Answer
$$y' = \frac{1}{{\sqrt {{x^2} + 1} }}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}x} \right] = \frac{1}{{\sqrt {{x^2} + 1} }} \cr
& {\text{Let }}y = {\sinh ^{ - 1}}x,{\text{ then}} \cr
& \sinh y = x \cr
& {\text{By implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {\sinh y} \right] = \frac{d}{{dx}}\left[ x \right] \cr
& \left( {\cosh y} \right)y' = 1 \cr
& {\text{Solve for }}y' \cr
& y' = \frac{1}{{\cosh y}} \cr
& {\text{By the Hyperbolic identity }}\cosh y = \sqrt {{{\sinh }^2}y + 1} \cr
& y' = \frac{1}{{\sqrt {{{\sinh }^2}y + 1} }} \cr
& y' = \frac{1}{{\sqrt {{x^2} + 1} }} \cr} $$
