## Calculus, 10th Edition (Anton)

$$y' = \frac{{\sinh x}}{{\left| {\sinh x} \right|}} = 1{\text{ for x > 0 and }}y' = \frac{{\sinh x}}{{\left| {\sinh x} \right|}} = - 1{\text{ for }}x < 0$$
\eqalign{ & y = {\cosh ^{ - 1}}\left( {\cosh x} \right) \cr & {\text{find the derivatvive}} \cr & y' = \left( {{{\cosh }^{ - 1}}\left( {\cosh x} \right)} \right)' \cr & {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr & \frac{d}{{dx}}\left[ {{{\cosh }^{ - 1}}u} \right] = \frac{1}{{\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}} \cr & y' = \frac{1}{{\sqrt {{{\left( {\cosh x} \right)}^2} - 1} }}\left( {\cosh x} \right)' \cr & {\text{so}} \cr & y' = \frac{1}{{\sqrt {{{\cosh }^2}x - 1} }}\left( {\sinh x} \right) \cr & {\text{hyperbolic identity }}{\cosh ^2}x - 1 = {\sinh ^2}x \cr & y' = \frac{1}{{\sqrt {{{\sinh }^2}x} }}\left( {\sinh x} \right) \cr & {\text{simplifying}} \cr & y' = \frac{1}{{\left| {\sinh x} \right|}}\left( {\sinh x} \right) \cr & {\text{From the definition of the absolute value we have }} \cr & y' = \frac{{\sinh x}}{{\left| {\sinh x} \right|}} \cr & 1{\text{ for x > 0 and }} - 1{\text{for }}x < 0 \cr}