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