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