Answer
$$y' = 6{\sinh ^2}\left( {2x} \right)\cosh \left( {2x} \right)$$
Work Step by Step
$$\eqalign{
& y = {\sinh ^3}\left( {2x} \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {{{\sinh }^3}\left( {2x} \right)} \right)' \cr
& {\text{chain rule}} \cr
& y' = 3{\sinh ^2}\left( {2x} \right)\left( {\sinh \left( {2x} \right)} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {\sinh u} \right] = \cosh u\frac{{du}}{{dx}} \cr
& y' = 3{\sinh ^2}\left( {2x} \right)\left( {\cosh \left( {2x} \right)\left( {2x} \right)'} \right) \cr
& y' = 3{\sinh ^2}\left( {2x} \right)\left( {\cosh \left( {2x} \right)\left( 2 \right)} \right) \cr
& {\text{simplifying}} \cr
& y' = 6{\sinh ^2}\left( {2x} \right)\cosh \left( {2x} \right) \cr} $$
