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

1. Apply the quotient rule, which is:

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

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

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

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

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

   1. Let $u = x - 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x - 1\right)$:

      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: $1$

      The result of the chain rule is:

      $$2 x - 2$$

   Now plug in to the quotient rule:

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

2. Now simplify:

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

The answer is:

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