Answer
$$\frac{{dy}}{{dx}} = \frac{4}{{\sqrt {16{x^2} + 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\sinh ^{ - 1}}\left( {4x} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}\left( {4x} \right)} \right] \cr
& {\text{Using the rules of theorem 5}}{\text{.20 }}\left( {{\text{Page 389}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\left( {4x} \right)}^2} + 1} }}\frac{d}{{dx}}\left[ {4x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {16{x^2} + 1} }}\left( 4 \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{4}{{\sqrt {16{x^2} + 1} }} \cr} $$
