Answer
$$y' = \frac{1}{{\sqrt {{x^2} - 1} {{\cosh }^{ - 1}}x}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {{{\cosh }^{ - 1}}x} \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {\ln \left( {{{\cosh }^{ - 1}}x} \right)} \right)' \cr
& y' = \frac{{\left( {{{\cosh }^{ - 1}}x} \right)'}}{{{{\cosh }^{ - 1}}x}} \cr
& y' = \frac{1}{{{{\cosh }^{ - 1}}x}}\left( {{{\cosh }^{ - 1}}x} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\cosh }^{ - 1}}u} \right] = \frac{1}{{\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}} \cr
& y' = \frac{1}{{{{\cosh }^{ - 1}}x}}\left( {\frac{1}{{\sqrt {{x^2} - 1} }}} \right)\left( x \right)' \cr
& y' = \frac{1}{{{{\cosh }^{ - 1}}x}}\left( {\frac{1}{{\sqrt {{x^2} - 1} }}} \right)\left( 1 \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{1}{{\sqrt {{x^2} - 1} {{\cosh }^{ - 1}}x}} \cr} $$
