Answer
$$y' = 4\cosh \left( {4x - 8} \right)$$
Work Step by Step
$$\eqalign{
& y = \sinh \left( {4x - 8} \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left[ {\sinh \left( {4x - 8} \right)} \right]' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {\sinh u} \right] = \cosh u\frac{{du}}{{dx}} \cr
& y' = \cosh \left( {4x - 8} \right)\left[ {4x - 8} \right]' \cr
& y' = \cosh \left( {4x - 8} \right)\left( 4 \right) \cr
& y' = 4\cosh \left( {4x - 8} \right) \cr} $$
