Answer
$$ y'= \frac{e^{\cosh^{-1}x} }{\sqrt{x^2-1}}.$$
Work Step by Step
Recall that $(\cosh^{-1} x)'=\dfrac{1}{\sqrt{x^2-1}}$
Recall that $(e^x)'=e^x$
Since $ y=e^{\cosh^{-1}x}$, then the derivative, by using the chain rule, is given by
$$ y'= e^{\cosh^{-1}x}({\cosh^{-1}x})'=\frac{e^{\cosh^{-1}x} }{\sqrt{x^2-1}}.$$
