Derivative of x^3/(x-1)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = x^{3}$ and $g{\left(x \right)} = x - 1$.

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

   1. Apply the power rule: $x^{3}$ goes to $3 x^{2}$

   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:

   $\frac{- x^{3} + 3 x^{2} \left(x - 1\right)}{\left(x - 1\right)^{2}}$

2. Now simplify:

   $$\frac{x^{2} \left(2 x - 3\right)}{\left(x - 1\right)^{2}}$$

The answer is: