$\displaystyle y=\sqrt{x^2+2x-1}$ find $\displaystyle y^,$ answer in book is $\displaystyle \frac{x+1}{y} $ but don't see how?

I did this...

$\displaystyle y^, = \frac{1}{2}(x^2+2x-1)^{-\frac{1}{2}}(2x+2)

\Rightarrow

\frac

{x+1}

{(x^{2}+2x-1)^{\frac{1}{2}}}

$