# Math Help - y=(x^2+2x-1)^(1/2) find y'

1. ## y=(x^2+2x-1)^(1/2) find y'

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

I did this...

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

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

2. never mind I see what is in the denominator is y