Derivative of (x-2)^2/(x+1)

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = \left(x - 2\right)^{2}$ and $g{\left(x \right)} = x + 1$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

   1. Let $u = x - 2$.

   2. Apply the power rule: $u^{2}$ goes to $2 u$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x - 2\right)$:

      1. Differentiate $x - 2$ term by term:

         1. The derivative of the constant $-2$ is zero.

         2. Apply the power rule: $x$ goes to $1$

         The result is: $1$

      The result of the chain rule is:

      $$2 x - 4$$

   To find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $x + 1$ term by term:

      1. The derivative of the constant $1$ is zero.

      2. Apply the power rule: $x$ goes to $1$

      The result is: $1$

   Now plug in to the quotient rule:

   $\frac{- \left(x - 2\right)^{2} + \left(x + 1\right) \left(2 x - 4\right)}{\left(x + 1\right)^{2}}$

2. Now simplify:

   $$\frac{\left(x - 2\right) \left(x + 4\right)}{\left(x + 1\right)^{2}}$$

The answer is:

$$\frac{\left(x - 2\right) \left(x + 4\right)}{\left(x + 1\right)^{2}}$$