find h'(x) if h(x) = $\displaystyle \ln(x+\sqrt{x^2-1}) $

$\displaystyle

h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2})(2x)

$

then I tried to simplify:

$\displaystyle h'(x) = \frac{(2x+x(x^2-1)^\frac{-1}{2})}{x+\sqrt{x^2-1}}$

Is this correct?