Answer
$$ g'(t)
=\frac{t\sinh^{-1}t}{\sqrt{t^2-1}}+\frac{\sqrt{t^2-1}}{\sqrt{t^2+1}}.$$
Work Step by Step
Since $ g(t)=\sqrt{t^2-1}\sinh^{-1}t $, then the derivative, by using the product rule, is given by
$$ g'(t)=\frac{(t^2-1)'}{2\sqrt{t^2-1}}\sinh^{-1}t+\sqrt{t^2-1}\frac{1}{\sqrt{t^2+1}}\\
=\frac{2t\sinh^{-1}t}{2\sqrt{t^2-1}}+\frac{\sqrt{t^2-1}}{\sqrt{t^2+1}}\\
=\frac{t\sinh^{-1}t}{\sqrt{t^2-1}}+\frac{\sqrt{t^2-1}}{\sqrt{t^2+1}}$$
