## Calculus (3rd Edition)

$$y' =-6 \sinh (9-3t) \cosh (9-3t).$$
Since $y=\cosh^2 (9-3t)$, then the derivative $y'$, using the chain rule, is given by $$y'= 2 \cosh (9-3t) ( \cosh (9-3t))'=2(-3) \sinh (9-3t) \cosh (9-3t) \\ =-6 \sinh (9-3t) \cosh (9-3t).$$