Question: User: $\frac{d}{dx}$ $x^2$ $\left(x-1\right)^{-1}$
1. **Problem:** Find the derivative of $x^2\left(x-1\right)^{-1}$ with respect to $x$.
2. Use the product rule. If $f(x)g(x)$ is a product, then $$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x).$$
3. Let $f(x)=x^2$ and $g(x)=(x-1)^{-1}$.
4. Differentiate each part:
- $f'(x)=2x$
- For $g(x)=(x-1)^{-1}$, use the chain rule:
$$g'(x)=-1(x-1)^{-2}\cdot 1=-(x-1)^{-2}$$
5. Substitute into the product rule:
$$\frac{d}{dx}[x^2(x-1)^{-1}] = 2x(x-1)^{-1} + x^2\bigl(-(x-1)^{-2}\bigr)$$
6. Rewrite with a common denominator:
$$2x(x-1)^{-1} = \frac{2x}{x-1} = \frac{2x(x-1)}{(x-1)^2}$$
$$x^2(x-1)^{-2} = \frac{x^2}{(x-1)^2}$$
7. Combine the terms:
$$\frac{2x(x-1)}{(x-1)^2} - \frac{x^2}{(x-1)^2} = \frac{2x(x-1)-x^2}{(x-1)^2}$$
8. Simplify the numerator:
$$2x(x-1)-x^2 = 2x^2-2x-x^2 = x^2-2x$$
9. Factor the numerator if desired:
$$x^2-2x = x(x-2)$$
10. Final answer:
$$\boxed{\frac{x(x-2)}{(x-1)^2}}$$