$\displaystyle h(x) = \ln(x + \sqrt{x^{2} - 9})$

$\displaystyle h(x) = \ln(x + (x^{2} - 9)^{1/2})$

$\displaystyle h'(x) = \ln(1 + [\dfrac{1}{2}(u)^{-1/2} du]$

$\displaystyle h'(x) = \ln(1 + [\dfrac{1}{2}(u)^{-1/2} 2x]$

$\displaystyle h'(x) = \ln(1 + [x (x^{2} - 9)^{1/2} ]$ ?? What now?