Answer
$$f'\left( x \right) = \frac{{\left| x \right|}}{{x\sqrt {{x^2} + 4} }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\operatorname{csch} ^{ - 1}}\left( {2/x} \right) \cr
& {\text{find the derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ - 1}}\left( {2/x} \right)} \right) \cr
& {\text{use derivatives of the inverse hyperbolic functions}} \cr
& f'\left( x \right) = - \frac{1}{{\left| {2/x} \right|\sqrt {1 + {{\left( {2/x} \right)}^2}} }}\frac{d}{{dx}}\left( {\frac{2}{x}} \right) \cr
& f'\left( x \right) = - \frac{1}{{\left| {2/x} \right|\sqrt {1 + {{\left( {2/x} \right)}^2}} }}\left( { - \frac{2}{{{x^2}}}} \right) \cr
& {\text{simplify}} \cr
& f'\left( x \right) = \frac{{\left| x \right|}}{{2\sqrt {\frac{{{x^2} + 4}}{{{x^2}}}} }}\left( {\frac{2}{{{x^2}}}} \right) \cr
& f'\left( x \right) = \frac{{\left| x \right|}}{{\frac{{\sqrt {{x^2} + 4} }}{x}}}\left( {\frac{1}{{{x^2}}}} \right) \cr
& f'\left( x \right) = \frac{{\left| x \right|}}{{x\sqrt {{x^2} + 4} }} \cr} $$
