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