Answer
$$y' = \frac{{{{\operatorname{sech} }^2}x}}{{\sqrt {1 + {{\tanh }^2}x} }}$$
Work Step by Step
$$\eqalign{
& y = {\sinh ^{ - 1}}\left( {\tanh x} \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {{{\sinh }^{ - 1}}\left( {\tanh x} \right)} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 + {u^2}} }}\frac{{du}}{{dx}} \cr
& y' = \frac{1}{{\sqrt {1 + {{\left( {\tanh x} \right)}^2}} }}\left( {\tanh x} \right)' \cr
& y' = \frac{1}{{\sqrt {1 + {{\tanh }^2}x} }}\left( {{{\operatorname{sech} }^2}x} \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{{{{\operatorname{sech} }^2}x}}{{\sqrt {1 + {{\tanh }^2}x} }} \cr} $$
