Find y' for y = ((x+2)²)/((x+1)³(x+3)^4)

Find $y'$ for $y = \dfrac{(x+2)^{2}}{(x+1)^{3}(x+3)^{4}}$.

Step 1: Take the natural logarithm

\[ \ln y = 2\ln(x+2) - 3\ln(x+1) - 4\ln(x+3). \]

Step 2: Differentiate

\[ \frac{y'}{y} = \frac{2}{x+2} - \frac{3}{x+1} - \frac{4}{x+3}. \]

Step 3: Combine the bracket over a common denominator

The common denominator is $(x+1)(x+2)(x+3)$: $\frac{2(x+1)(x+3) - 3(x+2)(x+3) - 4(x+1)(x+2)}{(x+1)(x+2)(x+3)}$.

Expanding each product: \[\begin{aligned} 2(x+1)(x+3) &= 2x^{2}+8x+6, \\ 3(x+2)(x+3) &= 3x^{2}+15x+18, \\ 4(x+1)(x+2) &= 4x^{2}+12x+8, \end{aligned}\]

so the numerator is

\[ \bigl(2x^{2}+8x+6\bigr)-\bigl(3x^{2}+15x+18\bigr)-\bigl(4x^{2}+12x+8\bigr) = -5x^{2}-19x-20. \]

Step 4: Assemble

Multiplying by $y$, one factor $(x+2)$ cancels and the remaining denominators combine: \[y' = \frac{(x+2)^{2}}{(x+1)^{3}(x+3)^{4}}\cdot\frac{-\bigl(5x^{2}+19x+20\bigr)}{(x+1)(x+2)(x+3)} = \frac{-\bigl(5x^{2}+19x+20\bigr)(x+2)}{(x+1)^{4}(x+3)^{5}}.\]

Step 5: Verification

\[x = 0:\quad y = \frac{4}{1\cdot 81} \approx 0.049383, \qquad y' = -\frac{20\cdot 2}{1\cdot 243} \approx -0.164609\]

\[\frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx -0.164609 \;\;\checkmark\]

\[\text{Check via the bracket: } y\left(\tfrac{2}{2}-\tfrac{3}{1}-\tfrac{4}{3}\right) = 0.049383\times(-3.3333) \approx -0.164609 \;\;\checkmark\]