Answer
$$\frac{{dy}}{{dx}} = \frac{3}{{\sqrt {9{x^2} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\cosh ^{ - 1}}\left( {3x} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\cosh }^{ - 1}}\left( {3x} \right)} \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\cosh }^{ - 1}}u} \right] = \frac{{u'}}{{\sqrt {{u^2} - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dx}}\left[ {3x} \right]}}{{\sqrt {{{\left( {3x} \right)}^2} - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{3}{{\sqrt {9{x^2} - 1} }} \cr} $$
