Find the value of $\displaystyle \frac{dy}{dx} $ at the point $\displaystyle (2 , \frac{1}{5} ) $ on the curve with equation $\displaystyle y = (x-1)(2x+1)^{-1} $

$\displaystyle u = x-1 $

$\displaystyle v = (2x+1)^{-1} $

I know you have to use the chain rule on v so this is what I get,

$\displaystyle \frac{dv}{dx} = -1 \times (2x+1)^{-2} $

but my books says $\displaystyle \frac{dv}{dx} = -2(2x+1)^{-2} $

How have they got -2? As I have been taught that you first bring down the power to the front of the expression and than minus 1 from it.

Is this not correct?