(x+3) + 4(x+1) = 3x + 9 + 4x + 4 = 7x + 13$$ - Multiply by $$(x+2)$$: $$ (x+2)(7x + 13) = 7x^2 + 13x + 14x + 26 = 7x^2 + 27x + 26 $$ 9. **Combine terms inside bracket:** $$ 2x^2 + 8x + 6 - (7x^2 + 27x + 26) = -5x^2 - 19x - 20 $$ 10. **Final derivative:** $$ y' = \frac{(x+1)^2 (x+3)^3 (x+2) (-5x^2 - 19x - 20)}{(x+1)^6 (x+3)^8} = \frac{(x+2)(-5x^2 - 19x - 20)}{(x+1)^4 (x+3)^5} $$ This is the simplified form of the derivative. **Answer:** $$ y' = \frac{(x+2)(-5x^2 - 19x - 20)}{(x+1)^4 (x+3)^5} $$