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