Differentiate $\frac{x^3}{{(x-1)}^2}$ 
Find $\frac{d}{dx}\frac{x^3}{{(x-1)}^2}$

I start by finding the derivative of the denominator, since I have to use the chain rule. 
Thus, I make $u=x-1$ and $g=u^{-2}$. I find that $u'=1$ and $g'=-2u^{-3}$. I then multiply the two together and substitute $u$ in to get:
$$\frac{d}{dx}(x-1)^{2}=2(x-1)$$
After having found the derivative of the denominator I find the derivative of the numerator, which is $3x^2$. With the two derivatives found I apply the quotient rule, which states that 
$$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v'u-vu'}{v^2}$$
and substitute in the numbers
$$\frac{d}{dx}\frac{x^3}{(x-1)^2}=\frac{3x^2(x-1)^2-2x^3(x-1)}{(x-1)^4}$$
Can I simplify this any further?Is the derivation correct?

You're mixing the product rule and the quotient rule. You can apply each of them, but not simultaneously.


*

*by the product rule: remember $\dfrac{\mathrm d}{\mathrm dx}\Bigl(\dfrac1{x^n}\Bigr)=-\dfrac n{x^{n+1}}$, so
\begin{align}
\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)&=\dfrac{\mathrm d}{\mathrm dx}(x^3)\cdot\dfrac1{(1-x)^2}+x^3\dfrac{\mathrm d}{\mathrm dx}\dfrac1{(1-x)^2} \\
&= \frac{3x^2}{(1-x)^2}+\frac{2x^3}{(1-x)^3} =\frac{x^2\bigl(3(1-x)+2x\bigr)}{(1-x)^3}\\&
=\frac{x^2(3-x)}{(1-x)^3}.
\end{align}

*by the quotient rule:
$$\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)=\frac{3x^2(1-x)^2+x^3\cdot2(1-x)}{(1-x)^4}=\frac{x^2\color{red}{(\not1-\not x)}\bigl(3(1-x)+2x\bigr)}{( 1-x)^{\color{red}{\not4}\,3}}=\dots$$

*Since there are exponents, logarithmic differentiation may help make it shorter. It's simpler to use it with Lagrange's notations: set $f(x)=\dfrac{x^3}{(1-x)^2}$. Then
$$\frac{f'(x)}{f(x)}=\frac 3x+\frac2{1-x}=\frac{3-x}{x(1-x)}, \quad\text{so }\;f'(x)=\frac{f'(x)}{f(x)}\cdot f(x)=\dotsm $$

