Given g(2)=36\displaystyle {g{{\left({2}\right)}}}={36}, g(2)=1\displaystyle {g}'{\left({2}\right)}={1} and f(x)=g(x)\displaystyle {f{{\left({x}\right)}}}=\sqrt{{{g{{\left({x}\right)}}}}}, find f(2)\displaystyle {f}'{\left({2}\right)}.