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

1. Apply the quotient rule, which is:

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

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

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

   1. Differentiate $x^{2} - x + 1$ term by term:

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

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

      3. The derivative of a constant times a function is the constant times the derivative of the function.

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

         So, the result is: $-1$

      The result is: $2 x - 1$

   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:

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

2. Now simplify:

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

The answer is: