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 $(-3u-2)$?

If you have some patience to do directly, you will get it.
$$
\frac{d}{dx}\left(\frac{x(1-3x)}{\sqrt{x-1}}\right) =\frac{\sqrt{x-1}(1-6x) - \frac{1}{2\sqrt{x-1}}(x-3x^2)}{x-1} \to \frac{-9x^2 +13x-2}{2(x-1)^\frac{3}{2}}
$$
