Derivative of $\frac {x\cdot\left(1 - 3x\right)}{\sqrt{x-1}}$ Problem. Find the first derivative of $$ \dfrac {x \left( 1 - 3x \right)}{\sqrt{x-1}} $$

Work. Let $u = x-1$ and $y = \dfrac {(u+1)(-3u-2)}{\sqrt{u}} $
Using the chain rule, I got$$\dfrac{(-9x^2-5x+2)}{(2(x-1)^\frac{3}{2})}$$
But the answer is $$\dfrac{(-9x^2+13x+2)}{\left(2(x-1)^\frac{3}{2}\right)}$$ 
I'm not sure what I did wrong, maybe something related to the $(-3u-2)$?

You don't have to use the Chain Rule. You can, instead, use the Product Rule. 
$ f(x) = \dfrac {x \cdot (1 - 3x)}{\sqrt{x-1}} = \dfrac {x-3x^2}{\sqrt{x-1}} $
$ f'(x) = \dfrac {\sqrt {x-1} \cdot (1-6x) - (x-3x^2) \cdot \left( \sqrt{x-1} \right)'}{x-1} $ 
$ f'(x) = \dfrac {\sqrt{x-1} \cdot (1-6x) - (x-3x^2) \cdot \dfrac {1}{2\sqrt{x-1}}}{x-1} $
I used the Chain Rule here only to find the derivative of $ \sqrt {x-1} $, which, it seems, you know how to do. I'll leave the rest as an exercise for you -- just simplify. You should get the right answer. Good luck! 
